Monty Hall Problem

[hgc]

I know most of you are familiar with this one already, but we must settle it once and for all...

Monty presents 3 doors; behind one is a car, and behind the other two are goats. You pick a door. But before Monty opens that door, he reveals behind another door that there is a goat. He then gives you the opportunity to switch your choice. What is the probability that you will get the car by switching, or by staying?

[drkitten]

Originally posted by hgc
I know most of you are familiar with this one already, but we must settle it once and for all...

Monty presents 3 doors; behind one is a car, and behind the other two are goats. You pick a door. But before Monty opens that door, he reveals behind another door that there is a goat. He then gives you the opportunity to switch your choice. What is the probability that you will get the car by switching, or by staying?

What makes you think that we can settle this once and for all? Anyone dumb enough not to know the answer and stubborn enough not to read the literally thousands of web pages that dissect this to death is not going to be swayed, even by my deathless prose.

[Rob Lister]

I agree with drkitten. How exactly is this not settled. Good God man, this has been proven to death. What is your point.

[rppa]

You will never settle it once and for all.

The answer depends on Monty's rules of behavior.

Want to create even more arguments? Here's another one that has the net bitterly divided:

You meet a woman with her son. She tells you she has two children. What is the probability the other child is a boy?

[hgc]

Well I say it's 50/50. Bite me, everybody!

[Iconoclast]

Oh God, it's happening again.

[Matabiri]

Originally posted by rppa
Want to create even more arguments? Here's another one that has the net bitterly divided:

You meet a woman with her son. She tells you she has two children. What is the probability the other child is a boy?

In this situation, 50%, because the child is specified by his presence.

[hgc]

Originally posted by Iconoclast
Oh God, it's happening again. Did we do this before? I've been hanging out here far too long.

[drkitten]

Originally posted by rppa

You meet a woman with her son. She tells you she has two children. What is the probability the other child is a boy?

Zero, because she's lying and has only one child.

[Iconoclast]

Originally posted by new drkitten
Zero, because she's lying and has only one child.
Yeah, that's what I said. She's running a social security scam by alternately dressing the kid in boy's and the girl's clothes so she can get extra benifits.

[Iconoclast]

Originally posted by hgc
Did we do this before? I've been hanging out here far too long. I think we've done it about seven or eight times since the forums started, but it may have been mostly in the puzzles forum that it appeared. I'll stop derailing now, these threads are always a blast to watch.

[Matabiri]

Originally posted by Iconoclast
Yeah, that's what I said. She's running a social security scam by alternately dressing the kid in boy's and the girl's clothes so she can get extra benifits.

Not to mention the huge variety of fake beards and noses.

[Lothian]

Originally posted by hgc
I know most of you are familiar with this one already, but we must settle it once and for all...
.... What is the probability that you will get the car by switching, or by staying? I am more than happy to settle it for once and all.

There is a 100% probability that you will get the car by switching or by staying.

[T'ai Chi]

Originally posted by hgc

Monty presents 3 doors; behind one is a car, and behind the other two are goats. You pick a door. But before Monty opens that door, he reveals behind another door that there is a goat. He then gives you the opportunity to switch your choice. What is the probability that you will get the car by switching, or by staying?

Here's (http://www.martini.nu/justin/lmad.htm) my take on it.

[richardm]

Well, he'd always make sure you got the car if you hadn't had your guns taken away by limp-wristed liberals.

[richardm]

Sorry, must have used the wrong reply template there.

[DaveW]

OK, sorry to keep this apparently unpopular thread going. I googled the problem, and one thing that was missing in my mind from all the "answer" pages was that, after the one wrong door is opened, why is the decision not now 1/2? You can either change, or you can keep your door. The prize can be behind either, so this new choice throws out the old choice of 1/3. This must obviously be the counterintuitive part of the answer, but my limited statistical knowledge makes me think it is more of a misunderstanding that there is now a new choice, not just a continuation of the original choice.

[hgc]

Originally posted by DaveW
OK, sorry to keep this apparently unpopular thread going. I googled the problem, and one thing that was missing in my mind from all the "answer" pages was that, after the one wrong door is opened, why is the decision not now 1/2? You can either change, or you can keep your door. The prize can be behind either, so this new choice throws out the old choice of 1/3. This must obviously be the counterintuitive part of the answer, but my limited statistical knowledge makes me think it is more of a misunderstanding that there is now a new choice, not just a continuation of the original choice. You are correct, sir.

[HarryKeogh]

Originally posted by DaveW
OK, sorry to keep this apparently unpopular thread going. I googled the problem, and one thing that was missing in my mind from all the "answer" pages was that, after the one wrong door is opened, why is the decision not now 1/2? You can either change, or you can keep your door. The prize can be behind either, so this new choice throws out the old choice of 1/3. This must obviously be the counterintuitive part of the answer, but my limited statistical knowledge makes me think it is more of a misunderstanding that there is now a new choice, not just a continuation of the original choice.

imagine there are 100 doors instead of just 3. now imagine Monty asks you after each door is opened do you want to switch. when he gets down to 2 doors do you still think your original door has a 50% chance of being right?

right, it has a 1% chance. The same as when you originally chose.

this analogy helped me understand this puzzle when I first saw it.

[hgc]

Originally posted by HarryKeogh
imagine there are 100 doors instead of just 3. now imagine Monty asks you after each door is opened do you want to switch. when he gets down to 2 doors do you still think your original door has a 50% chance of being right?

right, it has a 1% chance. The same as when you originally chose.

this analogy helped me understand this puzzle when I first saw it. Or how about with the 100 door scenario, you pick a door, Monty reveals 98 goats. Do you still think you have a 99% chance by switching? That is the apt analogy.

[Beleth]

Originally posted by hgc
Or how about with the 100 door scenario, you pick a door, Monty reveals 98 goats. Do you still think you have a 99% chance by switching? That is the apt analogy. It depends on whether Monty knows where the car is, and is purposefully avoiding it - or Monty doesn't know where the car is, and it just so happens that he has avoided revealing it this far. (It also depends on whether Monty must always reveal a door and allow you to change your choice or not, but that adds unneeded complexity.)

If he knows and is avoiding it, your original choice had a 1% chance of being right when you selected it, and it still does... which means that the other door has a 99% chance of having the car.

If he doesn't know and just got lucky, then there is a 50/50 chance.


Here's the analogy that convinced me.
Say you pick one of the three doors.
Instead of showing you that one of the two doors remaining has a goat behind it (which, to be honest, you already knew; you just didn't know which one), Monty gives you the option to select both remaining doors in exchange for the door you just picked.

What would be the odds then?
They would be exactly the same as in the original problem.

[DaveW]

Originally posted by HarryKeogh
imagine there are 100 doors instead of just 3. now imagine Monty asks you after each door is opened do you want to switch. when he gets down to 2 doors do you still think your original door has a 50% chance of being right?

right, it has a 1% chance. The same as when you originally chose.

this analogy helped me understand this puzzle when I first saw it.

Actually, if he narrowed it down to the door I originally picked and one more door, I do think it is now an 50% chance. How is this not so?

[DaveW]

Originally posted by Beleth
It depends on whether Monty knows where the car is, and is purposefully avoiding it - or Monty doesn't know where the car is, and it just so happens that he has avoided revealing it this far. (It also depends on whether Monty must always reveal a door and allow you to change your choice or not, but that adds unneeded complexity.)

If he knows and is avoiding it, your original choice had a 1% chance of being right when you selected it, and it still does... which means that the other door has a 99% chance of having the car.

If he doesn't know and just got lucky, then there is a 50/50 chance.


Here's the analogy that convinced me.
Say you pick one of the three doors.
Instead of showing you that one of the two doors remaining has a goat behind it (which, to be honest, you already knew; you just didn't know which one), Monty gives you the option to select both remaining doors in exchange for the door you just picked.

What would be the odds then?
They would be exactly the same as in the original problem.


I guess the big thing I am not getting is why are the two choices not seperated? The original guess is 1/3, yes, and then you get presented with a new choice (yes/no). Your analogy really doesn't make sense to me.

[HarryKeogh]

Originally posted by DaveW
Actually, if he narrowed it down to the door I originally picked and one more door, I do think it is now an 50% chance. How is this not so?

because he knows which door has the prize behind it. He has to have the prize behind one of those 2 after eliminating 98 other doors.

like Beleth mentioned above, if he didn't know and the prize wasnt revealed it would be a pretty good coincidence and a 50/50 chance

but the important part of the puzzle is that he does know.

I don't know any other way to explain it, hope this helps.

[DaveW]

Originally posted by HarryKeogh
because he knows which door has the prize behind it. He has to have the prize behind one of those 2 after eliminating 98 other doors.

like Beleth mentioned above, if he didn't know and the prize wasnt revealed it would be a pretty good coincidence and a 50/50 chance

but the important part of the puzzle is that he does know.

I don't know any other way to explain it, hope this helps.

I don't see how that follows. Whether he knows what door has the prize or not, in the end, I am still down to one of two doors. Sure, it'd look suspicious if I originally picked door 1 and he walked down all the other doors from 100 down, skipping just door 57, but, in the end, he could have just picked door 57 at random to not open to leave me to switch to when the prize is behind my door. The last choice I am making in this case is between door 1 and door 57, not door 1 and the other 99 doors (98 of which are open).

[rppa]

Originally posted by DaveW
Actually, if he narrowed it down to the door I originally picked and one more door, I do think it is now an 50% chance. How is this not so?

99 times out of a hundred you're going to have the wrong door on your first choice. 99 times out of a hundred that unopened door has the car behind it.

The door you chose and the door that's still closed are not randomly selected.

You make your first choice, picking one door out of a hundred. What are the chances you've got the prize there? 1/100. There is very little chance you got the prize the first time. You KNOW that's not a car behind your door. Why should Monty's shenanigans change that and sudenly make it 50/50 that you got it right the first time? What makes you think there's now a good chance the goat you were certain was there is now a car?

[Jorghnassen]

There's goats in Monty Haul now?

[Paul C. Anagnostopoulos]

Forget the 100 doors explanation. Here's an easier one:

Do you agree that if Monty offered you both of the other doors, you'd switch to them? Of course, since you'd have a 2/3 chance instead of your original 1/3 chance.

Well, that's exactly what Monty did. He silently offered you both doors, then opened the one that doesn't have the car, then asked if you'd like the other one.

~~ Paul

[DaveW]

Originally posted by rppa
99 times out of a hundred you're going to have the wrong door on your first choice. 99 times out of a hundred that unopened door has the car behind it.

The door you chose and the door that's still closed are not randomly selected.

You make your first choice, picking one door out of a hundred. What are the chances you've got the prize there? 1/100. There is very little chance you got the prize the first time. You KNOW that's not a car behind your door. Why should Monty's shenanigans change that and sudenly make it 50/50 that you got it right the first time? What makes you think there's now a good chance the goat you were certain was there is now a car?

OK, let's start with what little I know of statistics and choices. If a choice is independent, the probability of that choice does not change (think consecutive dice rolls). But, if my choice somehow effects the outcome, the probabilities change. So, I choose 1 door (1/3). Monty shows me one bad door. Now my new chances, if I start all over again with this new knowledge, is 1/2 (I won't choose the obviously bad open door). This is essentially what is happening in this example, as far as I can tell.

[hgc]

I just rethought the 100 door analogy. It is a 99% chance of getting the car if switching, because it was a 99% chance that the car was in that pool of 99 that you didn't pick, and since Monty was nice enough to reveal 98 goats (and he must, so goes the premise), then it's a 99% chance that the remaining one is the car.

It's 2/3. End of story (until I change my mind again).

[DaveW]

Originally posted by Paul C. Anagnostopoulos
Forget the 100 doors explanation. Here's an easier one:

Do you agree that if Monty offered you both of the other doors, you'd switch to them? Of course, since you'd have a 2/3 chance instead of your original 1/3 chance.

Well, that's exactly what Monty did. He silently offered you both doors, then opened the one that doesn't have the car, then asked if you'd like the other one.

~~ Paul

Woah! I see it more like: I get a non-sensical first choice. Then he shows me a bad one and I get to choose one out the remaining two. How did I ever get to select 2 doors?

[DaveW]

Originally posted by hgc
I just rethought the 100 door analogy. It is a 99% chance of getting the car if switching, because it was a 99% chance that the car was in that pool of 99 that you didn't pick, and since Monty was nice enough to reveal 98 goats (and he must, so goes the premise), then it's a 99% chance that the remaining one is the car.

It's 2/3. End of story (until I change my mind again).

Gah! Now I'm the only statistics idiot in this thread :p

[Paul C. Anagnostopoulos]

But you aren't starting over, because the bad door could have had the car, even though it does not.

~~ Paul

[Ipecac]

Or here's yet another way to think about it.

You had a 1/3 chance of picking the right door. You have a 2/3 chance of being wrong. When Monty reveals the goat behind one of the other doors, he isn't revealing anything you don't already know. Why? Because given any two doors, one of them MUST have a goat behind it. So there's still a 2/3 chance that the car is behind one of the pair of doors and a 1/3 chance that it's behind the door you picked. The odds don't change because he revealed which of the two doors had the goat.

[DaveW]

Originally posted by Paul C. Anagnostopoulos
But you aren't starting over, because the bad door could have had the car, even though it does not.

~~ Paul

I feel like I'm not making myself clear... the second choice's probability is dependent on the first choice...just like you seemed to confirm. That is why the statistics should change...

[DaveW]

Originally posted by Ipecac
Or here's yet another way to think about it.

You had a 1/3 chance of picking the right door. You have a 2/3 chance of being wrong. When Monty reveals the goat behind one of the other doors, he isn't revealing anything you don't already know. Why? Because given any two doors, one of them MUST have a goat behind it. So there's still a 2/3 chance that the car is behind one of the pair of doors and a 1/3 chance that it's behind the door you picked. The odds don't change because he revealed which of the two doors had the goat.

Ah, but he is! He lets me know one of the two doors that was wrong!

[Jorghnassen]

Any opinions on the Bayesian vs Frequentist point of view on the problem? :)

[Paul C. Anagnostopoulos]

DaveW said:
Woah! I see it more like: I get a non-sensical first choice. Then he shows me a bad one and I get to choose one out the remaining two. How did I ever get to select 2 doors?
You get to select two doors by virtue of (a) Monty eliminating one; (b) you getting to choose the other.

Let me restate it:

Do you agree that if Monty offered you both of the other doors, you'd switch to them? Of course, since you'd have a 2/3 chance instead of your original 1/3 chance.

Well, that's effectively what Monty did. He "silently" offered you both doors, then opened the one that doesn't have the car, then asked if you'd like the other one. You are effectively getting to choose both of the other doors.

~~ Paul

[DaveW]

Just to let everyone know... I'm not trying to bust anyone's chops, I really just don't get it, and I know I don't, but hashing this out like this will hopefully help me learn it. Thanks!

[DaveW]

Originally posted by Paul C. Anagnostopoulos
You get to select two doors by virtue of (a) Monty eliminating one; (b) you getting to choose the other.

Let me restate it:

Do you agree that if Monty offered you both of the other doors, you'd switch to them? Of course, since you'd have a 2/3 chance instead of your original 1/3 chance.

Well, that's effectively what Monty did. He "silently" offered you both doors, then opened the one that doesn't have the car, then asked if you'd like the other one. You are effectively getting to choose both of the other doors.

~~ Paul

Sure, but I only ever get one door at a time. I never "have" 2 doors.

[DaveW]

This post sums up my thinking.
http://www.cut-the-knot.org/terry.shtml

[hgc]

Originally posted by DaveW
This post sums up my thinking.
http://www.cut-the-knot.org/terry.shtml The problem is that it doesn't take into account that when I go into the 2nd choice, my probability of having picked the right door in the 1st choice remains the same. It's still a 1/3 probability that I chose the car the first time. That means that it's still a 2/3 probability that the remaining 2 doors contain the car.

[gnome]

It depends, as has been said before, on whether Monty will always open another door and offer to switch. If that's a given, you're better off switching, because there is a 2/3 chance that one of the remaining two, instead of the door you picked, has the car.

If you don't assume Monty will always open another door, it's back to 50-50.

Questions?

[Iconoclast]

Originally posted by DaveW
Sure, but I only ever get one door at a time. I never "have" 2 doors.
If you swap then you DO have two doors. What Monte is doing is saying "You can keep the first door you chose OR you can have either of the other two, and look here I've just opened one of them and it's not the right one, so if you switch you'll know which of the two doors to pick"

[rppa]

Originally posted by DaveW
Just to let everyone know... I'm not trying to bust anyone's chops, I really just don't get it, and I know I don't, but hashing this out like this will hopefully help me learn it. Thanks!

Well, we're all trying to get at your intuition. The demonstration is much easier.

Demonstration: Start 100 identical games in 100 identical studios, all with the car behind door 57. In each game the contestant picks a different door.
Now Monty is going to go from studio to studio. 99 of these people didn't pick door 57. For those he leaves them only the door they have, and door 57. In the other studio he leaves one other door unopened.

Now all contestants switch to door 57. 99 of them win.

If none of the contestants switch, only 1 wins.

So in 99 out of 100 identical games, switching led to a win. Thus, the probability of a win from a switch is 99%.

Now the intuition. Let's try this a different way. You pick a door. You're pretty sure you picked wrong, correct? So Monty is going to tease you by opening doors. How certain are you that the car is among the doors he's going to open? Pretty sure. You should be, there's a 99% chance of that.

But Monty is feeling cruel today and opens goat doors first. Now he's down to 10 doors. You know he controls the game, and you knew before he opened anything that he had a 99% chance of having the car, right? He's just teasing you, opening that one last.

And now he's down to one door, and he starts to open it... what are the chances that you're going to see a car there? Don't you think it's pretty likely? You thought it was 99% sure he had the car before opening doors, he's just being mean about the order of opening. So why if he waits to open the car last, is there suddenly a 50% chance that he didn't have the car at all? You KNOW he has the car.

Switch places. Monty is guessing, you know where the car is. Monty picks a (probably wrong door) and you're going to play the cruel open-the-car-last game. But you know where the car is. Don't you think it's pretty likely that in nearly every game, you're going to get to one last door that has the car behind it?

Play it out with cards or coins.

[Dragon]

Originally posted by DaveW
This post sums up my thinking.
http://www.cut-the-knot.org/terry.shtml Then think again, because that post is wrong, wrong, wrong.

Try this link (http://www.grand-illusions.com/monty.htm) - and have a go at the simulator.

[DaveW]

Originally posted by rppa
Well, we're all trying to get at your intuition. The demonstration is much easier.

Demonstration: Start 100 identical games in 100 identical studios, all with the car behind door 57. In each game the contestant picks a different door.
Now Monty is going to go from studio to studio. 99 of these people didn't pick door 57. For those he leaves them only the door they have, and door 57. In the other studio he leaves one other door unopened.

Now all contestants switch to door 57. 99 of them win.

If none of the contestants switch, only 1 wins.

So in 99 out of 100 identical games, switching led to a win. Thus, the probability of a win from a switch is 99%.

Now the intuition. Let's try this a different way. You pick a door. You're pretty sure you picked wrong, correct? So Monty is going to tease you by opening doors. How certain are you that the car is among the doors he's going to open? Pretty sure. You should be, there's a 99% chance of that.

But Monty is feeling cruel today and opens goat doors first. Now he's down to 10 doors. You know he controls the game, and you knew before he opened anything that he had a 99% chance of having the car, right? He's just teasing you, opening that one last.

And now he's down to one door, and he starts to open it... what are the chances that you're going to see a car there? Don't you think it's pretty likely? You thought it was 99% sure he had the car before opening doors, he's just being mean about the order of opening. So why if he waits to open the car last, is there suddenly a 50% chance that he didn't have the car at all? You KNOW he has the car.

Switch places. Monty is guessing, you know where the car is. Monty picks a (probably wrong door) and you're going to play the cruel open-the-car-last game. But you know where the car is. Don't you think it's pretty likely that in nearly every game, you're going to get to one last door that has the car behind it?

Play it out with cards or coins.

I played it out with Excel, 2000+ runs of the three door gag. According to what I have seen so far, there is not a doubling of probability by switching. There is a slight advantage, on average to switching, though (probably due to the whole effect of Monty having to pick a bad one for you, tipping the scales somewhat). But, it really appears to not support the doubling mantra. I can't post it here from work, so I'll try to do it from home a bit later.

[DaveW]

Originally posted by Dragon
Then think again, because that post is wrong, wrong, wrong.

Try this link (http://www.grand-illusions.com/monty.htm) - and have a go at the simulator.

I played with those simulators. Problem is, it would take a long time to get a good sample size.

[Yaotl]

Originally posted by DaveW
I played with those simulators. Problem is, it would take a long time to get a good sample size.

You can let it run itself.

[DaveW]

Originally posted by Yaotl
You can let it run itself.

Oops, I was thinking of the online simulators. I'll look into those.

[Ipecac]

Originally posted by DaveW
Ah, but he is! He lets me know one of the two doors that was wrong!

Believe me, I went through the same arguments you did when I first heard of this.

It doesn't matter which of the two doors is wrong. You already know that at least one of them is wrong. Whether it's B or C is irrelevant to the odds.

If he offered you the choice to pick A or B & C, you would pick B & C, right? And you would do so KNOWING that one of the two, B or C, must have a goat. So when Monty reveals that the goat is actually behind B, that doesn't change the odds one bit.

[DaveW]

I played the Excel sim I made out to about 8000 runs. I'll try to post it by tomorrow, but, if the theory is actually right, something is wrong with my spreadsheet (yeah, more likely, I know, but I want to know what!)

[Ipecac]

Assuming the programming is correct, this simulator: Montysim (http://www.grand-illusions.com/simulator/montysim.htm) demonstrates the odds very effectively. Over 1000 tests, and the odds came out exactly as they should. Changing your door wins the car twice as much as keeping your door.

Try it!

[Robin]

Look at it this way. Someone who consistently adopts the switching strategy will always be rewarded if they chose wrongly in the first place.

You have a 2/3 chance of being wrong in the first place so your odds of getting the prize are 2/3 by changing your choice.

If you try this game 100 times and always switch you should win approximately 66 times (not 50 times).

If you try this game 100 times and always stick you will win approximately 33 times.

[Rolfe]

It's déjà vu all over again (http://forums.randi.org/showthread.php?s=&threadid=41501).

I swear some of the posts are verbatim.

Rolfe.

[Mason]

I just wanted to say one small thing.

.999... = 1

Carry on.

[Yaotl]

Originally posted by Mason
I just wanted to say one small thing.

.999... = 1

Carry on.

2+2=5 for extremely large values of 2

[Robin]

Originally posted by Rolfe
It's déjà vu all over again (http://forums.randi.org/showthread.php?s=&threadid=41501).

I swear some of the posts are verbatim.

Rolfe.

Why are so many people still not getting it? In the original thread there are some correct explanations but they are all unnecessarily long.

[fishbob]

Wait a minute.

Your first choice with all 3 doors closed is not relevant, because the first pick was only for dramatic effect - it has no bearing on the outcome. Your option to choose a second time is the only real choice, so your odds are 50/50. The real choice is between the 2 closed doors and a goat. I guess there is a slight chance some kind of survivalist or pervert would pick the goat, leaving each door with a chance of 49.99 %.

Am I missing something?

[Ipecac]

Originally posted by fishbob
Wait a minute.

Your first choice with all 3 doors closed is not relevant, because the first pick was only for dramatic effect - it has no bearing on the outcome. Your option to choose a second time is the only real choice, so your odds are 50/50. The real choice is between the 2 closed doors and a goat. I guess there is a slight chance some kind of survivalist or pervert would pick the goat, leaving each door with a chance of 49.99 %.

Am I missing something?

Yes.

Go run the simulator posted above. It will demonstrate the results very effectively.

[Robin]

Originally posted by fishbob
Wait a minute.

Your first choice with all 3 doors closed is not relevant, because the first pick was only for dramatic effect - it has no bearing on the outcome. Your option to choose a second time is the only real choice, so your odds are 50/50. The real choice is between the 2 closed doors and a goat. I guess there is a slight chance some kind of survivalist or pervert would pick the goat, leaving each door with a chance of 49.99 %.

Am I missing something?

If the odds were 50/50 and you played the game 1000 times then you would win about 500 cars whichever strategy you took right?

But when you actually play the game 1000 times and never switch you win about 333 cars.

If you play the game 1000 times and always switch you win about 666 cars.

If you doubt this then try it. So yes, you are missing something.

Tell me if the following statement is true or false about the problem:

"Someone who always switches will always get the car if their initial guess was wrong"


If the above statement is true then the only other question is "what is the probability of the initial guess being wrong"

(edited to change the number tries)

[epepke]

Originally posted by gnome
It depends, as has been said before, on whether Monty will always open another door and offer to switch. If that's a given, you're better off switching, because there is a 2/3 chance that one of the remaining two, instead of the door you picked, has the car.

If you don't assume Monty will always open another door, it's back to 50-50.

Actually, the ultimate result is that it's whatever Monty feels like doing at the time. Even the real Monty Hall understood this. Which, of course, he should, because he's smarter than most of the 'tards here.

But I'm convinced from empirical evidence that it's not possible to explain this to people. Fortunately, it is possible and a lot more fun to win free drinks off of them.

[Robin]

Actually, the ultimate result is that it's whatever Monty feels like doing at the time. Even the real Monty Hall understood this. Which, of course, he should, because he's smarter than most of the 'tards here

It has nothing to do with what Monty feels like doing as long as he behaves as stated in the problem. I am convinced that all disagreement in this topic is based around misunderstanding the problem.

Here is how I understand it:


1. There are three doors

2. Behind one door is a car

3. Behind the two other doors are goats

4. You make an initial guess but the door you guessed is not opened yet

5. The host opens one of the doors you didn't pick revealing a goat.

6. You are given the chance to change your guess.

The question posed is do you improve your odds of winning the car by changing your guess and if so by how much?


If you disagree then please state the problem as you understand it.

Step 5 appears to be the sticking point. Some, like gnome, are suggesting that he can arbitrarily leave out this step. But if he can do this then it becomes another problem entirely.

If the problem is as stated then your chances of winning the car are 2/3 if you change, 1/3 if you don't for reasons that have been explained already.

[rppa]

Originally posted by fishbob
Your first choice with all 3 doors closed is not relevant, because the first pick was only for dramatic effect - it has no bearing on the outcome.

Huh? There are two possibilities on the first pick: You picked a car, or you picked a goat. In 1/3 of the games you are holding a car, in 2/3 a goat.

Do you agree with that much? Because that's the crux of it.

Your option to choose a second time is the only real choice, so your odds are 50/50. The real choice is between the 2 closed doors and a goat.

No, you had a first choice, and it was completely random. And in 2/3 of the games, that random choice will leave you with a goat.

At 2nd choice time, in 2/3 of the games you chose a goat first and Monty is showing you a door with a car. In 1/3 of the games you chose a car the first time and Monty is showing you a door with a goat.

Am I missing something?

Yes: your assertion that the first choice is irrelevant, since the first choice is all. If your first choice was a goat, then there is a 100% chance that Monty's door is a car.

[apoger]

Let me see if I can add some clarity to all this...

Proposition #1
We will flip an unbiased coin three times. What are the odds that heads will come up three times?

Answer:
There are eight patterns that may emerge
hhh
hht
hth
htt
ttt
tth
tht
thh

Thus the probability of three heads in a row is one in eight.

Experiment #1

We flip a coin ----> Heads


Experiemnt #2

We flip a coin ----> Heads


Proposition #2
We will flip an unbiased coin one time. What are the odds that heads will come up this time?

Answer:
There are only two possibilities. Heads or tails. The odds of either are one in two.

Proposition #1 does not affect proposition #2.
The prior coin flips do not affect the odds of proposition #2.

This is a common example used in teaching statistics, and as such will be familiar to most people here.

As far as I can tell the problem at hand is almost the same situation.


Proposition #3
Monty offers you the choice of three doors. One has a car behind it, the other two have goats.

You can think of this as:

Car Goat Goat

Experiment #3
You pick one.
It does not matter what you pick. If you pick "car" then mentally eliminate the left "goat". If you pick a "goat", then mentally eliminate the other goat.

You are left with two doors. One has a car. One has a goat. One was your choice, the other not.


Proposition #4
Monty offers you the choice of two doors. One has a car behind it, the other has a goat. One door was chosen by you ealier. You can keep "your door" or just as easily change to the other. What is the odds of getting the "car"?

I submit that proposition #3 has no bearing on proposition #4. They are different choices. The fact that you picked a door earlier has no relevance to proposition #4. It's exactly the same as coin flips that happened prior to proposition #2.

All that's important is:
A> There are two choices
B> You are equally able to choose either
C> One choice is victory and the other defeat

This is exactly the same as a coin flip.
The odds are 1 in 2, or 50%.

Many people here seem intent on adding bigger numbers, as if that changes anything. It does not. If there are a BILLION doors and I pick one, and only one is a winner, then my odds are one in a billion. However if God comes by and eliminates all but two choices, one of which is the winner, and I can choose either... it's back to one in two. The prior one in a billion choice has no bearing on the subsequent one in two choice.


I hope that this helps.... ;)

[Robin]

Proposition #4
Monty offers you the choice of two doors. One has a car behind it, the other has a goat. One door was chosen by you ealier. You can keep "your door" or just as easily change to the other. What is the odds of getting the "car"?

I submit that proposition #3 has no bearing on proposition #4. They are different choices. The fact that you picked a door earlier has no relevance to proposition #4. It's exactly the same as coin flips that happened prior to proposition #2.

All that's important is:
A> There are two choices
B> You are equally able to choose either
C> One choice is victory and the other defeat

This is exactly the same as a coin flip.
The odds are 1 in 2, or 50%.


Proposition #3 has no bearing only if you ignore the information you got from it and randomly choose between staying and changing.

But you don't make a random choice about this you make an informed choice about this.

Your logic would go "If my initial guess was wrong then the door Monty is offering me must have the car. There is a 0.667 probability of my initial guess being wrong, therefore there is an 0.667 chance of Monty's door having the car."

[apoger]

Originally posted by Robin
Proposition #3 has no bearing only if you ignore the information you got from it and randomly choose between staying and changing.

What info transfers?
I submit none. Just like coin flips from prior experiments. They already happened and do not influence the current flip.


But you don't make a random choice about this you make an informed choice about this.

Your logic would go "If my initial guess was wrong then the door Monty is offering me must have the car. There is a 0.667 probability of my initial guess being wrong, therefore there is an 0.667 chance of Monty's door having the car."


How do you get the .667 probability? It seems to me that this is the statistic from the first proposition. However new info has been obtained (one goat exposed) and the odds must be recalculated for the new proposition which is just between two choices.

You are making the mistake from the coin example. By your thinking a thrid coin flip would not be 50/50 just becuase the last two experiments came up both heads.

[Robin]

You are making the mistake from the coin example. By your thinking a thrid coin flip would not be 50/50 just becuase the last two experiments came up both heads.

No you are making the mistake of assuming the examples are the same. You cannot control how a coin comes down but you can control the decision to stay or switch.

Just tell me, as I asked before, is the following statement true or false?

"If I change my selection and my initial guess was wrong then I will definitely get the car"

If you accept that statement as true then this is the information that transfers from your proposition #3.

If you don't accept that statement then please give an example of a case where your initial guess is wrong, you change your choice and you don't get the car.

[rppa]

Let me see if I can add some clarity to all this...

There is no clarity in all this. Monty Hall discussions will go on until the heat death of the universe.

Proposition #4
Monty offers you the choice of two doors. One has a car behind it, the other has a goat. One door was chosen by you ealier. You can keep "your door" or just as easily change to the other. What is the odds of getting the "car"?

I submit that proposition #3 has no bearing on proposition #4.

This is the fundamental error. Monty's actions depend on yours. Unlike the coin toss, it is not an independent action, but is completely deterministic.

In probability terms, let A = you chose a car on your first trial
~A = you chose a goat
Let B = Monty chooses a car after his actions,
~B = Monty's door is a goat

If your actions and Monty's are independent, then P(B|A) = P(B).
That is, the probability that Monty chooses a goat is independent of what you do. But that's not true.

In fact, the true model is:

P(B|A) = 0
P(B|~A) = 1

If you chose a goat, Monty's door is a car. P(B|~A) = 1.
If you chose a car, Monty's door is a goat. P(B|A) = 0.

Now the question we want to know is, what is P(B)? When all the actions are over, what is the probability that Monty's door holds a car?

P(B) = P(B|A) P(A) + P(B|~A) P(~A)
= 0 * (1/3) + 1 * (2/3)
= 2/3.


They are different choices. The fact that you picked a door earlier has no relevance to proposition #4. It's exactly the same as coin flips that happened prior to proposition #2.

Absolutely untrue. The doors Monty is offering you depend on what you chose the first time. Hence the conditional probabilities are different than the unconditioned probability.

All that's important is:
A> There are two choices
B> You are equally able to choose either
C> One choice is victory and the other defeat

Nope. And it's because Monty's actions are conditioned on yours.

Here are the two possibilities for that fateful 2nd choice, before we open the two remaining doors:

Door 1 2
Goat Car
Car Goat

Now, you are saying that Monty has arranged things so that these are equally likely outcomes. But they aren't. If Door 1 is your original
choice and Door 2 is the one Monty offers after opening the rest, then the probability that the arrangement is (Goat, Car) is (N-1)/N where N = the number of original doors. The probability that the final arrangement is (Goat,Car) is exactly the probability that the first choice was Goat.

The first coin flip was NOT a fair coin. You were NOT equally able to choose Goat or Car on the first choice, and so you aren't equally likely to be based with (Goat, Car) and (Car, Goat) after Monty's move.

If Monty had you choose "your" door after he opened all but 2, then we'd be faced with a 50/50 choice. Then we'd have P(A) = P(B) = 1/2. But that's not the situation. P(A) is not 1/2, it's 1/N.

Many people here seem intent on adding bigger numbers, as if that changes anything. It does not. If there are a BILLION doors and I pick one, and only one is a winner, then my odds are one in a billion.

Precisely.

However if God comes by and eliminates all but two choices, one of which is the winner, and I can choose either... it's back to one in two. The prior one in a billion choice has no bearing on the subsequent one in two choice.

Also correct. However, if God lets you choose first, so that there is only a 1 in a bililion chance you've got the car, then there is a 999,999,999 in a billion chance that God's door is the car. Those outcomes are all the ones in which you didn't pick the car first.

I hope that this helps.... ;)

It helped me sort out the thinking of the 50-percenters.

[apoger]

After rereading all this insanity, suddenly I "got it".
It still seems counter-intuitive... but you (er... all of you) are correct. The first choice does transfer information that can be used to improve the odds in the second pick.

Now the trick will be to figure out a clear way to communicate it. ;)

[epepke]

Originally posted by Robin
It has nothing to do with what Monty feels like doing as long as he behaves as stated in the problem. I am convinced that all disagreement in this topic is based around misunderstanding the problem.

Here is how I understand it:

If you disagree then please state the problem as you understand it.

As I said, I have empirical evidence that this cannot be explained. It just bounces off people's foreheads. That's why I prefer to win drinks off of them.

However, with full expectation that few people will read this, and those who read it will fight it, I feel comfortable in saying the following. Nothing in the original question to vos Savant indicated that Monty is in any way obligated to offer a choice. For all you know, he could only offer you the choice if you had picked the curtain with the car. Which means that a strategy to pick the other curtain would fail 100% of the time.

But people are just so terribly impressed with themselves for understanding what is essentially the kind of math that should reasonably be expected from a 14-year-old that they work terribly hard to avoid the obvious conclusion. Which means they suck at being skeptics. Which is why I don't feel bad for winning free drinks off of them and humiliating them in front of their peers. Because skeptics should think about these things.

[Robin]

Originally posted by epepke
However, with full expectation that few people will read this, and those who read it will fight it, I feel comfortable in saying the following. Nothing in the original question to vos Savant indicated that Monty is in any way obligated to offer a choice. For all you know, he could only offer you the choice if you had picked the curtain with the car. Which means that a strategy to pick the other curtain would fail 100% of the time.


Well as I said please state the problem as you understand it or offer a link to it.

I have not seen the version where Monty can decide arbitrarily whether or not to offer the second choice.

If this was the case then you couldn't have a strategy to pick the other curtain 100% of the time because you wouldn' t be offered the choice 100% of the time.

[hgc]

Look at it this way...

Each door has a 1/3 chance of being the car. So with the first choice your door has a 1/3 chance and the other 2 doors together have a 2/3 chance. But then one of those other 2 doors is knocked out. But still the other 2 doors have their 2/3 chance. The revealed door went to zero chance and transfered its 1/3 to the switch choice door, giving it 2/3.

[Robin]

Originally posted by gnome
It depends, as has been said before, on whether Monty will always open another door and offer to switch. If that's a given, you're better off switching, because there is a 2/3 chance that one of the remaining two, instead of the door you picked, has the car.

If you don't assume Monty will always open another door, it's back to 50-50.

Questions?

I asked epepke and I will ask you - please state the problem or show a link to the version where the host can decide arbitrarily whether or not to open another door and offer to switch.

This is the way I saw it:

Suppose you're on a game show, and you're given the choice of three doors; Behind one door is a car; behind the others, goats. You pick a door, say No.1 and the host, who knows what's behind the doors, opens another door, say No.3, which has a goat. He then says to you, "Do you want to pick door No.2?" Is it to your advantage to switch your choice? -Craig F. Whitaker, Columbia, Md.

Let me emphasise this :

...the host, who knows what's behind the doors, opens another door, say No.3, which has a goat. He then says to you, "Do you want to pick door No.2?"

Revealing the goat and offering a choice is part of the conditions of the problem, and so the probability, if you always change selections, is 0.667.

If the host does not complete these steps then certainly that changes the odds. It also changes the odds if the host does not turn up in the first place and has no goats or cars.

[apoger]

And now an attempt to clarify for all those that haven't made the connection...

First thing is we drop Monty, doors, car, and the goats, as they are dragging too much emotional baggage at this point.


Let us say that you and I are sitting at a table.
I place 10 paper cups on the table.
Under one cup I hide a walnut.

I move five cups to the right side of the table.
I move five cups to the left side of the table.

The odds of the walnut being on the left side is 50%. Same goes for the right side. I think we can all see that.

Now I move one cup from left to right. The odds of the walnut are being on the right side is now 60%. The left is down to 40%.

Then I move three more cups from the left to the right. The nine cups on the right represent a 90% chance of having a wlanut. The single cup on the left has a mere 10% chance.

Here comes the trick...
I now turn over eight cups that have no walnut. All eight are on the right side. Now we have new information. If the 90% chance of the walnut being on the right side is valid, then there is a 100% chance that the walnut is under the last cup on the right side. However this does not change the 90% chance the the walnut is on the right side. Thus there is a 90% chance of the right cup having the walnut and a 10% chance of the left cup having it.

Given a choice between the two cups, you would of course gamble on the right cups 90% chance.


This visualization helped me focus on the probem. I hope it's simplicity has helped someone else. :)

[TeaBag420]

Originally posted by epepke
Actually, the ultimate result is that it's whatever Monty feels like doing at the time. Even the real Monty Hall understood this. Which, of course, he should, because he's smarter than most of the 'tards here.

But I'm convinced from empirical evidence that it's not possible to explain this to people. Fortunately, it is possible and a lot more fun to win free drinks off of them.

I think that your use of the word "'tards" is offensive.

[Robin]

Originally posted by apoger
And now an attempt to clarify for all those that haven't made the connection...

First thing is we drop Monty, doors, car, and the goats, as they are dragging too much emotional baggage at this point.


Let us say that you and I are sitting at a table.
I place 10 paper cups on the table.
Under one cup I hide a walnut.

I move five cups to the right side of the table.
I move five cups to the left side of the table.

The odds of the walnut being on the left side is 50%. Same goes for the right side. I think we can all see that.

Now I move one cup from left to right. The odds of the walnut are being on the right side is now 60%. The left is down to 40%.

Then I move three more cups from the left to the right. The nine cups on the right represent a 90% chance of having a wlanut. The single cup on the left has a mere 10% chance.

Here comes the trick...
I now turn over eight cups that have no walnut. All eight are on the right side. Now we have new information. If the 90% chance of the walnut being on the right side is valid, then there is a 100% chance that the walnut is under the last cup on the right side. However this does not change the 90% chance the the walnut is on the right side. Thus there is a 90% chance of the right cup having the walnut and a 10% chance of the left cup having it.

Given a choice between the two cups, you would of course gamble on the right cups 90% chance.


This visualization helped me foucus on the probem. I hope it's simplicity has helped someone else. :)

I think that is a very good way of putting it. When I first heard the problem I said 50% and it took me quite a long time to change my mind.

[Robin]

Originally posted by TeaBag420 in response to epepke
I think that your use of the word "'tards" is offensive.
Not only that but inaccurate because the result has nothing to do with how the host feels.

The host can be malicious, fair or helpful but it does not affect the outcome one iota.

So it would be offensive if he was right. If he is wrong then it blows right back in his face and is rather amusing.

Epepke, feel free to answer my last post and state the version of the problem where the host has the option of not offering the second choice.

[TeaBag420]

Originally posted by epepke
As I said, I have empirical evidence that this cannot be explained. It just bounces off people's foreheads. That's why I prefer to win drinks off of them.

However, with full expectation that few people will read this, and those who read it will fight it, I feel comfortable in saying the following. Nothing in the original question to vos Savant indicated that Monty is in any way obligated to offer a choice. For all you know, he could only offer you the choice if you had picked the curtain with the car. Which means that a strategy to pick the other curtain would fail 100% of the time.

But people are just..... a 14-year-old..... terribly hard to avoid ....... suck .....Which is why I don't feel bad for ..... humiliating them in front ...... think about these things.

And if Monty followed the strategy outlined above, most contestants would go home winners, because they wouldn't switch...WHETHER THEY KNEW ABOUT THE RESTRICTION OR NOT.

Why don't you back up your assertion about the original question by citing your source?

[69dodge]

Originally posted by DaveW
This post sums up my thinking.
http://www.cut-the-knot.org/terry.shtmlHere's an excerpt from that web page:<blockquote>What's being said is that if I pick #1 and monty shows #3, then 2/3 of the time #2 will win. So If I had picked #2 to start with and monty opens #3, then the odds go 2/3 to #1. Why would they change? What if I don't have to tell monty? What if I could write it on a secret ballot?</blockquote>If you write your initial door choice on a secret ballot and don't tell Monty what it is, he might end up opening the same door you chose. If you use a secret ballot, 50/50 is indeed the correct answer, even in those instances where Monty happens not to open your door.

If Monty is guaranteed to open a goat door, and if he's guaranteed not to open your door, then the extra car odds go to the door that he might have opened but in fact didn't. This makes perfect sense: probably, the reason he didn't open that door is precisely that it has a car behind it. On the other hand, you knew from the beginning that he wasn't going to open your door, so the fact that he didn't open it doesn't tell you anything you didn't already know about what's behind it.

It's not seeing the goat that's important; it's seeing Monty's door choice. In particular, it's seeing which door he chooses to open when he is constrained (a) to open one door, (b) not to open your door, and (c) not to open the car door. If he is not so constrained, the odds will be different, even if you happen to end up seeing the same goat behind the same door as before.

[TeaBag420]

Nevermind, I'll do it....

The Monty Hall Dilemma was discussed in the popular "Ask Marylin" question-and-answer column of the Parade magazine. Details can also be found in the "Power of Logical Thinking" by Marylin vos Savant, St. Martin's Press, 1996.

Marylin received the following question:
Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?
Craig. F. Whitaker
Columbia, MD

[Robin]

What if I could write it on a secret ballot?

It would save a lot of time and space if everyone would agree to address the problem as stated, and not if it were another totally unrelated problem.

[Robin]

Originally posted by TeaBag420
Nevermind, I'll do it....

The Monty Hall Dilemma was discussed in the popular "Ask Marylin" question-and-answer column of the Parade magazine. Details can also be found in the "Power of Logical Thinking" by Marylin vos Savant, St. Martin's Press, 1996.

Marylin received the following question:
Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?
Craig. F. Whitaker
Columbia, MD

I quoted this a few posts back but it is worth seeing again. The conditions of the problem clearly state that (i) the host opens another door and that (ii) the door he opens has a goat behind it and (iii) that he offers a second choice.

The answer does not depend on the intentions of the host.

Or am I being a 'tard?

[insomneac]

Marilyn vos Savant popularized this problem decades ago. One of her books (can't remember its title) contains her solution to the problem and the results of nationwide school trials. In trial after trial, switching resulted in winning the car almost twice as often as staying.

[rppa]

Originally posted by 69dodge
Here's an excerpt from that web page:<blockquote>What's being said is that if I pick #1 and monty shows #3, then 2/3 of the time #2 will win. So If I had picked #2 to start with and monty opens #3, then the odds go 2/3 to #1. Why would they change? What if I don't have to tell monty? What if I could write it on a secret ballot?</blockquote>

DaveW hasn't told us his spreadsheet formulas so we don't know where his simulation went wrong. I think I can address this excerpt though.

We're not saying that 2/3 of the time #2 will win. We're saying that 2/3 of hte time Monty's door will win.

If I pick door #1, there are three equally likely possibilities:

1. I've got the winner already. Monty picks #2 or #3 with equal probability.
2. I've got a goat and the car is in #2. Monty opens #3.
3. I've got a goat and the car is in #3. Monty opens #2.

In 2 of these 3 possibilities, the other door is the winner. The odds aren't 2/3 that #2 is the winner, they're 2/3 that the door Monty didn't open is the winner. 1/3 on #2, 1/3 on #3.

If I picked #2 to begin with, there is still a 1/3 chance of the car being in #1, #2, or #3. But in the case that it is in #1 or #3, the winner will be the door left after Monty's move.

If I don't tell Monty what I picked, then his move is no longer dependent on mine, and we're in the 50/50 situation. In fact, he might choose to open the door I already picked, something that can't happen in our version of the game.

[69dodge]

Originally posted by Robin
It has nothing to do with what Monty feels like doing as long as he behaves as stated in the problem.epepke is interpreting the problem statement simply as a description of what Monty does in one particular instance. You are interpreting the problem statement not only as a description of what Monty does, but also as a description of what the contestant, before playing, knew Monty was going to do. You're both right, given your respective interpretations.

[Robin]

Originally posted by 69dodge
epepke is interpreting the problem statement simply as a description of what Monty does in one particular instance. You are interpreting the problem statement not only as a description of what Monty does, but also as a description of what the contestant, before playing, knew Monty was going to do. You're both right, given your respective interpretations.

No, it does not matter if the contestant knows what Monty is going to do, or what Monty's intentions are. It makes no difference.

If the conditions as set out in the problem are satisfied then the answer is "Yes - you double your chances of winning by switching".

Somebody please show me a case where Monty can affect the outcome.

[insomneac]

Edited because Robin said it better.

[CurtC]

Originally posted by Robin
I quoted this a few posts back but it is worth seeing again. The conditions of the problem clearly state that (i) the host opens another door and that (ii) the door he opens has a goat behind it and (iii) that he offers a second choice.

The answer does not depend on the intentions of the host.

Or am I being a 'tard?You're not being a 'tard, but you're wrong nonetheless. The answer depends greatly on the intentions of the host, because in the original problem statement (as given in the OP) all you know is that the host switched this time, not whether he had to.

My favorite illustration of this is one I came up with myself. Imagine that you're walking down the street and come upon a street hustler offering a game of find-the-ball-under-a-cup. You plop down your $20, and the hustler mixes the three cups around very quickly, so you get lost. You make a guess at random, and the hustler then shows you one of the empty cups, and asks if you'd like to switch your guess to the other one. Would you switch?

You'd be a fool if you did, because the street hustler wouldn't be offering you the choice at all had you guessed wrong initially. If you switch, you have a 100% chance of losing.

The only difference between this problem statement and the OP is who the person hosting is, and what his intentions can be assumed to be.

[69dodge]

Originally posted by Robin
No, it does not matter if the contestant knows what Monty is going to do, or what Monty's intentions are. It makes no difference.Suppose you know, before starting your game, that Monty intends to allow you to switch only if you initially pick the car. If you pick a goat, he'll just give you the goat.

Now, you start playing the game. You pick a door. Monty opens a different door, revealing a goat, and asks if you'd like to switch to the third door. Would you switch?

Of course you wouldn't. The fact that he offered to let you switch tells you that you picked the car.

Monty's actions are exactly as described in the opening post of this thread. What you know about his intentions makes all the difference.

[insomneac]

Originally posted by 69dodge
Suppose you know, before starting your game, that Monty intends to allow you to switch only if you initially pick the car. If you pick a goat, he'll just give you the goat.

Now, you start playing the game. You pick a door. Monty opens a different door, revealing a goat, and asks if you'd like to switch to the third door. Would you switch?

Of course you wouldn't. The fact that he offered to let you switch tells you that you picked the car.

Monty's actions are exactly as described in the opening post of this thread. What you know about his intentions makes all the difference.

But you don't know Monty's intentions at all. Therefore you don't take them into consideration.

[Robin]

Originally posted by 69dodge
Suppose you know, before starting your game, that Monty intends to allow you to switch only if you initially pick the car. If you pick a goat, he'll just give you the goat.

Now, you start playing the game. You pick a door. Monty opens a different door, revealing a goat, and asks if you'd like to switch to the third door. Would you switch?

Of course you wouldn't. The fact that he offered to let you switch tells you that you picked the car.

Monty's actions are exactly as described in the opening post of this thread. What you know about his intentions makes all the difference.

OK, from the last two posts (yours and CurtC) I see where you are coming from. Let me sleep on it.

[TeaBag420]

What Monty thinks, wants, or is required to do is IRRELEVANT.

What matters is what he does.

It's so tempting to call folks retarts, but I struggled with this problem for about an hour before someone beat the correct understanding into my head. So my heart goes out to the retarts.

[Cabbage]

Originally posted by Robin
I quoted this a few posts back but it is worth seeing again. The conditions of the problem clearly state that (i) the host opens another door and that (ii) the door he opens has a goat behind it and (iii) that he offers a second choice.

The answer does not depend on the intentions of the host.
Actually, the intentions of the host do matter. Here are a couple of examples:

Suppose sometimes the host opens a door after you make your selection, and sometimes he doesn't. Now, when it's your turn to play, he opens a door to reveal a goat. Should you switch? In this example, I don't think it's so clear what the probabilities of switching vs. not switching are. It depends on the motivations of the host--Maybe he only opens a door and offers the switch when you've already picked the correct door. Maybe he just likes to be tricky like that.

On the other hand, say the host always opens a door other than yours, but suppose the host doesn't know where the car is himself. Say you pick door 1, and the host says, "Just to make it interesting, let's see what's behind one of the doors you didn't pick...Door 3. If there's a car behind that door, you lose; if it's a goat, I'll give you a chance to switch!"

He opens the door, revealing a goat. Should you switch?

Now's it's 50/50 whether you switch or not. To calculate it:

First, what were the chances of a goat being revealed in the first place? The host's door is picked at random, so there's a 2/3 chance the goat is revealed. (Note that if the host always shows a goat door, this probability would be 100%. This note to be continued to contrast the two problems).

Now, what is the probability that both 1. A goat was revealed, and 2. You picked the right door?

The probability that picked right is 1/3. Given that has happened, the probability a goat is revealed is 100%. So the probability of them both happening is 1/3. (Note, if a goat is always revealed, the chance of these two events happening is still 1/3).

Now, given that a goat was revealed (2/3 chance), what is the probabilty that both a goat was revealed and you picked the right door? (1/3) divided by the 2/3 were normalizing on, (i.e., given that the goat was revealed). So the probability you picked correctly is 1/2--It doesn't matter if you switch or not.

(Note that when the goat is always revealed, probability becomes (1/3) / 1 = 1/3 chance of picking correctly initially (so 2/3 chance of winning if you switch, which answers the question as it's generally intended to be stated--that you know Monty always reveals a goat, (intentionally, of course)).

[Robin]

By Cabbage
...but suppose the host doesn't know where the car is himself?


Unfortunately I was looking at another version of the problem, as asked of Marilyn Vos Savant in 1990. The version stipulated that he does. As I said before, the problem is in differing understandings of the question.

[Cabbage]

Originally posted by Robin
Unfortunately I was looking at another version of the problem, as asked of Marilyn Vos Savant in 1990. The version stipulated that he does. As I said before, the problem is in differing understandings of the question.
OK, I wasn't sure about that.

On the other hand, it still makes a difference in that version of the problem if the host sometimes opens a door, and sometimes doesn't.

[insomneac]

Originally posted by Cabbage

Suppose sometimes the host opens a door after you make your selection, and sometimes he doesn't. Now, when it's your turn to play, he opens a door to reveal a goat. Should you switch? In this example, I don't think it's so clear what the probabilities of switching vs. not switching are. It depends on the motivations of the host--Maybe he only opens a door and offers the switch when you've already picked the correct door. Maybe he just likes to be tricky like that.



There's that same mistake. The stated problem doesn't say squat about Monty's intentions or his knowledge about what lies behind each door. So stop making unwarranted assumptions when solving the problem.

[Robin]

Cabbage
On the other hand, say the host always opens a door other than yours, but suppose the host doesn't know where the car is himself. Say you pick door 1, and the host says, "Just to make it interesting, let's see what's behind one of the doors you didn't pick...Door 3. If there's a car behind that door, you lose; if it's a goat, I'll give you a chance to switch!"

He opens the door, revealing a goat. Should you switch?

Now's it's 50/50 whether you switch or not. To calculate it:

First, what were the chances of a goat being revealed in the first place? The host's door is picked at random, so there's a 2/3 chance the goat is revealed. (Note that if the host always shows a goat door, this probability would be 100%. This note to be continued to contrast the two problems).

Now, what is the probability that both 1. A goat was revealed, and 2. You picked the right door?

But in this case the odds are still 0.667 if you switch, even if the host didn't know where the car is. The fact that you now see a goat means that if you originally guessed wrong then the remaining door contains the car. And the odds of you getting it wrong are 0.667

[Cabbage]

So stop making unwarranted assumptions when solving the problem.
Actually, you are the one making assumptions about the problem--you're assuming he is operating in a "fair" manner, and that he always reveals a goat.

I'm saying I don't know--maybe he reveals a goat sometimes, maybe sometimes he doesn't. Maybe he's operating in a deceptive manner--Always revealing a goat and offering the switch when I'm initially right, and not revealing a goat when I'm wrong.

[Robin]

Cabbage,

However you (and others) are right about the matter of whether the host sometimes offers you a second chance and sometimes doesn't. Here I was just assuming given that it was a game show that the problem was stating the rules rather than his actions on one occasion.

In this case it is a problem of imprecision of definition.

[Cabbage]

Originally posted by Robin
But in this case the odds are still 0.667 if you switch, even if the host didn't know where the car is. The fact that you now see a goat means that if you originally guessed wrong then the remaining door contains the car. And the odds of you getting it wrong are 0.667
The problem is that going in, a full 1/3 of the time you are going to lose the game off the bat--just because the host revealed the car instead. There's no chance to switch in those situations.

Given that a goat has been revealed, we can now restrict down to the remaining 2/3 of the times when you do get a chance to switch.

Out of those times you get to switch, which occur with only a 2/3 frequency, 1/3 of the time you will have chosen correctly. 1/3 out of 2/3 of the time, which is a 1/2 chance of having chosen correctly, given that a goat was revealed.

[CurtC]

Originally posted by insomneac
But you don't know Monty's intentions at all. Therefore you don't take them into consideration.As the problem is stated, there is not enough information to solve it. The only way to solve it in the classic understanding of the problem is to assume that Monty must always show an empty door and offer the switch. If that's not stated explicitly, the problem can't be formally solved without some assumptions.

[TeaBag420]

Originally posted by Cabbage
Actually, you are the one making assumptions about the problem--you're assuming he is operating in a "fair" manner, and that he always reveals a goat.

I'm saying I don't know--maybe he reveals a goat sometimes, maybe sometimes he doesn't. Maybe he's operating in a deceptive manner--Always revealing a goat and offering the switch when I'm initially right, and not revealing a goat when I'm wrong.

GODDAM, HOW RETARDED CAN YOU BE? IF HE DOESN'T REVEAL A GOAT, IT'S STILL TO YOUR ADVANTAGE TO SWITCH.

[TeaBag420]

Originally posted by CurtC
As the problem is stated, there is not enough information to solve it. The only way to solve it in the classic understanding of the problem is to assume that Monty must always show an empty door and offer the switch. If that's not stated explicitly, the problem can't be formally solved without some assumptions.

Okay, you too may be a retard. The problem says that Monty opens a door. The problem does not say you do multiple trials. The problem can be solved as stated. You don't ASSUME jackcensoredbyMrsGrundysvag, and you don't care what Monty ALWAYS does. You are presented with one trial and you have to make the most advantageous possible decision. Switcheroonimo, mcvalta arooni-mo.

[TeaBag420]

Originally posted by Cabbage
The problem is that going in, a full 1/3 of the time you are going to lose the game off the bat--just because the host revealed the car instead. There's no chance to switch in those situations.

Now you are just making scensoredhit up.


Given that a goat has been revealed, we can now restrict down to the remaining 2/3 of the times when you do get a chance to switch.

Out of those times you get to switch, which occur with only a 2/3 frequency, 1/3 of the time you will have chosen correctly. 1/3 out of 2/3 of the time, which is a 1/2 chance of having chosen correctly, given that a goat was revealed.

Well, son, you kin use them fancy italicks all ye want, but in this part of the country, 1/3 of 2/3 is 2/9, not 1/2..... now gowon! Squeal! Squeal lahk a pig! Did I mention you got a purty mouth?

[Cabbage]

TeaBag420, if I felt you were here to learn, or even teach, as you may feel the case to be, I would attempt to help clear your confusion. As your intention instead seems simply to be to **** with people (either that, or your reading comprehension just sucks), I won't waste anymore time in that attempt for your sake.

[69dodge]

Originally posted by Robin
But in this case the odds are still 0.667 if you switch, even if the host didn't know where the car is. The fact that you now see a goat means that if you originally guessed wrong then the remaining door contains the car. And the odds of you getting it wrong are 0.667The probability that you guessed wrong was initially 2/3. But probabilities can change if you get additional relevant information. The relevant information here is that you saw a goat instead of the car. You were more likely to see a goat if you picked the car than if you didn't. (If you picked the car, you were certain to see a goat; if you didn't, you had a 50:50 chance of seeing a goat.) So, since you did in fact see a goat, that increases the probability that you picked the car. Before, the probability you picked the car was 1/3; now, it's 1/2.

(Ok, that was the Bayesian version ... now for the frequentist version ... )

It's true that 2/3 of all your guesses will be wrong. But that 2/3 includes some cases where the host subsequently reveals the car. Since we now know that in this particular case the host didn't reveal the car, we shouldn't still be counting the whole 2/3. We should only be counting that part of the 2/3 where the host doesn't reveal the car, i.e., half of it. So we're left with 1/3 where you guessed wrong, and 1/3 where you guessed right. In the cases that are consistent with the current one, you guessed right as often as you guessed wrong. Therefore, the probability is now 1/2.

[TeaBag420]

Originally posted by Cabbage
TeaBag420, if I felt you were here to learn, or even teach, as you may feel the case to be, I would attempt to help clear your confusion. As your intention instead seems simply to be to (censored) (http://forums.randi.org/showthread.php?s=&postid=1870531930#rule8) with people (either that, or your reading comprehension just sucks), I won't waste anymore time in that attempt for your sake.

So you REALLY think that one third of two thirds is one half? You're willing to accept the inescapable conclusion that one and one half equals two thirds?

While your use of bad language is disturbing, I am even more distressed, nay, perplexed by your ignorance of grade school arithmetic.

Thank you for your magnanimous offer to clear up my confusion, but 1/3 of 2/3 is NOT and NEVER WILL BE 1/2.

[Cabbage]

Originally posted by TeaBag420
So you REALLY think that one third of two thirds is one half? You're willing to accept the inescapable conclusion that one and one half equals two thirds?

While your use of bad language is disturbing, I am even more distressed, nay, perplexed by your ignorance of grade school arithmetic.

Thank you for your magnanimous offer to clear up my confusion, but 1/3 of 2/3 is NOT and NEVER WILL BE 1/2.
Okay, one more time (just on the offhand chance you actually do have a brain):

If you reread, you'll notice I didn't say "1/3 of 2/3"; I said "1/3 out of 2/3" (there's that reading comprehension I was referring to earlier). 1/3 out of 2/3 is 1/2. Just like 1 out 2 is 1/2. Or 3 out of 6 is 1/2. Or 1/4 out of 1/2 is 1/2. Or, in general, just like x out of 2x is 1/2 (if x is any nonzero real number).

[Matabiri]

Originally posted by hgc
I know most of you are familiar with this one already, but we must settle it once and for all...

You were right; this is always fun to watch.

[Lothian]

I thought the best strategy was to fire up in the air.

[gnome]

Originally posted by 69dodge
The probability that you guessed wrong was initially 2/3.

I'm not sure where you get this.

[pgwenthold]

Originally posted by Robin
No, it does not matter if the contestant knows what Monty is going to do, or what Monty's intentions are. It makes no difference.

If the conditions as set out in the problem are satisfied then the answer is "Yes - you double your chances of winning by switching".

Somebody please show me a case where Monty can affect the outcome.

Easy. If Monty only offers the switch when you have chosen the car, then you never win by switching.

This is a legitimate criticism. The probability calculations only work if Monty offers the choice indiscriminately.

[BillHoyt]

Originally posted by gnome


I'm not sure where you get this. [/B]

gnome,

There are three doors on the stage. Only one of them has the real prize. The probability you got it wrong the first time is, therefore, 2/3.

[BillHoyt]

Ladies and gentlemen,

This argument pops up everywhere on the internet, and has done so for many years. It mostly comes from people never having watched the show. Monty always, always, always, always, always opened up a klunker door on the "reveal" part of the game. It was quite deliberate. It was with knowledge. It is the only way there is any game there at all. Why are you nattering on about assumed intentions, etc., when it is quite clear there would be no tease, and no game if he didn't reveal a klunker every time?

The contestant was always asked to pick one of three doors. Monty always revealed one of the other doors. It always had a klunker behind it. "A year's supply of disposable nosehair trimmers!" He could only have done this having knowledge of where the real prize is. When he removes the known door, the probability space collapses onto the remaining two doors, leaving the contestant's original choice with a 2/3 chance of being wrong. The real fun was watching so many contestants familiar with the game still not get that it was in their best interest to switch to the other door.

[JamesM]

Originally posted by BillHoyt
It mostly comes from people never having watched the show. Monty always, always, always, always, always opened up a klunker door on the "reveal" part of the game. It was quite deliberate.
The Wikipedia entries on The Monty Hall Problem (http://en.wikipedia.org/wiki/Monty_Hall_problem), Let's Make a Deal (http://en.wikipedia.org/wiki/Let%27s_Make_a_Deal) and Monty Hall (http://en.wikipedia.org/wiki/Monty_Hall) contradict you:
Because of his work on Let's Make a Deal, Hall's name is used in a popular probability puzzle known as the Monty Hall problem. He himself gave a pretty good explanation of the solution to that problem, and why the solution did not apply to the case of the actual show, in an interview with New York Times reporter John Tierney in 1991
In the Big Deal, the two contestants were allowed to make a simple choice between three curtains. The top winner in the Big Deal had first choice. One curtain hid the day's Big Deal, which was often multiple cars, a large cash prize, or multiple trips, and typically valued around $10,000.
As stated, the problem is an extrapolation from the game show: contestants on Let's Make a Deal were not allowed to switch. As Monty Hall wrote to Selvin [1],

And if you ever get on my show, the rules hold fast for you -- no trading boxes after the selection.

[DaveW]

Sorry I didn't get around to posting my spreadsheet, but I realized last night that there was an error in it (though I am not sure exactly what yet, I know there is one) and didn't have time to fix it. And reading some of the new posts since my last led me to this line of thinking, which is the important point I think I was missing: it's not so much which door you pick first, it's that Monty's pick of door is constrained if you pick wrongly to begin with (which you likely will). This makes the second choice not entirely independent of the first.

[BillHoyt]

Originally posted by JamesM
The Wikipedia entries on The Monty Hall Problem (http://en.wikipedia.org/wiki/Monty_Hall_problem), Let's Make a Deal (http://en.wikipedia.org/wiki/Let%27s_Make_a_Deal) and Monty Hall (http://en.wikipedia.org/wiki/Monty_Hall) contradict you:

Read your own entries, James, to see that they contradict one another. You seem to ignore the inconsistencies in the entries.

[BillHoyt]

I think JREF posters should be fully aware of exactly what "Wikipedia" is. This disclaimer surely should make its unreliability abundantly clear:

"Wikipedia is an online open-content encyclopedia, that is, a voluntary association of individuals and groups who are developing a common resource of human knowledge. Its structure allows anyone with an Internet connection and World Wide Web browser to alter the content found here.Therefore, please be advised that nothing found here has necessarily been reviewed by professionals with the expertise necessary to provide you with complete, accurate or reliable information.

That's not to say that you won't find valuable and accurate information at Wikipedia, however please be advised that Wikipedia cannot guarantee, in any way whatsoever, the validity of the information found here.It may recently have been changed, vandalized or altered by someone whose opinion does not correspond with the state of knowledge in the particular area you are interested in learning about. "

[JamesM]

Bill, I've read over the three entries again, and have failed to spot where they contradict each other. I've never seen the show, which might explain my difficulties, so perhaps you (or someone else) can help me.

I'm also well aware of the potential unreliability of wikipedia. When I first read about the Monty Hall problem one of the things that struck me was a mention that it couldn't be applied to the real Let's Make a Deal, so your claim that it could be (and indeed was part of the enjoyment of the show) rather struck me.

I didn't mean to suggest you were wrong (you are of course, an equally (un)reliable internet source), only that there was disagreement on that point - apologies for inadvertantly implying otherwise. If you have any evidence that the Monty Hall problem set-up really does apply to Let's Make a Deal, I would like to see it.

[JamesM]

Here's the official Let's Make a Deal (http://www.letsmakeadeal.com/) website. In the Show Info (http://www.letsmakeadeal.com/showinfo.htm) section, there's a description of the final game:
Sometimes when a Trader had decided to “take the Curtain,� Monty offered to buy it back again… $1,000… $2,000… $3,000 not to take the Curtain! Traders never knew how high he would go.
There's no mention here of Monty revealing one of the other curtains or offering a swap.

There is also a section on The Monty Hall Problem (http://www.letsmakeadeal.com/problem.htm), which reprints the letter that Hall allegedly sent to Steve Selvin. Again, Monty states:
And if you ever get on my show, the rules hold fast for you -- no trading boxes after the selection.
Bill, do you have anything to say on this? I will admit that this is not necessarily much more reliable than wikipedia...

[BillHoyt]

Originally posted by JamesM
Here's the official Let's Make a Deal (http://www.letsmakeadeal.com/) website. In the Show Info (http://www.letsmakeadeal.com/showinfo.htm) section, there's a description of the final game:

There's no mention here of Monty revealing one of the other curtains or offering a swap.

There is also a section on The Monty Hall Problem (http://www.letsmakeadeal.com/problem.htm), which reprints the letter that Hall allegedly sent to Steve Selvin. Again, Monty states:

Bill, do you have anything to say on this? I will admit that this is not necessarily much more reliable than wikipedia...

Uh-huh. You need to read this article (http://www25.brinkster.com/ranmath/marlright/montynyt.htm) . It is a copy of a New York Times article on the subject:

"Mr. Hall said he was not surprised at the experts' insistence that the probability was 1 out of 2. "That's the same assumption contestants would make on the show after I showed them there was nothing behind one door," he said.

"They'd think the odds on their door had now gone up to 1 in 2, so they hated to give up the door no matter how much money I offered. By opening that door we were applying pressure. We called it the Henry James treatment. It was 'The Turn of the Screw.' "

Mr. Hall said he realized the contestants were wrong, because the odds on Door 1 were still only 1 in 3 even after he opened another door. Since the only other place the car could be was behind Door 2, the odds on that door must now be 2 in 3."

Further down the page, the author effectively puts Hall's quip to Selvin in context:

"according to the rules of the show, ... he did have the option of not offering the switch, and he usually did not offer it."
Hall did not have to allow the contestant to switch. This is what he meant in his letter to Selvin; that Selvin would not be offered the switch.

[JamesM]

Thanks Bill, much appreciated!

[JamesM]

However, I note in the article it says he "usually did not offer" the switch. How does this square with "he always, always, always, always opened up a klunker door" - would he have opened the door if he wasn't offering a switch? What was the point?

And isn't the Wikipedia entry still correct? The Monty Hall problem doesn't apply to Let's Make a Deal, because the article says Monty was not compelled to offer a switch, and in fact he usually didn't.

Where am I going wrong?

edit: I was of course, completely wrong about why the MHP doesn't apply to Let's Make a Deal: I thought it was because Monty never offered a swap, when in fact it was because he only sometimes offered the swap.

[BillHoyt]

Originally posted by JamesM
However, I note in the article it says he "usually did not offer" the switch. How does this square with "he always, always, always, always opened up a klunker door" - would he have opened the door if he wasn't offering a switch? What was the point?

And isn't the Wikipedia entry still correct? The Monty Hall problem doesn't apply to Let's Make a Deal, because the article says Monty was not compelled to offer a switch, and in fact he usually didn't.

Where am I going wrong?

edit: I was of course, completely wrong about why the MHP doesn't apply to Let's Make a Deal: I thought it was because Monty never offered a swap, when in fact it was because he only sometimes offered the swap.

I wrote that "he always ... opened up a klunker door in the reveal part of the game." What I meant by "reveal part" is the part of the game where he revealed a door as a prelude to offering the switch. (Read the article a bit more; it discusses the psychological gamesmanship in LMAD.)

Go back to the Wikipedia entry to see that I've already edited it again. Cute, eh? The last edit I saw did not say "not compelled to offer a switch," but "not allowed to offer a switch." Wikipedia is a hardly a reliable source of information.

[gnome]

Originally posted by BillHoyt
gnome,

There are three doors on the stage. Only one of them has the real prize. The probability you got it wrong the first time is, therefore, 2/3.

Sorry. Was thinking backwards. My bad. You're correct.

[JamesM]

Originally posted by BillHoyt
I wrote that "he always ... opened up a klunker door in the reveal part of the game." What I meant by "reveal part" is the part of the game where he revealed a door as a prelude to offering the switch.
Just to make sure I've got this clear:

He didn't always offer a switch.

When he offered a switch, he always opened another door.

That door was NEVER the door that had the 'real' prize in.

Correct?

Go back to the Wikipedia entry to see that I've already edited it again. Cute, eh?
Hmm, looks like it's been edited back...

The last edit I saw did not say "not compelled to offer a switch," but "not allowed to offer a switch."
You are quite correct, the MHP entry was in error (the author appears to have made the same erroneous assumption I did). I was actually thinking of the Let's Make a Deal entry - it still seems to me that, as it states, the MHP doesn't apply to the real game, except when Monty offered the swap. And even then, we would have to assume that, whenever Monty Hall offered the swap, he did so regardless of what the contestant had chosen. Given the psychological aspects that Monty Hall mentions in the article you linked to, that doesn't seem necessarily obvious to me.

[Robin]

Originally posted by pgwenthold
Easy. If Monty only offers the switch when you have chosen the car, then you never win by switching.

This is a legitimate criticism. The probability calculations only work if Monty offers the choice indiscriminately.

As I said yesterday, there are two interpretations of the problem. If the second choice is part of the game then Monty's intentions have no bearing on the result.

If the second choice is not part of the game then there is no mathematical answer to the problem.

[hgc]

Originally posted by Robin
As I said yesterday, there are two interpretations of the problem. If the second choice is part of the game then Monty's intentions have no bearing on the result.

If the second choice is not part of the game then there is no mathematical answer to the problem. The 2nd interpretation is no interpretation at all, at least not to the problem I presented. The scenario is quite clear: Monty reveals a door and it is a goat. He then gives the option to switch.

[Robin]

Originally posted by hgc
The 2nd interpretation is no interpretation at all, at least not to the problem I presented. The scenario is quite clear: Monty reveals a door and it is a goat. He then gives the option to switch.

It is not clear it is ambiguous. What you don't say is whether this step is part of the game or whether it is something that the host decided to do off the bat.

That makes all the difference. If this step is part of the game then the answer is "yes - you double your chances by switching your choice"

If this step is optional then there is no mathematical answer.

[Yaotl]

Originally posted by Robin
It is not clear it is ambiguous. What you don't say is whether this step is part of the game or whether it is something that the host decided to do off the bat.

That makes all the difference. If this step is part of the game then the answer is "yes - you double your chances by switching your choice"

If this step is optional then there is no mathematical answer.

The Monty Hall Problem is as stated though, not as it was on the show. You get the second choice after he reveals a goat.

[Robin]

Originally posted by Yaotl
The Monty Hall Problem is as stated though, not as it was on the show. You get the second choice after he reveals a goat.
That is not what I am asking - I can see that from the statement. What I am asking is what is not made clear -

is the step to reveal the goat and offer the switch always done or is it only done sometimes?

If it is always done then there is a specific answer to the question.

If this step is only done sometimes there is no answer to the question.

The actual game show is irrelevant, I am asking about the problem as stated.

[Yaotl]

Originally posted by Robin
That is not what I am asking - I can see that from the statement. What I am asking is what is not made clear -

is the step to reveal the goat and offer the switch always done or is it only done sometimes?

If it is always done then there is a specific answer to the question.

If this step is only done sometimes there is no answer to the question.

Yes, it is always done. I don't see the ambiguity in the problem. He does open a goat door and you are offered the second choice. There is no may or may not in the actual problem.

[hgc]

Originally posted by Robin
That is not what I am asking - I can see that from the statement. What I am asking is what is not made clear -

is the step to reveal the goat and offer the switch always done or is it only done sometimes?

If it is always done then there is a specific answer to the question.

If this step is only done sometimes there is no answer to the question. Always and sometimes are not relevant. I gave a single scenario and described exactly what happened (except for specifying door numbers). No need to extrapolate how it's going to be different the next time.

If the thought experiment or real experiment of a large sample of occurences is useful, then I suggest it be the exact same every time: Monty reveals a door with a goat and offers a switch.

[Robin]

Originally posted by hgc
Always and sometimes are not relevant. I gave a single scenario and described exactly what happened (except for specifying door numbers). No need to extrapolate how it's going to be different the next time.

If the thought experiment or real experiment of a large sample of occurences is useful, then I suggest it be the exact same every time: Monty reveals a door with a goat and offers a switch.

Still two different problems:

1. If you are describing a single scenario with no information about the conduct of the game then there is no answer to the question. Not enough information has been provided.

2.If it is exactly the same every time then the answer is , "yes you double your chances by switching"

So it really has been settled, all that was required was for the problem to be made more precise.

[hgc]

Originally posted by Robin
Still two different problems:

1. If you are describing a single scenario with no information about the conduct of the game then there is no answer to the question. Not enough information has been provided.

2.If it is exactly the same every time then the answer is , "yes you double your chances by switching"

So it really has been settled, all that was required was for the problem to be made more precise. I have no idea what information is missing, or why saying it's multiple occurrences has any effect (remember, I'm asking about probability, not about results). Please tell me how you would phrase the problem to clear up the ambiguity.

[Drooper]

I think one of the reasons why this problem always causes so much controversy is because it is being analysed in the wrong way.

Most people tend to use probability theory alone, but this can only be done in this case with full information about the participants (as many posters are pointing out - e.g. is Monty trying to double bluff you?)

Maybe, if we tried to apply game theory, we could get a more general solution (although under game theory there may not be a stable solution).

[Robin]

Originally posted by Yaotl
Yes, it is always done. I don't see the ambiguity in the problem. He does open a goat door and you are offered the second choice. There is no may or may not in the actual problem.

This is where the misunderstanding has occurred. Nobody spotted the ambiguity in the question.

If the question described the way the game is played then it is reasonable to assume that this step is always carried out. If on the other hand the question describes just one single scenario then we have no way of knowing if any particular step is required or if it has just been done on this occasion.

Some people read it one way, some people the other. No one was right or wrong but we should proceed on the basis that there are now two versions of the problem.

[hgc]

Originally posted by Drooper
I think one of the reasons why this problem always causes so much controversy is because it is being analysed in the wrong way.

Most people tend to use probability theory alone, but this can only be done in this case with full information about the participants (as many posters are pointing out - e.g. is Monty trying to double bluff you?)

Maybe, if we tried to apply game theory, we could get a more general solution (although under game theory there may not be a stable solution). Not necessary. This is a pure, sterile logic problem. The reference to "Monty Hall," "doors," "car," "goats," and other particulars is only to give it flavor.

[Robin]

Originally posted by Drooper
I think one of the reasons why this problem always causes so much controversy is because it is being analysed in the wrong way.

Most people tend to use probability theory alone, but this can only be done in this case with full information about the participants (as many posters are pointing out - e.g. is Monty trying to double bluff you?)

Maybe, if we tried to apply game theory, we could get a more general solution (although under game theory there may not be a stable solution).

You can use probability if the offer to switch is a consistent part of the game. In this case it makes no difference if Monty is trying to bluff you.

If the offer to switch is not always done then you can't use probability. I am unable to say whether game theory would be useful.

[Yaotl]

Originally posted by Robin
This is where the misunderstanding has occurred. Nobody spotted the ambiguity in the question.

If the question described the way the game is played then it is reasonable to assume that this step is always carried out. If on the other hand the question describes just one single scenario then we have no way of knowing if any particular step is required or if it has just been done on this occasion.

Some people read it one way, some people the other. No one was right or wrong but we should proceed on the basis that there are now two versions of the problem.

No, it's quite clear. There is no real game here, just this one instance. You are given all the facts that you need to come up with the answer.

[hgc]

Originally posted by Robin
You can use probability if the offer to switch is a consistent part of the game. In this case it makes no difference if Monty is trying to bluff you.

If the offer to switch is not always done then you can't use probability. I am unable to say whether game theory would be useful. Once again I ask: what's always got to do with it? I am asking about a single scenario, where all the relevant data (in my opinion) is given, and a probability of 2 outcomes is requested.

[Robin]

Originally posted by hgc
I have no idea what information is missing, or why saying it's multiple occurrences has any effect (remember, I'm asking about probability, not about results). Please tell me how you would phrase the problem to clear up the ambiguity.

OK here are the two versions:

Version 1 - The game is conducted as follows:
1. There are three doors
2. Behind one door is a car
3. Behind the other two doors are goats
4. You guess which door contains the car but the door is not opened yet
5. The host opens one of the other doors revealing a goat and asks if you would like to make a new choice
If you were a contestant on the above game would making a new choice increase your chances of picking the car?


Version 2 - The game is conducted as follows:
1. There are three doors
2. Behind one door is a car
3. Behind the other two doors are goats
4. You guess which door contains the car but the door is not opened yet
5. The host will either open one of the other doors revealing a goat and ask you if you want to switch, or he may just go with your first choice
If you were a contestant on the above game would making a new choice increase your chances of picking the car?


Two different version, two different answers.

[hgc]

Originally posted by Robin
OK here are the two versions:





Two different version, two different answers. Version 1 is the question I posed. Version 2 is not. There is no way Version 2 can be construed as my scenario. Like I said before. Don't imagine other scenarios. The question provides and requires exactly one scenario, and your Version 1 is that scenario.

[Drooper]

Originally posted by Robin
You can use probability if the offer to switch is a consistent part of the game. In this case it makes no difference if Monty is trying to bluff you.

If the offer to switch is not always done then you can't use probability. I am unable to say whether game theory would be useful.

I believe game theory is the correct way to analyse the problem in its general form, meaning the choice made (to reveal or not reveal) by the Monty Hall is not known in advance. It deals with such choices under imperfect information.

[Robin]

Originally posted by hgc
Once again I ask: what's always got to do with it? I am asking about a single scenario, where all the relevant data (in my opinion) is given, and a probability of 2 outcomes is requested.

As explained in my earlier post 'always' has everything to do with it. The reason is that there is an additional element of uncertainty and you don't know whether it was random or whether it was dependent on some other event. That makes a probabilistic answer impossible.

[Robin]

Originally posted by Yaotl
No, it's quite clear. There is no real game here, just this one instance. You are given all the facts that you need to come up with the answer.

Again you are misunderstanding, I have never considered the real game relevant.

But the question of whether Monty had the option of not offering the second choice is very much relevant. Here is an event that we cannot assign a probability to.

I have set out the two versions above in reply to hgc. If you are speaking of version 1 then the answer is "p=0.333 if you don't switch, p=0.667 if you do"

If you are speaking of version 2 then there is no probabilistic answer.

(edited to fix typo)

[hgc]

Originally posted by Robin
As explained in my earlier post 'always' has everything to do with it. The reason is that there is an additional element of uncertainty and you don't know whether it was random or whether it was dependent on some other event. That makes a probabilistic answer impossible. Aye Carumba! The only random event is what is the initial selection, and if that was the car, which of the other 2 doors will be revealed. There are no other random events. What on Earth could effect the outcome that is not accounted for in the original question?

[pgwenthold]

Originally posted by hgc
Once again I ask: what's always got to do with it? I am asking about a single scenario, where all the relevant data (in my opinion) is given, and a probability of 2 outcomes is requested.

Because you don't know why you have been given the option.

As I noted (and Robin has been saying), if you were only given the option because you guessed correctly, then you would lose when you switched.

This is why always is important (actually, to be fair, it can also be indiscriminately, as opposed to always). Unless you know that Monty gives you the option indiscriminately, then you cannot know that he has not done it only when you have chosen the correct door (or it could be only when you chose the wrong one, for that matter).

Nothing in the original description requires that the option was given indiscriminately. Given that, you do not have enough information to make the decision.

Of course, it seems pretty clear to me that the intent of the problem is that it is supposed to be an assumption of offering the option indiscriminately, but it is certainly not required.

Regardless, the answer is not that your chance of winning is 1/2.

[hgc]

Originally posted by Robin
Again you are misunderstanding, I have never considered the real game relevant.

But the question of whether Monty had the option of not offering the second choice is very much relevant. Here is an event that we cannot assign a probability to.

I have set out the two versions above in reply to hgc. If you are speaking of version 1 then the answer is "p=0.333 if you don't switch, p=0.667 if you do"

If you are speaking of version 2 then there is no probabilistic answer.

(edited to fix typo) There is zero question of what Monty will do. I described what he did. There is a single scenario. Everything you need to know is known. What Monty will do next time doesn't matter, because the problem posits no next time.

[Robin]

Originally posted by hgc
Aye Carumba! The only random event is what is the initial selection, and if that was the car, which of the other 2 doors will be revealed. There are no other random events. What on Earth could effect the outcome that is not accounted for in the original question?

Please think about this for a while before answering. If offering the second choice was a decision that the host made then that decision would affect the outcome. For example if the host only offers a second choice if you chose right there would be a different outcome than if the host offered a second choice if you chose wrong.

If the second choice was not a decision then it does not affect the outcome.

[pgwenthold]

Originally posted by hgc
Aye Carumba! The only random event is what is the initial selection, and if that was the car, which of the other 2 doors will be revealed. ?

and whether to reveal what is behind the door. Actually, it has to be either random or always. If it is only when you choose the car, then you are going to lose by switching.

However, you cannot tell from only one sample whether it is an indiscriminate thing or whether it was revealed because you picked the car.

[Robin]

Originally posted by Drooper
I believe game theory is the correct way to analyse the problem in its general form, meaning the choice made (to reveal or not reveal) by the Monty Hall is not known in advance. It deals with such choices under imperfect information.

However if Monty did not have a choice, ie he was required to offer the switch as part of the rules of the game then game theory is not required, it becomes a relatively simple probability problem.

[Robin]

Originally posted by hgc
Version 1 is the question I posed. Version 2 is not. There is no way Version 2 can be construed as my scenario. Like I said before. Don't imagine other scenarios. The question provides and requires exactly one scenario, and your Version 1 is that scenario.

I didn't imagine it, epepke and gnome pointed out this interpretation. I disagreed with them initially but on thinking about it I can see that you could read the question that way - ie there are 3 possible decisions to consider:

1. You initial choice
2. Monty's decision to reveal the goat and offer the switch
3. Your decision whether or not to switch

If version 1 is the question then the answer is as I stated, the probability of picking the car is 0.333 if you dont switch, 0.667 if you do.

[pgwenthold]

Originally posted by hgc
There is zero question of what Monty will do. I described what he did.

But you do not know why, and knowing why is critical, because:

1) If Monty only reveals a door after the contestent choses the car, then it means you have selected the car and chosing means you will lose

2) If Monty only reveals a door after the contestant choses wrong, then it means you have selected the wrong door and should switch.

3) If Monty always reveals doors when someone picks one, or does it indiscriminately, then switching means you will win 2/3 of the time

If you have a single instance, and you are shown a door, then you don't know if it is because Monty knows you have chosen a car and goes by (1), Monty knows you have chosen a goat and goes by (2), or this is just the way he always does it, or this is the way it happens sometimes.

[Robin]

Originally posted by pgwenthold
But you do not know why, and knowing why is critical, because:

1) If Monty only reveals a door after the contestent choses the car, then it means you have selected the car and chosing means you will lose

2) If Monty only reveals a door after the contestant choses wrong, then it means you have selected the wrong door and should switch.

3) If Monty always reveals doors when someone picks one, or does it indiscriminately, then switching means you will win 2/3 of the time

If you have a single instance, and you are shown a door, then you don't know if it is because Monty knows you have chosen a car and goes by (1), Monty knows you have chosen a goat and goes by (2), or this is just the way he always does it, or this is the way it happens sometimes.

Thank you you put it better than me. The only other option I would add is:

4) Monty was obliged to reveal the goat and offer the switch as part of the rules of the game in which case switching means you will win 2/3 of the time.

[rppa]

Originally posted by Robin
This is where the misunderstanding has occurred. Nobody spotted the ambiguity in the question.

If the question described the way the game is played then it is reasonable to assume that this step is always carried out. If on the other hand the question describes just one single scenario then we have no way of knowing if any particular step is required or if it has just been done on this occasion.

Some people read it one way, some people the other. No one was right or wrong but we should proceed on the basis that there are now two versions of the problem.

I've seen the statement that it always pays to switch except in the malicious-host assumption: An offer sto switch only comes when he knows you have the car. Another behaivor, which has been raised here, is that sometimes he offers to switch and sometimes he doesn't.

But of course you don't know whether Monty is feeling malicious or fair today, so somehow you have to make an optimum choice across multiple hypotheses.

[Hellbound]

Have to agree that "knowing why" is not a part of the original problem, but an extension of the problem to a generalized case, rather than the specific.

The question stated that the goat was shown. Period.

What you're doing in trying to make scenario two and trying to analyze why is adding in more trials, essentialyl adding a probability to the goat step. As such, you are changing the original problem (it then becomes "Monty sometimes opens a door" not "Monty opens a door"). You are going from a single scenario to a multiple grouping of scenarios.

Instead of thinking of this as a game show, think about it as a single, one-time only, never-to-be-repeated special event (like at a football game or something). He does offer you the choice, you are shown a goat.

The given answer (2/3 chance for switching) is ALWAYS correct for the problem as stated. To include the idea that you only get offered the choice if your door is the car is to assume multiple trials and to violate the statement that you are shown a goat. You have now added an additional problem to the original, and changed the outcomes.

For a single event, analyzed, the answer is correct for the problem as stated. Only if you look at multipl trials, and only iff the open option is not done always, does the probability change. It requires you to go beyond the original problem to develop an alternative answer. You say it requires more information, and that is the point. Getting more information takes the problem outside it's original boundaries and changes it into a new problem.

[Johnny Pneumatic]

What were the odds that this thread would have over 150 posts after one day?!

[Robin]

Originally posted by rppa
I've seen the statement that it always pays to switch except in the malicious-host assumption: An offer sto switch only comes when he knows you have the car. Another behaivor, which has been raised here, is that sometimes he offers to switch and sometimes he doesn't.

But of course you don't know whether Monty is feeling malicious or fair today, so somehow you have to make an optimum choice across multiple hypotheses.

As I said, if the host has a choice at all there is no probabilistic solution.

If the host has no choice but is obliged to offer the second choice then malicious or fair makes no difference - switching will always double your chances of picking the car.

[Paul C. Anagnostopoulos]

I may be repeating someone up above. If so, I apologize. A couple of points:

The problem does not necessarily have anything to do with the TV show.
The problem states that Monty opens a door he knows has a goat.


~~ Paul

[Robin]

Originally posted by Paul C. Anagnostopoulos
I may be repeating someone up above. If so, I apologize. A couple of points:

The problem does not necessarily have anything to do with the TV show.
The problem states that Monty opens a door he knows has a goat.


~~ Paul

But does he have to offer a second choice, or can he not offer the second choice and stay with the first?

[Paul C. Anagnostopoulos]

Robin said:
But does he have to offer a second choice, or can he not offer the second choice and stay with the first?
I'm not sure why this matters, since he did offer the contestant the chance to switch. There is no suggestion in the problem statement that his offer was conditional.

~~ Paul

[Cabbage]

Originally posted by Paul C. Anagnostopoulos
I'm not sure why this matters, since he did offer the contestant the chance to switch. There is no suggestion in the problem statement that his offer was conditional.
The problem is, there is no suggestion in the problem statement that his offer was unconditional, either.

I agree that his offer being unconditional is one interpretation of the problem--In this case, you do double your chances if you switch.

However, I also agree that the offer being conditional is another interpretation--In this case, the probabilities are indeterminate; they depend on that condition on which Monty chooses whether or not to open a door and offer a switch.

It's this ambiguity that's a flaw in the problem as it is often stated.

If we rephrased the problem, replacing Monty Hall with some con man running a sleazy carnival, would you still agree that it's safe to assume his offer to switch is unconditional?

[TeaBag420]

Paul Anagnowhatchamacallit is right.

He makes the offer. So the problem (which I posted) states. Therefore there is no conditionality. The situation before you is:

Contestant chooses one of three doors.

Monty offers one of the remaining two doors as an alternative selection, after opening the other of the two remaining doors.

Contestant either switches or doesn't switch.

[TeaBag420]

And then Cabbage whipped out his big ten inch ........ Crayola:

Originally posted by Cabbage
Okay, one more time (just on the offhand chance you actually do have a brain):

If you reread, you'll notice I didn't say "1/3 of 2/3"; I said "1/3 out of 2/3" (there's that reading comprehension I was referring to earlier). 1/3 out of 2/3 is 1/2. Just like 1 out 2 is 1/2. Or 3 out of 6 is 1/2. Or 1/4 out of 1/2 is 1/2. Or, in general, just like x out of 2x is 1/2 (if x is any nonzero real number).

I did reread it, and my reading comprehension is quite good, thank you very much.

You have introduced a novel usage of the phrase "out of" by applying it to fractions. ARE YOU TRYING TO SAY THAT 1/3 DIVIDED BY 2/3 IS ONE HALF? If so, you are correct, but I must give you no marks at all for clarity of expression.

Plus, you're wrong about the problem itself.

[Cabbage]

Goodbye, TeaBag420; I have neither the time, patience, nor inclination to educate you.

To anyone else still interested, say you encounter someone on the street offering to play this game with you, where you're trying to locate a pea hidden under a shell, or something like that.

You pick one of the shells, he lifts another one to show you the pea isn't underneath it. You either switch or don't switch, and you either win or you don't win; I don't care, that's not the point.

Now, after playing the game you run in to another person. He says, "You look like a smart man, I'll bet you know that once he offers the switch, you've got a 2/3 chance of winning when you do switch, and only a 1/3 chance when you don't switch. Let's say the two of us hang out here for a while, watch this guy playing his game with other passersby, and make a bet. Everytime a person switches and ends up winning, I'll give you $50, but if he switches and loses, you'll owe me $50. You can't lose--You ought to win twice as many times as I do! Whaddayasay?"

Do you take the bet?

[TeaBag420]

Originally posted by Cabbage
Goodbye, TeaBag420; I have neither the time, patience, nor inclination to educate you.



You've already played that card.



To anyone else still interested, say you encounter someone on the street offering to play this game with you, where you're trying to locate a pea hidden under a shell, or something like that.

You pick one of the shells, he lifts another one to show you the pea isn't underneath it. You either switch or don't switch, and you either win or you don't win; I don't care, that's not the point.

Now, after playing the game you run in to another person. He says, "You look like a smart man, I'll bet you know that once he offers the switch, you've got a 2/3 chance of winning when you do switch, and only a 1/3 chance when you don't switch. Let's say the two of us hang out here for a while, watch this guy playing his game with other passersby, and make a bet. Everytime a person switches and ends up winning, I'll give you $50, but if he switches and loses, you'll owe me $50. You can't lose--You ought to win twice as many times as I do! Whaddayasay?"

Do you take the bet?

No, because you will probably lose every time. The reason is that this is a con, not a game show. He's expecting you to be stupid enough not to know that there is no pea under any of the three shells at the time the player is given the choice to switch. Just like Cabbage is.

[Robin]

Shell game is exactly the same as the Monty Hall game. As long as the rules state that an offer to change must be made and as long as the rules are followed then it is advantageous to change your choice.

The following article contains a demonstration of this:

http://www.willamette.edu/cla/math/articles/marilyn.htm

So if you offer me 100 games and guarantee me that each time you will reveal one empty cup when I have put my finger on the choice and guarantee that I will be given the chance to switch, and I win the $50 on best of 100 then certainly I will take the bet. As long as you don't cheat it does not matter if you try to make me lose.

[Robin]

Monty Hall Problem - General Solution

A few people have suggested that this has been settled and I would be interested to see their solutions.

Here is my solution and it incorporates all the objections that have been raised. I have only used one unjustified assumption - that Monty never reveals the prize, mainly to simplify, but also because nobody has suggested that he does.

The problem is that as stated the problem might be read as a specific scenario and not a description of the game. This means that we cannot assume that Monty is required to open a door and offer a second guess. So we cannot assign a probability of 1 to this event and we cannot assign an even probability to the event because we do not know what basis he makes the decision on.

So clearly we need a little basic game theory and I have devised the following categories:

Contestant strategy:
1. Always switch when given the chance
2. Never switch when given the chance
3. Toss a coin if given the chance

Host behaviour:
1. Good Monty - always offers a second choice when initial guess was wrong
2. Bad Monty - always offers a second choice when initial guess was right
3. Indifferent Monty - tosses a coin about offering a second choice
4. No Choice Monty - is required by the rules of the game to offer a second choice

Method:
Simulate contestants in all possible combinations through 100,000 iterations each, create a grid of probabilities

Premises:
1. Host never reveals prize (unjustified)
2. The probability of picking the car is 0.33
3. The probability of picking a donkey is 0.67
4. If Monty offers a second choice AND I accept the offer AND I have initially picked a donkey THEN I will win the prize.
5. We do not have any knowledge about Monty's motivation
6. We do not know if Monty is obliged by the rules of the game to offer a second guess

Result:
A risk averse person will always stick to their original choice when given a chance to change - p min/p_mean/p_max=0.33/0.33/0.33
A risk neutral person will toss a coin to decide whether to change when given the chance - p_min/p_mean/p_max=0.17/0.44/0.67
A risk happy person will always change - p_min/p_mean/p_max=0.00/0.54/1.00

Summary:
<html>
<table border=1>
<tr><td rowspan=2 valign=center>Contestant Strategy</td><td align=center colspan=4>Host Behaviour</td></tr>
<tr><td>Good Monty</td><td>Bad Monty</td>
<td>Indifferent Monty</td><td>No Choice Monty</td></tr>
<tr><td>1. Always switch if offered</td><td>
100.00%
</td><td>
0.00%
</td><td>
49.98%
</td><td>
66.84%
</td></tr>
<tr><td>2. Never switch if offered</td><td>
33.16%
</td><td>
33.16%
</td><td>
33.18%
</td><td>
33.16%
</td></tr>
<tr><td>3. Toss a coin if offered a switch</td><td>
66.69%
</td><td>
16.77%
</td><td>
41.68%
</td><td>
50.02%
</td></tr>
</table>
</html>


That is all.

[hgc]

Originally posted by pgwenthold
But you do not know why, and knowing why is critical, because:

1) If Monty only reveals a door after the contestent choses the car, then it means you have selected the car and chosing means you will lose

2) If Monty only reveals a door after the contestant choses wrong, then it means you have selected the wrong door and should switch.

3) If Monty always reveals doors when someone picks one, or does it indiscriminately, then switching means you will win 2/3 of the time

If you have a single instance, and you are shown a door, then you don't know if it is because Monty knows you have chosen a car and goes by (1), Monty knows you have chosen a goat and goes by (2), or this is just the way he always does it, or this is the way it happens sometimes. Your logic is sound. The problem should be reworded to exclude the behavioral ambiguity -- for the simple version.

[Robin]

(Deleted)

[Cabbage]

Originally posted by Robin
Shell game is exactly the same as the Monty Hall game. As long as the rules state that an offer to change must be made and as long as the rules are followed then it is advantageous to change your choice.

The following article contains a demonstration of this:

http://www.willamette.edu/cla/math/articles/marilyn.htm

So if you offer me 100 games and guarantee me that each time you will reveal one empty cup when I have put my finger on the choice and guarantee that I will be given the chance to switch, and I win the $50 on best of 100 then certainly I will take the bet. As long as you don't cheat it does not matter if you try to make me lose.
I don't know if you were responding to me or TeaBag420 with this, but the point of my example with the shell game is that you don't know how the man running the shell game is playing (and he is playing it fair, so there is a pea somewhere, and the person playing the shell game can certainly win).

At the point you make the bet with the second man, you've only see the game played once--when you played. Should you take the bet without knowing how the shell game man plays from game to game?

[Robin]

Originally posted by Cabbage
I don't know if you were responding to me or TeaBag420 with this, but the point of my example with the shell game is that you don't know how the man running the shell game is playing (and he is playing it fair, so there is a pea somewhere, and the person playing the shell game can certainly win).

At the point you make the bet with the second man, you've only see the game played once--when you played. Should you take the bet without knowing how the shell game man plays from game to game?

In practice no. In theory if he could guarantee beforehand that he would offer you the chance to switch whatever happened then you would still have a greater chance by switching.

[Cabbage]

By the way, Robin, I wasn't actually addressing that shell game example to you, since you already agree that Monty's motivation as to whether or not he offers the switch actually does make a difference in the original problem. I was addressing it to those who think otherwise.

[TeaBag420]

Originally posted by Robin
Shell game is exactly the same as the Monty Hall game. As long as the rules state that an offer to change must be made and as long as the rules are followed then it is advantageous to change your choice.

The following article contains a demonstration of this:

http://www.willamette.edu/cla/math/articles/marilyn.htm

So if you offer me 100 games and guarantee me that each time you will reveal one empty cup when I have put my finger on the choice and guarantee that I will be given the chance to switch, and I win the $50 on best of 100 then certainly I will take the bet. As long as you don't cheat it does not matter if you try to make me lose.

Let me explain to you how the shell game works.

There IS NO PEA by the time the shells stop moving. You will NEVER win.

The person who offers you the bet is working with the person manipulating the shells and the pea.

Again, IT IS A CON.

If you run into someone who tries to run it like a legitimate game, then once again, it is nothing like the Monty Hall puzzle because you have a legitimate chance of knowing exactly where the pea is.

You guyses are way overthinking this.

[BillHoyt]

Originally posted by SkepticJ
What were the odds that this thread would have over 150 posts after one day?!

Do you want to stay with that question or take what's behind door number 2?

[BillHoyt]

Originally posted by JamesM
Just to make sure I've got this clear:

He didn't always offer a switch.

When he offered a switch, he always opened another door.

That door was NEVER the door that had the 'real' prize in.

Correct?
Exactly so.

[Iconoclast]

Originally posted by hgc
Your logic is sound. The problem should be reworded to exclude the behavioral ambiguity -- for the simple version. I told you these threads are always a blast.

[hgc]

Originally posted by Iconoclast
I told you these threads are always a blast. I know! And people can be so gosh-darn dense. :D

[CurtC]

Originally posted by Drooper
Maybe, if we tried to apply game theory, we could get a more general solution (although under game theory there may not be a stable solution).Yes, there is a stable solution. Assuming, in game theory style, that Monty wants to hang on to his prize and you want to win it, then Monty would never offer the chance to switch if you picked the wrong door initially - he would just take his prize and your turn is over. Now lets's say that he sometimes offers the switch when you picked the right door - you, the player, would know that he is only offering you the choice because you picked the right door initially, so you would never switch. Since you never would switch, Monty, in this game theory world, would never bother to even offer it.

In theory, practice is the same as theory. In practice, it isn't. The difference in this case is that Monty wants an exciting game show, and wants to give prizes away sometimes. He would sometimes offer the switch, and sometimes wouldn't. The problem statement in the OP does not constrain him to always offering it, so we have no way of knowing what criteria he uses to decide whether to offer the switch, so the problem statement has no solution.

It looks like we have three holdouts here, maintaining that his motivations don't matter, that everything you need to know is that it was offered this time. This is wrong. You do need to know his motivation for why he offered you the switch. I first heard this problem 15 years ago, and have been active in debating it for most of those. It took me a while too to realize that his motivations matter, and that the problem can't be solved as it's stated in the OP. Guys, re-read the explanations. This is a settled matter. And TeaBag420, you really don't come across too well when you're insulting, belligerent, and wrong.

[Paul C. Anagnostopoulos]

1) If Monty only reveals a door after the contestent choses the car, then it means you have selected the car and chosing means you will lose

2) If Monty only reveals a door after the contestant choses wrong, then it means you have selected the wrong door and should switch.

3) If Monty always reveals doors when someone picks one, or does it indiscriminately, then switching means you will win 2/3 of the time
Sorry, I didn't pay enough attention to these possibilities. However, I see nothing in the wording of the problem that would make me consider the first two scenarios. There is no hint of a conditional nature to the problem. It simply says I pick and door and Monty reveals a goat, no ifs, ands, or buts.

Here's a bunch of skeptics reading a child's word problem:

S1: Okay, it says "Johnny has 7 apples and gives 3 to his best friend Sally. How many apples does he have left?"

S2: Hmm. There's an obvious answer, but how do we know that one of his remaining apples isn't made of wax? Then he would only have 3 real apples left.

S1: Or he could give 3 real apples and one wax apple to Sally. That would still be "giving 3 apples," but now he would only have 3 apples left.

S2: How do we know that he gives her 3 apples only if she doesn't already have apples? If she does, he would have all 7 left.

S1: Yes, and perhaps Sally refuses apples if Johnny is wearing a red shirt. Then he'd still have all 7 apples.

~~ Paul

P.S.: I'm all for clarifying the wording.

[Yaotl]

Originally posted by Paul C. Anagnostopoulos
Sorry, I didn't pay enough attention to these possibilities. However, I see nothing in the wording of the problem that would make me consider the first two scenarios. There is no hint of a conditional nature to the problem. It simply says I pick and door and Monty reveals a goat, no ifs, ands, or buts.

Here's a bunch of skeptics reading a child's word problem:

S1: Okay, it says "Johnny has 7 apples and gives 3 to his best friend Sally. How many apples does he have left?"

S2: Hmm. There's an obvious answer, but how do we know that one of his remaining apples isn't made of wax? Then he would only have 3 real apples left.

S1: Or he could give 3 real apples and one wax apple to Sally. That would still be "giving 3 apples," but now he would only have 3 apples left.

S2: How do we know that he gives her 3 apples only if she doesn't already have apples? If she does, he would have all 7 left.

S1: Yes, and perhaps Sally refuses apples if Johnny is wearing a red shirt. Then he'd still have all 7 apples.

~~ Paul

P.S.: I'm all for clarifying the wording.

That also doesn't clarify "give". Give permanently or give to just hold. Then it begs the question of what "have" means in this sentence. Have altogether or just have in his immediate possession. How do we know if he's not stashing more apples somewhere else? Do those count?

This is fun :D

[pgwenthold]

Originally posted by Paul C. Anagnostopoulos
Sorry, I didn't pay enough attention to these possibilities. However, I see nothing in the wording of the problem that would make me consider the first two scenarios.

But how do you rule them out?

There is nothing in the original wording that would make you consider them, but there is nothing in the wording that prevents them, either.

[Cabbage]

Well, since no one seems to be biting on the example I gave, where the spectator makes a side bet with you (if the person switches and wins, you win the side bet; if the person switches and loses, you lose the side bet), I'll go ahead and finish it off, with the point I was going to make.

Let's say you decide to take the bet. I'm under the impression that everyone here, even those that claim Monty's motivations are irrelevant, would agree that this is a chancy bet. Whether you win or lose will depend greatly on how the operator plays the shell game in the long run, and whether or not he offers the switch consistently (and for clarity, I will state again--this shell game is a fair game, and the player can certainly win by finding the pea. Neither the operator can cheat (by not placing the pea under any shell), nor can the player cheat (by catching a glimpse of where the pea is, or following it in the shuffle)). This shell game is in every sense probabilistically equivalent to the Monty Hall game, with the stipulation that Monty may or may not always offer the switch.

Now that you've taken the bet, the next person comes up to play the shell game. You and the gentleman with whom you have made the wager wait to see what happens.

The contestant makes his selection, the shell game man reveals an empty shell, and offers the contestant the opportunity to switch.

Now, according to those who claim Monty's motivation makes no difference, the contestant believes his chances of winning will be 2/3 if he switches, and 1/3 if he doesn't switch. After all, this is only one game, and it's identical in every sense to the Monty Hall cars and goats game.

On the other hand, you, having made the side bet with the spectator, realize is not so clear as 2/3 vs. 1/3. You still have yet to learn how the shell game operator is operating--does he offer the switch consistently or not? Having seen the game only twice, it's far too early to tell.

So here we have two different people (you, having made the side bet, and the contestant, currently playing the game) judging the probabilities differently! Furthermore, you are both making those judgments with the exact same knowledge. The only difference between the two of you is that you have somewhat different bets riding on the outcome.

Is it possible for both people, with the exact same knowledge of the events, to come to different conclusions about the probabilities, and have both of them be correct? I don't see how.

[Paul C. Anagnostopoulos]

I would rule them out on the basis that these are fun math problems for normal people designed to point out that our instincts about probability are often incorrect. Also on the basis that it mentions absolutely nothing about any conditions on Monty's opening the goat door. If there were any conditions, they would have to be stated so we would have the complete problem. No conditions stated, no conditions implied.

Is that mathematically rigorous enough for ya? :D

~~ Paul

[rppa]

Originally posted by Paul C. Anagnostopoulos
Sorry, I didn't pay enough attention to these possibilities. However, I see nothing in the wording of the problem that would make me consider the first two scenarios. There is no hint of a conditional nature to the problem.

Sorry, but I disagree. And I'm a hardcore two-thirder.

It simply says I pick and door and Monty reveals a goat, no ifs, ands, or buts.

That's right. But in order to assess expectation values, you need to come up with a probability model for how Monty will behave under all possible scenarios, and all you have is a single data point: whatever prize I have behind my door, Monty is showing me a goat.

You've got to make SOME assumption about his behavior, or there's no way to calculate the probabilities. There is absolutely more than a hint of a conditional nature to the problem. The assumptions about the conditional nature are key to your calculation of expectation value.

However, somebody (Robin?) has done a very nice job of summarizing several different hypotheses about what the single observation we have might mean in general. And it bears out what I read in at least one website: The only time it doesn't pay to switch is if I think "Monty shows me a goat" implies "I have the car already".

Here's a bunch of skeptics reading a child's word problem:

Sorry, this is a strawman. As soon as you start working out the potential payoffs, you are going to make assumptions about Monty's rules of operation. It's impossible to calculate otherwise. The information stated is not sufficient to answer the question. Assumptions are needed.

What makes you think the information given is sufficient to know what Monty does under all circumstances?

[Robin]

rppa
You've got to make SOME assumption about his behavior, or there's no way to calculate the probabilities.
This is absolutely right. You could make a meal of what you are not told. Are you a contestant on the game show? Or is this an "ask the audience" segment where your choice gives you no advantage or disadvantage? What is the nature of the show, what are the rules? The car might be an old clunker and the donkey's might have $100,000 under their saddles.

Or it might be theatre sports where the advantage depends in how well you play the role.

[hgc]

Originally posted by Robin
This is absolutely right. You could make a meal of what you are not told. Are you a contestant on the game show? Or is this an "ask the audience" segment where your choice gives you no advantage or disadvantage? What is the nature of the show, what are the rules? The car might be an old clunker and the donkey's might have $100,000 under their saddles.

Or it might be theatre sports where the advantage depends in how well you play the role. Now it's everything AND the kitchen sink, eh? Please tell me how the relative value of the car and the donkeys affects the outcome.

[Robin]

--------------------------------------------------------------------------------
Originally posted by Paul C. Anagnostopoulos
Sorry, I didn't pay enough attention to these possibilities. However, I see nothing in the wording of the problem that would make me consider the first two scenarios.
--------------------------------------------------------------------------------

--------------------------------------------------------------------------------
Originally posted by pgwenthold:
But how do you rule them out?

There is nothing in the original wording that would make you consider them, but there is nothing in the wording that prevents them, either.
--------------------------------------------------------------------------------

It depends entirely on how you read it. It is like one of those pictures that looks like a young woman to some and an old woman to others.

If you, as I did, read it as "this is how the game is played" then the reasonable interpretation is that the host always offers the switch.

If you read it, as others have, as "this is a specific scenario within the game" then you don't have enough information to know.

The wording supports both readings.

However I still hold out that the motivations of the host (or director or whatever) is not the key question.

The key question is "does the host always offer a switch?".

Without that piece of information the problem does not have an answer.

[Yaotl]

Originally posted by hgc
Monty presents 3 doors; behind one is a car, and behind the other two are goats. You pick a door. But before Monty opens that door, he reveals behind another door that there is a goat. He then gives you the opportunity to switch your choice. What is the probability that you will get the car by switching, or by staying?

It's What is the probability that you will get the car by switching, or by staying?

Nowhere does it state what is the probably overall in the game, just this instance. Whether he does it later on or not for another contestant or the audience or whatever is irrelevant. You have to figure out your odds right then and there.

[hgc]

Originally posted by Yaotl
It's

Nowhere does it state what is the probably overall in the game, just this instance. Whether he does it later on or not for another contestant or the audience or whatever is irrelevant. You have to figure out your odds right then and there. Agreed. Did you think I thought anything different?

[Yaotl]

Originally posted by hgc
Agreed. Did you think I thought anything different?

Nope, I just wanted to restate what you said since I doubt anyone's looking at the original problem anymore. What with all the assumptions being bandied about and all.

[hgc]

Originally posted by Yaotl
Nope, I just wanted to restate what you said since I doubt anyone's looking at the original problem anymore. What with all the assumptions being bandied about and all. Thanks. I can see that the question needs to be reworded to exclude ambiguity about Monty's reason for revealing the donkey and offering the switch, but there is a lot of other irrelevancy flying around.

[CurtC]

hgc wrote:
Thanks. I can see that the question needs to be reworded to exclude ambiguity about Monty's reason for revealing the donkey and offering the switch...Thank you. That was my point. It needs to be specified, in the problem description, that Monty is constrained to offer you the choice to switch no matter what you picked or what mood he was in at the time.

[hgc]

Originally posted by CurtC
Thank you. That was my point. It needs to be specified, in the problem description, that Monty is constrained to offer you the choice to switch no matter what you picked or what mood he was in at the time. And not that Monty's intentions don't make for interesting discussion, but it won't work too well in a bar bet situation.

[Robin]

Originally posted by hgc
Now it's everything AND the kitchen sink, eh? Please tell me how the relative value of the car and the donkeys affects the outcome.

The wording of the original MHP (which was brought into play by epepke and gnome) said "Is it to your advantage to switch?". Clearly in this case the relative retail values are very relevant.

OK so the OP said "what is the probability that you will get the car by switching?". You still have a problem - it is not stated that you will get the car by guessing which door it is behind.

Maybe the point of the game is to eliminate prizes you don't want. Maybe like "Sale of the Century" picking the car but in the wrong order means you don't get it.

The point I am making is that you have to assume something, so what we want to know is what is it reasonable to assume?.

[Robin]

So what is actually being debated? Is it this?

Monty presents 3 doors; behind one is a car, and behind the other two are goats. You pick a door. But before Monty opens that door, he reveals behind another door that there is a goat. He then gives you the opportunity to switch your choice. What is the probability that you will get the car by switching, or by staying?

If so then the answer is easy. The answer is that there is not enough information to calculate a probability. There is not even enough information to state whether or not Monty's intentions are relevant. Even when making reasonable assumptions.

It is silly to spend so much time debating when we have not even decided what is being debated.

[hgc]

Originally posted by Robin
The wording of the original MHP (which was brought into play by epepke and gnome) said "Is it to your advantage to switch?". Clearly in this case the relative retail values are very relevant.

OK so the OP said "what is the probability that you will get the car by switching?". You still have a problem - it is not stated that you will get the car by guessing which door it is behind.

Maybe the point of the game is to eliminate prizes you don't want. Maybe like "Sale of the Century" picking the car but in the wrong order means you don't get it.

The point I am making is that you have to assume something, so what we want to know is what is it reasonable to assume?. Yes, when people communicate via human language, they make assumptions about meanings of words and phrases based on shared cultural experiences. If you want to engage in pedantic nitpickery you can uncover all kinds of bogus ambiguity.

Who didn't know, unquestionably, that the picking of the door was for receiving the prize behind that door? Raise you hands. I can understand that some puzzles are constructed with tricky loopholes hidden in the precise construct of the question. No reason to think that this is such a puzzle. This is a puzzle about probability.

[Cabbage]

Originally posted by Yaotl
Nowhere does it state what is the probably overall in the game, just this instance. Whether he does it later on or not for another contestant or the audience or whatever is irrelevant. You have to figure out your odds right then and there.
I'm reading this as implying you think these long term odds (as more and more contestants play the game) are somehow different from the short term odds (when you play a single game).

How can that be so?

Read my example where two bets are going on--One being the game as it is being currently played by a contestant (short term, only one game), while the other is a a wager made by a person on the outcome of the games (over some long term) whenever the current contestant switches.

How could you analyze such a situation in a way that gives different probabilities from the two different perspectives?

[Robin]

Originally posted by hgc
Yes, when people communicate via human language, they make assumptions about meanings of words and phrases based on shared cultural experiences. If you want to engage in pedantic nitpickery you can uncover all kinds of bogus ambiguity.

Who didn't know, unquestionably, that the picking of the door was for receiving the prize behind that door? Raise you hands. I can understand that some puzzles are constructed with tricky loopholes hidden in the precise construct of the question. No reason to think that this is such a puzzle. This is a puzzle about probability.

You are utterly determined to miss the point aren't you? The point I was making is that we have to assume.

I originally made the assumption that the second choice was part of the rules of the game and everyone jumped all over me and said "but it depends on the intentions of the host" because the problem does not explicitly state that Monty was obliged to offer the choice.

I had assumed that all information required answer the puzzle had been supplied and if the offer of second choice was only optional that would have been stated.

So my assumption that Monty had to offer the second choice is being called unreasonable.

But the alternative assumption that Monty did not have to offer the second choice is not regarded as unreasonable.

So my question is what is the criteria for any assumption being reasonable? That was the point I was making.

This is a puzzle about probability.

Incidentally and let's be crystal clear about this:

It is a probability problem if and only if Monty was obliged to reveal the goat and offer the second choice

In which case it was settled long ago and why are we arguing?

[hgc]

Originally posted by Robin
...

In which case it was settled long ago and why are we arguing? I have no idea. I think it had something to do with hitting 150 posts in this thread in a single day. :)

[TeaBag420]

To sum up: Cabbage presents a twist on the Monty Hall problem by introducing the shell game (which is a notorious con; check with Penn if you think this assertion is wrong), but with a twist. Robin then asserts that the shell game is exactly the same as the Monty Hall problem. He is incorrect about this in the real world, but it's not his puzzle so I ignore him. Then Cabbage says the grifter in the shell game IS playing fair in that the pea is really there somewhere, so I accept the redefinition of "shell game" solely for this discussion.

Then CurtC offered his game theory interpretation, which even he admitted was a blind alley (my phrase). CurtC then demonstrates that he doesn't understand that the OP was for one trial, and that the problem states what Monty actually DOES, not what he's required to do.

Cabbage clarifies that his version of the shell game is Platonian in its refinement, in that not only can the con not cheat by removing the pea, the player (mark) [b]can't try to keep track of the location of the pea.[b] Which makes it pointless to introduce this new game where trying to win is cheating. Oh well.

Now Cabbage explicitly told us on page five of this thread that there are two circumstances under which you, the passerby could win or lose. One, if the player (mark) switches and win: you win. Two, if the player switches and loses: you lose. If he doesn't switch, you neither win nor lose. No matter whether he does or not. Any redefinition of the bet for the "not switching" case or for the odds (you're betting even money--$50 to win, $50 to lose) at this point will go down in the book as puzzle changing, and puzzle changers are just one step above kid touchers in my book.

So, if you're inclined to gamble, you take the bet, because the chances of winning or breaking even are better than 50/50. Of course if the guy who offered you the bet wins the first time and walks away, you're screwed.

Class dismissed. What's next, Mrs. Landingham?

[TeaBag420]

Originally posted by Robin
You are utterly determined to miss the point aren't you? The point I was making is that we have to assume.

I originally made the assumption that the second choice was part of the rules of the game and everyone jumped all over me and said "but it depends on the intentions of the host" because the problem does not explicitly state that Monty was obliged to offer the choice.

I had assumed that all information required answer the puzzle had been supplied and if the offer of second choice was only optional that would have been stated.

So my assumption that Monty had to offer the second choice is being called unreasonable.

But the alternative assumption that Monty did not have to offer the second choice is not regarded as unreasonable.

So my question is what is the criteria for any assumption being reasonable? That was the point I was making.



Incidentally and let's be crystal clear about this:

It is a probability problem if and only if Monty was obliged to reveal the goat and offer the second choice

In which case it was settled long ago and why are we arguing?

No, it's a probability in cases where, as the OP states, Monty does in fact r.t.g.a.o.t.s.c. Otherwise it's one out of three. But since the original puzzle does in fact state that Monty does in fact r.t.g.a.o.t.s.c. it's a probability problem.

[Cabbage]

Originally posted by TeaBag420
To sum up: Cabbage presents a twist on the Monty Hall problem by introducing the shell game (which is a notorious con; check with Penn if you think this assertion is wrong), but with a twist. Robin then asserts that the shell game is exactly the same as the Monty Hall problem. He is incorrect about this in the real world, but it's not his puzzle so I ignore him. Then Cabbage says the grifter in the shell game IS playing fair in that the pea is really there somewhere, so I accept the redefinition of "shell game" solely for this discussion.

Then CurtC offered his game theory interpretation, which even he admitted was a blind alley (my phrase). CurtC then demonstrates that he doesn't understand that the OP was for one trial, and that the problem states what Monty actually DOES, not what he's required to do.

Cabbage clarifies that his version of the shell game is Platonian in its refinement, in that not only can the con not cheat by removing the pea, the player (mark) [b]can't try to keep track of the location of the pea.[b] Which makes it pointless to introduce this new game where trying to win is cheating. Oh well.

Now Cabbage explicitly told us on page five of this thread that there are two circumstances under which you, the passerby could win or lose. One, if the player (mark) switches and win: you win. Two, if the player switches and loses: you lose. If he doesn't switch, you neither win nor lose. No matter whether he does or not. Any redefinition of the bet for the "not switching" case or for the odds (you're betting even money--$50 to win, $50 to lose) at this point will go down in the book as puzzle changing, and puzzle changers are just one step above kid touchers in my book.

So, if you're inclined to gamble, you take the bet, because the chances of winning or breaking even are better than 50/50. Of course if the guy who offered you the bet wins the first time and walks away, you're screwed.

Class dismissed. What's next, Mrs. Landingham?
I must say I'm pleasantly surprised by the civility of your response, and so I will continue to respond as well. A few points:

The only reason I replaced the car and goats game with the shell game in my example was that's a more realistic setting. If you're walking down the street, it's reasonable to expect to see a shell game, but seeing a stage set up with three doors, each concealing a car or goat, would be a bit unusual. It makes no difference either way, since one of the axioms in my problem are that these two games are in every way equivalent; if you prefer, replace the shell game with the cars and goats.

You are correct, if the player doesn't switch, then you neither win nor lose your side bet. The bet is only in effect when the current player switches.

If you're worried that the person you make the side bet with may win the first bet and then walk away, we can add an addtional stipulation--that you both agree to play until, say, 50 people have played the game and switched (of course, the actual number of people who play the game during this period will most likely be greater than 50, since many of them may choose not to switch, voiding the bet for that round).

The key issue, however, is that you have no knowledge how the shell game operator plays (or how the host plays the car and goats game). You've only seen him play once --the game in which you yourself played, and were offered the switch. Will he continue to offer the switch every game? Who knows?...To answer that with certainty one way or the other would require knowing what the shell game operator intends, and only he knows that.

[Robin]

Originally posted by TeaBag420
No, it's a probability in cases where, as the OP states, Monty does in fact r.t.g.a.o.t.s.c. Otherwise it's one out of three. But since the original puzzle does in fact state that Monty does in fact r.t.g.a.o.t.s.c. it's a probability problem.

No, and this is absolutely crucial to the misunderstanding:

It is a probability problem if and only if Monty has to reveal the goat offer the choice.

If all you know is that on this occasion he revealed the goat and offered the choice then it is not a probability problem.

Incidentally neither is it a matter of Monty's intentions, not at least until we know whether or not it he had to make the offer.

[Paul C. Anagnostopoulos]

rppa said:
That's right. But in order to assess expectation values, you need to come up with a probability model for how Monty will behave under all possible scenarios, and all you have is a single data point: whatever prize I have behind my door, Monty is showing me a goat.
I agree that if this were the statement of a puzzle for which we were offering a $1 million prize for a solution, it would be important to explicitly state all the parameters of the problem. I'm simply saying that some of us are taking it a bit too seriously. It is generally the case with logic problems that unstated conditions can be assumed to be absent. Otherwise you could conjure up dozens of possible conditions: What if Monty uses different procedures on different days of the week?

You've got to make SOME assumption about his behavior, or there's no way to calculate the probabilities. There is absolutely more than a hint of a conditional nature to the problem. The assumptions about the conditional nature are key to your calculation of expectation value.
Where is the hint about any conditions?

Sorry, this is a strawman.
And a joke, too. :D Well, except for my remark that there are many conditions we could assume if we wanted to.

What makes you think the information given is sufficient to know what Monty does under all circumstances?
It's not, if we are being formal about it.

Here's Craig Whitaker's letter to "Ask Marilyn":

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, always opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?
I have edited it to eliminate the ambiguity. Well, unless this is only the procedure on Tuesday.

For someone who is merely objecting to our taking this too formally, I'm being awfully serious, ain't I?

~~ Paul

[TeaBag420]

Originally posted by Cabbage
I must say I'm pleasantly surprised by the civility of your response, and so I will continue to respond as well. A few points:

The only reason I replaced the car and goats game with the shell game in my example was that's a more realistic setting. If you're walking down the street, it's reasonable to expect to see a shell game, but seeing a stage set up with three doors, each concealing a car or goat, would be a bit unusual. It makes no difference either way, since one of the axioms in my problem are that these two games are in every way equivalent; if you prefer, replace the shell game with the cars and goats.

You are correct, if the player doesn't switch, then you neither win nor lose your side bet. The bet is only in effect when the current player switches.

If you're worried that the person you make the side bet with may win the first bet and then walk away, we can add an addtional stipulation--that you both agree to play until, say, 50 people have played the game and switched (of course, the actual number of people who play the game during this period will most likely be greater than 50, since many of them may choose not to switch, voiding the bet for that round).

The key issue, however, is that you have no knowledge how the shell game operator plays (or how the host plays the car and goats game). You've only seen him play once --the game in which you yourself played, and were offered the switch. Will he continue to offer the switch every game? Who knows?...To answer that with certainty one way or the other would require knowing what the shell game operator intends, and only he knows that.

No. If he doesn't offer the switch you break even. So if he offers the switch once you have a 2/3 chance of winning $50. If he offers it twice you have a 1/9 chance of losing $100, a 4/9 chance of winning $100, and a 4/9 chance of breaking even.

Because of the nature of the bet, you don't even have to consider cases where he doesn't offer the switch. If he never offers the switch, it's as if you never made the bet.
As long as he offers a switch at least once, the odds are in favor of your at least breaking even. If he never offers the switch at all, you lose nothing. It doesn't matter what his plan or strategy is, nor what you know or don't know about it; this paragraph encompasses all possibilities.

Unless, of course, there's collusion between him and the man you're betting with. In which case we're back to this being a con.

Let me reiterate: Given the problem as presented, there is no strategy the grifter could employ short of collusion that would give an advantage to the man with whom you are betting. It does not matter.

[TeaBag420]

Originally posted by Robin
No, and this is absolutely crucial to the misunderstanding:

It is a probability problem if and only if Monty has to reveal the goat offer the choice.

If all you know is that on this occasion he revealed the goat and offered the choice then it is not a probability problem.



Then what field of mathematics would it fall under?

[Cabbage]

It is generally the case with logic problems that unstated conditions can be assumed to be absent. Otherwise you could conjure up dozens of possible conditions:
In fact, that's exactly what I've been trying to say!

The problem never stated the condition that a door is always revealed; therefore, that conditioned is assumed to be absent.

The ones claiming Monty always reveals a door are the ones making the assumption. That assumption, obviously, is that Monty always opens a door.

The one claiming Monty may or may not always open a door are operating without assumptions. Of course, that situation (making no assumptions) makes the problem unsolvable--We can't assign a realistic probability to the situation in this case.

[Cabbage]

Originally posted by TeaBag420
No. If he doesn't offer the switch you break even. So if he offers the switch once you have a 2/3 chance of winning $50. If he offers it twice you have a 1/9 chance of losing $100, a 4/9 chance of winning $100, and a 4/9 chance of breaking even.

But here you are assuming that you know how the man operates the shell game. Maybe he likes to be tricky, offering the switch only when the initial guess is correct, just to tempt them. It's still a perfectly fair game--The person playing the game has every right to choose not to switch, and therefore win the game. But if the man operates the game this way, the player will lose every time he switches

Because of the nature of the bet, you don't even have to consider cases where he doesn't offer the switch. If he never offers the switch, it's as if you never made the bet.
As long as he offers a switch at least once, the odds are in favor of your at least breaking even. If he never offers the switch at all, you lose nothing. It doesn't matter what his plan or strategy is, nor what you know or don't know about it; this paragraph encompasses all possibilities.

Unless, of course, there's collusion between him and the man you're betting with. In which case we're back to this being a con.

Let me reiterate: Given the problem as presented, there is no strategy the grifter could employ short of collusion that would give an advantage to the man with whom you are betting. It does not matter.

There's no collusion between the shell game man and the man you're betting with; he's every bit in the dark as you are as to how the shell game man plays his game. However, the fact that there is no collusion doesn't mean the odds are clear cut--That, once again, depends on how the man operates his shell game.

[Robin]

Originally posted by TeaBag420
Then what field of mathematics would it fall under?

It does not come under any field of mathematics because it is incomplete.

For example look at the following:
<pre>
1. x=4
2. y=3
3. z=redistribute(x,y)
</pre>
What is the value of z?

This looks like a mathematical problem, but is not because we don't know what the hell line 3 means.

[Robin]

Originally posted by Cabbage
The one claiming Monty may or may not always open a door are operating without assumptions.

No, they are operating with one less assumption.

Of course, that situation (making no assumptions) makes the problem unsolvable--We can't assign a realistic probability to the situation in this case. [/B]

It makes it unsolvable as a probability problem at least.

[Robin]

Originally posted by Cabbage
The ones claiming Monty always reveals a door are the ones making the assumption. That assumption, obviously, is that Monty always opens a door.


The ones claiming that the results depend on the hosts intentions are also making an assumption, ie that Monty optionally opens a door.

[Cabbage]

Originally posted by Robin
The ones claiming that the results depend on the hosts intentions are also making an assumption, ie that Monty optionally opens a door.
No, we're not.

I certainly agree it's possible that Monty always opens a door. I am also aware of the possibility that Monty may open a door only occasionally. I'm not assuming, I'm simply admitting that I don't know.

[Robin]

Originally posted by Cabbage
No, we're not.

I certainly agree it's possible that Monty always opens a door. I am also aware of the possibility that Monty may open a door only occasionally. I'm not assuming, I'm simply admitting that I don't know.

Exactly my point.

If you say that the result depends on the intentions of the host then you are assuming that Monty has a choice.

If you don't know whether Monty has a choice then you don't know if it depends on the behaviour of the host.

For example you said:

Actually, the intentions of the host do matter

(your emphasis) which makes the assumption that he has a choice. More correctly you should have said.

Actually, the intentions of the host may matter

(edited for clarity)

[Cabbage]

Originally posted by Robin
Exactly my point.

If you say that the result depends on the intentions of the host then you are assuming that Monty has a choice.

If you don't know whether Monty has a choice then you don't know if it depends on the behaviour of the host.

For example you said:



(your emphasis) which makes the assumption that he has a choice. More correctly you should have said.



(edited for clarity)
I'm afraid I don't see the distinction.

The point I'm trying to make is that the host's intentions always matter.

If he always opens a door, then his intentions matter, simply because his intention is to always open a door.

If he sometimes opens a door, then his intentions matter, simply because his intention is to sometimes open a door.

Any way you look at it, his intentions matter.

[TeaBag420]

Originally posted by Cabbage
In fact, that's exactly what I've been trying to say!

The problem never stated the condition that a door is always revealed; therefore, that conditioned is assumed to be absent.

The ones claiming Monty always reveals a door are the ones making the assumption. That assumption, obviously, is that Monty always opens a door.

The one claiming Monty may or may not always open a door are operating without assumptions. Of course, that situation (making no assumptions) makes the problem unsolvable--We can't assign a realistic probability to the situation in this case.

The OP makes none of these claims. It also doesn't claim that Monty discriminates against black people. It is about a ONE-TIME incident where you, the contestant are offered a choice. Nothing else. These other lines of questioning are not part of the original problem, which I posted. The original submitter of the problem has posted here... nope, got the name wrong... Whitaker?I was thinking CFLarsen (or maybe I misunderstood).

[TeaBag420]

Originally posted by Cabbage
I must say I'm pleasantly surprised by the civility of your response, and so I will continue to respond as well. A few points:

The only reason I replaced the car and goats game with the shell game in my example was that's a more realistic setting. If you're walking down the street, it's reasonable to expect to see a shell game, but seeing a stage set up with three doors, each concealing a car or goat, would be a bit unusual. It makes no difference either way, since one of the axioms in my problem are that these two games are in every way equivalent; if you prefer, replace the shell game with the cars and goats.

You are correct, if the player doesn't switch, then you neither win nor lose your side bet. The bet is only in effect when the current player switches.

If you're worried that the person you make the side bet with may win the first bet and then walk away, we can add an addtional stipulation--that you both agree to play until, say, 50 people have played the game and switched (of course, the actual number of people who play the game during this period will most likely be greater than 50, since many of them may choose not to switch, voiding the bet for that round).

The key issue, however, is that you have no knowledge how the shell game operator plays (or how the host plays the car and goats game). You've only seen him play once --the game in which you yourself played, and were offered the switch. Will he continue to offer the switch every game? Who knows?...To answer that with certainty one way or the other would require knowing what the shell game operator intends, and only he knows that.

First, I don't think it matters how many games are played, so let's agree to throw the stipulation out.

Second, and I'm starting to get this, if the grifter has the option of only offering the switch when you've got the right shell, then in those cases switching gives you a zero chance of winning. If the grifter isn't required to follow rules that you, the bettor know, then this isn't solvable for any arbitrary number of games, which is what I suspect is what you were getting at. Unless you consider it a solution to say "Don't take the bet". Which gets us back to it being a con, and was my original solution.

Because if the grifter doesn't follow known rules, why would someone offer you the bet? And if he doesn't follow known rules, why offer you the bet, absent collusion?

It's a con.

Robin's assertion that the shell game is the same as the Monty Hall problem appears defective under this analysis and I will let you take that up with him.

[TeaBag420]

Originally posted by Robin
It does not come under any field of mathematics because it is incomplete.

For example look at the following:
<pre>
1. x=4
2. y=3
3. z=redistribute(x,y)
</pre>
What is the value of z?

This looks like a mathematical problem, but is not because we don't know what the hell line 3 means.


I have rephrased your example:

<pre>
1. x=4
2. y=3
3. z=Idon'tknow(x,y)
</pre>

Equally helpful, don't you think? Your argument seems to boil down to "I don't know." But in a bad way.

If it's not a math problem, maybe it should be moved to another board. Why don't you liaise with the moderators on that?

[Robin]

Originally posted by Cabbage
I'm afraid I don't see the distinction.

The point I'm trying to make is that the host's intentions always matter.

If he always opens a door, then his intentions matter, simply because his intention is to always open a door.

If he sometimes opens a door, then his intentions matter, simply because his intention is to sometimes open a door.

Any way you look at it, his intentions matter.

Oh come on, now you are being silly. The word 'intention' implies a choice, it makes no sense to say someone has an intention if they don't have a choice. When we say he did not have a choice we are saying that it is a premise of the problem that he always offers a choice.

Are you seriously - I mean seriously - saying you cannot see the distinction?

[TeaBag420]

Originally posted by Cabbage
I'm afraid I don't see the distinction.

The point I'm trying to make is that the host's intentions always matter.

If he always opens a door, then his intentions matter, simply because his intention is to always open a door.

If he sometimes opens a door, then his intentions matter, simply because his intention is to sometimes open a door.

Any way you look at it, his intentions matter.

NO. And this is the problem I have with conflating the idealized shell game and the Monty Hall problem. The OP stated that Monty opens a door--on one trial. The problem doesn't really begin until he opens the door.

If you don't know his intentions there is no reason to try to take them into account. If you DO know them, THEN it ceases to be a probability problem.

"If he always opens a door, then his intentions matter, simply because his intention is to always open a door."

Does he do the show unintentionally, or does that matter?

[Robin]

Originally posted by TeaBag420
I have rephrased your example:

<pre>
1. x=4
2. y=3
3. z=Idon'tknow(x,y)
</pre>

Equally helpful, don't you think? Your argument seems to boil down to "I don't know." But in a bad way.



What happened did they declare "miss the point day" and not tell me? This is exactly what I mean't.

If you know that the host had to offer the second choice then it is a probability problem.

If you know that the host didn't have to offer the second choice then it is not a probability problem but may be some other kind of maths problem.

If you don't know whether the host had a choice or not then it is not a problem at all, it is just some incomplete bit of text.

My position all along has been "we need to precisely define the problem before we can find and answer". I did not think that it was particularly controversial.

By the way, is there a "good way" and a "bad way" not to know something? I thought you either knew something or didn't.

If it's not a math problem, maybe it should be moved to another board. Why don't you liaise with the moderators on that?
Did I start this thread? News to me.

I think that you really need to come to the point. Please state what you think the solution to the problem is.

[Cabbage]

Originally posted by Robin
Oh come on, now you are being silly. The word 'intention' implies a choice, it makes no sense to say someone has an intention if they don't have a choice. When we say he did not have a choice we are saying that it is a premise of the problem that he always offers a choice.

Are you seriously - I mean seriously - saying you cannot see the distinction?
Yes, that's exactly what I'm saying--I don't see the distinction.

How is "always offering a switch" not a choice?

Perhaps you've misunderstood. When I've referred to the host's intentions, or choice, I'm not referring to whether or not he intends, or chooses, to reveal the door in a particular game. I'm referring to his choice on how to run the game in the first place.

It is most certainly a choice on how to run the game:

Let's go back to the original problem, focusing on a game show. How was this game show formulated in the first place? I can imagine the host to be sitting around drinking coffee. He thinks to himself, "Here's the deal, the contestant will try to pick the door containing the prize. Only one of the three doors will contain the prize!"

Then he thinkgs, "Wait a minutel, let's put a little more action into it. Let's say that once the contestant makes the selection, I may reveal one of the doors, and offer the contestant a chance to switch then.

"OK, I might be able to go somewhere with that. Do I want to always offer the switch, or just sometimes?"

He thinks for a moment, and says, "I think I'll do it all the time!"

How can you possibly say that is not a choice on how to run the game?

[Cabbage]

Originally posted by TeaBag420
First, I don't think it matters how many games are played, so let's agree to throw the stipulation out.

Second, and I'm starting to get this, if the grifter has the option of only offering the switch when you've got the right shell, then in those cases switching gives you a zero chance of winning. If the grifter isn't required to follow rules that you, the bettor know, then this isn't solvable for any arbitrary number of games, which is what I suspect is what you were getting at. Unless you consider it a solution to say "Don't take the bet". Which gets us back to it being a con, and was my original solution.

Because if the grifter doesn't follow known rules, why would someone offer you the bet? And if he doesn't follow known rules, why offer you the bet, absent collusion?

It's a con.

Robin's assertion that the shell game is the same as the Monty Hall problem appears defective under this analysis and I will let you take that up with him.
It's not a con; that was a clearly stated condition of my problem.

Why would the stranger offer the side bet? He may want to offer it because he has faith that some one offering such a game is probably going to be good at it. The shell game man may be expected to be good at running the game, offering the switch sometimes, not offering at others, and doing it in such a way to mislead people into making the wrong decision. He may also be good at reading people, knowing by their behavior what to expect or not when they are present with a switch. Of course, none of this is to say that he cheats--Again, it is most certainly a fair game, the man running it may just be damn good with his strategy of offering a switch at some times, and not offering it at others.

One particular comment you made:If the grifter isn't required to follow rules that you, the bettor know, then this isn't solvable for any arbitrary number of games, which is what I suspect is what you were getting at. Unless you consider it a solution to say "Don't take the bet". Which gets us back to it being a con, and was my original solution.
Again, it's not a con.

But yes, this is exactly what I'm getting at--The shell game man's strategy is unknown to you.

I wasn't looking for a solution with respect to either taking the bet or not taking the bet; I have been trying to use this example to illustrate that the issue of how the host plays the game makes a huge difference in the odds.

Finally, I don't see how this shell game is any different from the original Monty Hall problem. In the original problem, you are only given information about a single game. Not enough information is given regarding the host's strategy--He opened a door for you, certainly, but maybe that was just your particular game.

[Robin]

Cabbage
How can you possibly say that is not a choice on how to run the game?

Is there anybody else here that thinks that the choice on how to run the game n the first place is relevant? Can we assume that the rules don't change during the game?

[Cabbage]

Originally posted by Robin
Is there anybody else here that thinks that the choice on how to run the game n the first place is relevant? Can we assume that the rules don't change during the game?
Well, a simpler word for "choice on how to run the game" would be "strategy".

It all boils down to the host's strategy.

Maybe he has a complex strategy, offering the switch some times, not offering it at others, all in an attempt to deceive and mislead people into making the wrong decisions.

Or maybe he has a very simple strategy, in which he doesn't even bother to be tricky--he just always opens a door and offers a chance to switch.

The information in the original problem only pertains to the single game in which you are playing. Hardly enough to conclude anything about the host's strategy. But I think it should be clear that the strategies of both you and the host are vitally important in the game, and will affect the probabilities.

[Robin]

Originally posted by Cabbage
But yes, this is exactly what I'm getting at--The shell game man's strategy is unknown to you.


Can we make a distinction between strategy and rules.

Teabag is saying that the shell game man will break the rules and palm the pea. But we must assume for the problem that he will not - we must assume that he follows all rules.

Now his strategy is unknown to us - that is OK. But we would not get into any game without the rules being explained to us.

So say we have ruleset 1.

1. There are three cups
2. Shell man puts a pea under the cup
3. The cups are shuffled so that you don't know where the pea is
4. You put your finger on the cup where you think the pea is
5. The shell man lifts one of the other cups and reveals no pea
6. He asks us if we would like to switch
7. Depending on whether you accept the new choice you leave your finger on the cup or move it to the last remaining cup
8. You lift the cup your finger is on and find a pea then you win.

In this case it advantages you to change your choice as long as he follows the rules but the shell man's strategy is irrelevant.

So say we have ruleset 2

1. There are three cups
2. Shell man puts a pea under the cup
3. The cups are shuffled so that you don't know where the pea is
4. You put your finger on the cup where you think the pea is
5. The shell man may lift one of the other cups and reveals no pea
6. If he has lifted the cup in step 5 asks us if we would like to switch
7. If you have been offered a choice you may decide to move your finger to the last remaining cup or leave it where it is. If you have not been offered a choice then you leave the finger where it is.
8. You lift the cup your finger is on and find a pea then you win.

In this case it depends on the shell man's strategy.

[Robin]

Originally posted by Cabbage
Well, a simpler word for "choice on how to run the game" would be "strategy".

It all boils down to the host's strategy.

Maybe he has a complex strategy, offering the switch some times, not offering it at others, all in an attempt to deceive and mislead people into making the wrong decisions.

Or maybe he has a very simple strategy, in which he doesn't even bother to be tricky--he just always opens a door and offers a chance to switch.

The information in the original problem only pertains to the single game in which you are playing. Hardly enough to conclude anything about the host's strategy. But I think it should be clear that the strategies of both you and the host are vitally important in the game, and will affect the probabilities.

See my previous post, there are strategies and rules. If the rules state that he must reveal a goat and offer a choice then his strategy is irrelevant.

[Cabbage]

Originally posted by Robin
Can we make a distinction between strategy and rules.

Teabag is saying that the shell game man will break the rules and palm the pea. But we must assume for the problem that he will not - we must assume that he follows all rules.

Now his strategy is unknown to us - that is OK. But we would not get into any game without the rules being explained to us.

So say we have ruleset 1.

1. There are three cups
2. Shell man puts a pea under the cup
3. The cups are shuffled so that you don't know where the pea is
4. You put your finger on the cup where you think the pea is
5. The shell man lifts one of the other cups and reveals no pea
6. He asks us if we would like to switch
7. Depending on whether you accept the new choice you leave your finger on the cup or move it to the last remaining cup
8. You lift the cup your finger is on and find a pea then you win.

In this case it advantages you to change your choice as long as he follows the rules but the shell man's strategy is irrelevant.

So say we have ruleset 2

1. There are three cups
2. Shell man puts a pea under the cup
3. The cups are shuffled so that you don't know where the pea is
4. You put your finger on the cup where you think the pea is
5. The shell man may lift one of the other cups and reveals no pea
6. If he has lifted the cup in step 5 asks us if we would like to switch
7. If you have been offered a choice you may decide to move your finger to the last remaining cup or leave it where it is. If you have not been offered a choice then you leave the finger where it is.
8. You lift the cup your finger is on and find a pea then you win.

In this case it depends on the shell man's strategy.
I agree 100%.

[Robin]

Originally posted by Cabbage
I agree 100%.

But do you agree that with ruleset 1 the shell man's strategy does not matter?

[Cabbage]

Originally posted by Robin
But do you agree that with ruleset 1 the shell man's strategy does not matter?
I'm not sure what you mean. "Does not matter" with respect to what?

[Cabbage]

Sorry, I think I see what you mean now. No it doesn't matter, in the sense that I don't see how the rules allow for any choice or strategy on the host's part.

[Robin]

Originally posted by Cabbage
I'm not sure what you mean. "Does not matter" with respect to what?

OK I suspect you have gone into troll mode.

Under ruleset 1 in my earlier post, the shell man's strategy does not affect the outcome of the game as long as he follows the rules as laid out.

Do you agree.

[Robin]

Originally posted by Cabbage
Sorry, I think I see what you mean now. No it doesn't matter, in the sense that I don't see how the rules allow for any choice or strategy on the host's part.

OK, so what is your problem with my earlier post in saying that the hosts intentions (or strategy) may not affect the outcome of the game because he may be obliged to offer the second guess under the rules.

[Cabbage]

Originally posted by Robin
OK, so what is your problem with my earlier post in saying that the hosts intentions (or strategy) may not affect the outcome of the game because he may be obliged to offer the second guess under the rules.
Maybe we misunderstand each other? If there was, and the misunderstanding was my fault, I apologize.

If he is obliged to offer the second guess, I agree, his strategy will not affect the outcome of the game.

However, let's look back at the original problem: Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what's behind the doors, opens another door, say number 3, which has a goat. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?
Now, at the point the host reveals another door and offers a switch, how do we know whether we are operating under your ruleset 1 or ruleset 2? It is not given in the problem, and cannot be known without additional information (or assumptions).

[CurtC]

Well, this is entertaining. So far, I think everyone gets it except Teabag420. Paul C. has some reservations, that even saying you don't know the host's strategy is making an assumption, which I think is being a little obtuse. The problem statement doesn't give us any information on the host's strategy, but his strategy would be necessary to know in order to solve for the probability. Saying "there's not enough information" is not making an assumption, it's admitting that we can't solve it unless we make an assumption beyond what the problem says.

I'll take one more shot at explaining it to the gentler, kinder TeaBag420. Let's consider these two opposing preludes to the problem statement. In situation A, Monty is an evil guy and will offer a guest the chance to switch his guess only if the guest made the correct guess initially. In situation B, Monty is a brain-dead robot and always reveals a non-winning door after the initial guess, giving the guest the chance to switch to the remaining door.

Now consider the problem statement, as it's given in the OP, following each of these situation statements. You now find yourself in the situation where you've made your initial guess and Monty has offered you the choice to switch. What do you do?

I think it's pretty clear that if he's operating under situation A, you will always lose by switching, and if he's operating by situation B, you'll win more often by switching. Both of these scenarios are completely compatible with what's described in the OP, and we have no idea which is the right one, but the best choice for you to make completely depends on which one it is. Do you want to assume one or the other? Why? As Dirty Harry said, "Feelin' lucky, punk?"

[TeaBag420]

Originally posted by Robin
Can we make a distinction between strategy and rules.

Teabag is saying that the shell game man will break the rules and palm the pea. But we must assume for the problem that he will not - we must assume that he follows all rules.



GODDAMMIT YOU MOTHERLOVIN' CHANUKAH SUCKER! NO I AM NOT SAYING THAT. I HAVE ACCEPTED CABBAGE'S "CLARIFICATION" (BECAUSE IT IS HIS NEW PUZZLE) WHICH FORBIDS THE GRIFTER FROM PALMING OR OTHERWISE REMOVING THE PEA.

TRY TO FIRETRUCKING READ.

I know you wrote some other stuff, but I didn't read it because you effed up so big, which is not to say I won't respond to it later.

[TeaBag420]

Originally posted by Robin
What happened did they declare "miss the point day" and not tell me? This is exactly what I mean't.

If you know that the host had to offer the second choice then it is a probability problem.

If you know that the host didn't have to offer the second choice then it is not a probability problem but may be some other kind of maths problem.

So you offer two possibilities, both of which say it's a math problem, but previously you said it's not a math problem. Care to jack, um, I mean explain?

[TeaBag420]

Originally posted by CurtC
Well, this is entertaining. So far, I think everyone gets it except Teabag420. Paul C. has some reservations, that even saying you don't know the host's strategy is making an assumption, which I think is being a little obtuse. The problem statement doesn't give us any information on the host's strategy, but his strategy would be necessary to know in order to solve for the probability. Saying "there's not enough information" is not making an assumption, it's admitting that we can't solve it unless we make an assumption beyond what the problem says.

I'll take one more shot at explaining it to the gentler, kinder TeaBag420. Let's consider these two opposing preludes to the problem statement. In situation A, Monty is an evil guy and will offer a guest the chance to switch his guess only if the guest made the correct guess initially. In situation B, Monty is a brain-dead robot and always reveals a non-winning door after the initial guess, giving the guest the chance to switch to the remaining door.

Now consider the problem statement, as it's given in the OP, following each of these situation statements. You now find yourself in the situation where you've made your initial guess and Monty has offered you the choice to switch. What do you do?

I think it's pretty clear that if he's operating under situation A, you will always lose by switching, and if he's operating by situation B, you'll win more often by switching. Both of these scenarios are completely compatible with what's described in the OP, and we have no idea which is the right one, but the best choice for you to make completely depends on which one it is. Do you want to assume one or the other? Why? As Dirty Harry said, "Feelin' lucky, punk?"

You summed up your argument when you said "preludes to the OP". There are no preludes to the OP. You might as well say, "What if the Tooth Fairy offers Monty a really big bribe to throw the game?" The OP is the OP and you can string all the pearl necklaces you want on it, but it doesn't change.

[Cabbage]

Originally posted by TeaBag420
You summed up your argument when you said "preludes to the OP". There are no preludes to the OP. You might as well say, "What if the Tooth Fairy offers Monty a really big bribe to throw the game?" The OP is the OP and you can string all the pearl necklaces you want on it, but it doesn't change.
This isn't really a fair comparison. Adding the comment about the tooth fairy skews the problem, and leaves it fundamentally changed.

However, the additions regarding the strategy of the host are actually relevant, since the original problem, as stated, leaves out vitally important information regarding the host's strategy. In other words, these comments clarify a previously vague problem.

[Cabbage]

And maybe I'm beating this point into the ground now, but here's another illustrative example.

Say you're playing a game of chess. Just after you make a particular move, you realize you've made a blunder--your opponent can now force checkmate in three moves. You cannot, however, take back this blundered move.

What is the probability that you will lose the game?

(This problem is not analagous to the original Monty Hall problem in a probabilistic sense, but it is analogous in the sense they are both vague).

Who can say what the probability is? If your opponent is an experienced grand master, you will most certainly lose the game. On the other hand, if your opponent is merely a beginner, it's quite possible he will miss the opportunity for mate, make his own blunder later on, costing himself the game. I have personally been in this situation, blundered into a losing position, yet still managed to take a victory simply because my opponent missed his opportunity.

The probability of you losing your chess game depends on the quality of your opponent.

Similarly, in the Monty Hall game, any advantage you may or may not gain by switching is unclear. It fully depends on Monty's strategy (or lack thereof, if you will), which is an unknown variable.

[TeaBag420]

Cabbage wrote:

"This isn't really a fair comparison. Adding the comment about the tooth fairy skews the problem, and leaves it fundamentally changed."

Godfrey Daniels, Cabbage. Adding anything skews the problem. That's the problem with adding to the problem, which has already been posted to this board.

[TeaBag420]

Originally posted by Cabbage
And maybe I'm beating this point into the ground now, but here's another illustrative example.

Say you're playing a game of chess. Just after you make a particular move, you realize you've made a blunder--your opponent can now force checkmate in three moves. You cannot, however, take back this blundered move.

What is the probability that you will lose the game?

(This problem is not analagous to the original Monty Hall problem in a probabilistic sense, but it is analogous in the sense they are both vague).

Who can say what the probability is? If your opponent is an experienced grand master, you will most certainly lose the game. On the other hand, if your opponent is merely a beginner, it's quite possible he will miss the opportunity for mate, make his own blunder later on, costing himself the game. I have personally been in this situation, blundered into a losing position, yet still managed to take a victory simply because my opponent missed his opportunity.

The probability of you losing your chess game depends on the quality of your opponent.

Similarly, in the Monty Hall game, any advantage you may or may not gain by switching is unclear. It fully depends on Monty's strategy (or lack thereof, if you will), which is an unknown variable.

I agree with everything you say above. Additionally, this is a quintessential example of problem changing. Problem changers are one step above kid touchers in my book. In the OP, you don't know and shouldn't care about Monty's strategy, but you know what he did.

(This problem is not analagous to the original Monty Hall problem in a probabilistic sense, but it is analogous in the sense they are both vague).

Game, set, match. Anything vague is analogous to the original Monty Hall problem in a way you think is important and/or relevant.

You go now. No trouble.

[Cabbage]

Originally posted by TeaBag420
Cabbage wrote:

"This isn't really a fair comparison. Adding the comment about the tooth fairy skews the problem, and leaves it fundamentally changed."

Godfrey Daniels, Cabbage. Adding anything skews the problem. That's the problem with adding to the problem, which has already been posted to this board.
The addition doesn't skew the problem if its purpose is actually to clarify the problem. In fact, it clarifies a problem which was initially unsolvable.

[Cabbage]

Originally posted by TeaBag420
I agree with everything you say above. Additionally, this is a quintessential example of problem changing. Problem changers are one step above kid touchers in my book.
Where did I change the problem? I merely pointed out the fact that having a single example of the game in which the host offers the switch is insufficient information to deduce what the host's strategy in the game may be. Your probability of winning or losing the game depends in a vital manner on that host's strategy.

[Paul C. Anagnostopoulos]

CurtC said:
Well, this is entertaining. So far, I think everyone gets it except Teabag420. Paul C. has some reservations, that even saying you don't know the host's strategy is making an assumption, which I think is being a little obtuse. The problem statement doesn't give us any information on the host's strategy, but his strategy would be necessary to know in order to solve for the probability. Saying "there's not enough information" is not making an assumption, it's admitting that we can't solve it unless we make an assumption beyond what the problem says.
My point is an informal one, because I agree that if this were on the National Math Test, clarity would be critical. I just don't see why we need to worry about conditions that are not explicitly stated in this problem, when considering it informally in order to make the point about peoples' probability intuitions. Isn't it reasonable to assume that, by default, unstated conditions do not exist? The point, after all, is to allow someone to solve the problem!

Also, folks, this problem statement is all over the Net, and sometimes it does make it clear that he always offers a choice:

http://www.remote.org/frederik/projects/ziege/

The candidate chooses one of the doors. But it is not opened; the host (who knows the location of the sports car) opens one of the other doors instead and shows a goat. The rules of the game, which are known to all participants, require the host to do this irrespective of the candidate's initial choice.

Of course, if we persist, this is still ambiguous. Monty may be required to offer a choice regardless of the player's initial selection, but it doesn't say he has to do it if the player is wearing red pants.

~~ Paul

[Drooper]

Originally posted by Paul C. Anagnostopoulos
M I just don't see why we need to worry about conditions that are not explicitly stated in this problem,
~~ Paul

What? and spoil all the fun?;)

[BillHoyt]

Originally posted by Paul C. Anagnostopoulos
My point is an informal one, because I agree that if this were on the National Math Test, clarity would be critical. I just don't see why we need to worry about conditions that are not explicitly stated in this problem, when considering it informally in order to make the point about peoples' probability intuitions. Isn't it reasonable to assume that, by default, unstated conditions do not exist? The point, after all, is to allow someone to solve the problem!

Also, folks, this problem statement is all over the Net, and sometimes it does make it clear that he always offers a choice:

http://www.remote.org/frederik/projects/ziege/



Of course, if we persist, this is still ambiguous. Monty may be required to offer a choice regardless of the player's initial selection, but it doesn't say he has to do it if the player is wearing red pants.

~~ Paul

No, no, no, Paul. Good grief! Can't you understand the absolute need to twist a simple probability problem into a game-theoretic one while introducing hidden assumptions about motivations, the exchange value of the US$ and how many speckled dwarves are hiding in Carol Merrill's mini-skirt?

[pgwenthold]

Originally posted by BillHoyt
No, no, no, Paul. Good grief! Can't you understand the absolute need to twist a simple probability problem into a game-theoretic one while introducing hidden assumptions about motivations,

Any calculation of the probability requires some sort of assumption. The 2/3 probability that is the common answer requires the assumption that the option is allowed indiscriminately. If you assume that the option is only allowed when you have initially chosen the car, then the probabilty that you will win when you switch is 0. If you assume that the option is only allowed when you have chosen wrong, then the answer is 1. However, the short answer is that any answer requires an assumption. Since you don't have any information regarding his strategy, the question can't be answered absolutely.

It's actually very similar to geometry. The deniers sound like those defending Euclidian geometry, trying to claim that the 5th postulate is obvious. Others keep saying, yeah but it's not required. Within the constraints of the problem, you can create other scenerios where the probability is much different from 2/3. It all depends on your initial assumptions.

As Robin points out, without any assumptions, the probability cannot be calculated.