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 th