Pepsi, Apple and Probability

[HarryKeogh]

I was never good at this so if someone could figure it out for me I'd appreciate it...

ok, Pepsi is running a promotion where if the bottle cap has a message written under it you win a free song download from Apple. The funny part is if you tilt the bottle just right you could read the winning (or non-winning) message. The odds of getting a winning cap are one in three.

the question: if I'm at the convenience store and tilt the first bottle and see that it's not a winner and I put it back and grab a second bottle (from among dozens of bottles) and pay for it (without checking the cap) what are the chances the second bottle is a winner? my guess is 1 in 2 but it's pretty much that...a guess. Can someone provide the correct answer along with the math behind it?

[Rolfe]

Is this like the two goats and the car all over again?

Because my brain swallowed itself trying to make sense of that one.

Rolfe.

[Jon_in_london]

Originally posted by HarryKeogh
the question: if I'm at the convenience store and tilt the first bottle and see that it's not a winner and I put it back and grab a second bottle (from among dozens of bottles) and pay for it (without checking the cap) what are the chances the second bottle is a winner? my guess is 1 in 2 but it's pretty much that...a guess. Can someone provide the correct answer along with the math behind it?

The odds of getting it will awlays be one in three.

[Dragon]

Originally posted by Jon_in_london


The odds of getting it will awlays be one in three.

No, they won't.
Think about it.
(Hint:- There is only a finite number of bottles.)

[patnray]

The odds would still be one in three since the the dozens of bottles on the shelf were all placed randomly and are independant of each other.

If you knew beforehand that there were 12 bottles on the self and you knew that there were 4 winners among them , then your odds would improve to 4 out of 11 after removing one non-winner. But you do not have this knowledge after checking just one bottle.

[wollery]

Let me expand on Jons' answer. If there are only three bottles and you can tell that one of them is not a winner then the probability that one of the other two is a winner is 1/2.

However there are millions of bottles. This means that knowing that one of them isn't a winner does not affect the odds of the next bottle you pick up. (It may be that none of the bottles on that shelf are winners.)

[patnray]

Originally posted by Dragon


No, they won't.
Think about it.
(Hint:- There is only a finite number of bottles.)

But you don't know how many winners are in the collection on the shelf. And the bottles are independant of each other.

If you checked every bottle on the shelf except one, and all the bottles you check were losers, what are the odds that the last bottle is a winner? Still 1 in 3.

[The Don]

[pedant mode] If there are exactly 1/3 of each and you have eliminated one, even if there are millions haven't the odds shifted, albeit in a miniscule fashion, towards 1/2[/pedant mode]

If there are millions of bottles then you have made no MATERIAL difference to your odds of winning.

[Valmorian]

Originally posted by wollery
Let me expand on Jons' answer. If there are only three bottles and you can tell that one of them is not a winner then the probability that one of the other two is a winner is 1/2.

Huh? So you're saying that if you can tell that two of them are not winners then the probability that the last one is a winner is 1/1, or 1?

I don't think so. ;)

EDIT: Unless you mean there are only 3 bottles in the entire contest (which there is not).

[(S)]

I think the basis of the 1-in-2 guess is the idea that, after three tries, you're /guaranteed/ to win, because the odds are 1-in-3. As a matter of fact, you only have a 19/27 chance of winning after three tries. [A 2/3 chance of not winning on any particular draw. 2/3 * 2/3 * 2/3 = 8 / 27 chance of not winning at all after three. 1 - 8 / 27 = 19 / 27 chance of winning.] That's only 70%, for those of us who don't like fractions.

[HarryKeogh]

Originally posted by The Don
[pedant mode] If there are exactly 1/3 of each and you have eliminated one, even if there are millions haven't the odds shifted, albeit in a miniscule fashion, towards 1/2[/pedant mode]

If there are millions of bottles then you have made no MATERIAL difference to your odds of winning.

so if there are 900,000 bottles and 300,000 winners are you saying that my odds have changed from 300,000 in 900,000 to 300,000 in 899,999?

[(S)]

Yes.

[Ladewig]

No, they won't.
Think about it.
(Hint:- There is only a finite number of bottles.)

I give special recognition to the first post that ever made me laugh and curse simultaneously. It's very hard to breathe when one does that.

.. As a matter of fact, you only have a 19/27 chance of winning after three tries. [A 2/3 chance of not winning on any particular draw. 2/3 * 2/3 * 2/3 = 8 / 27 chance of not winning at all after three. 1 - 8 / 27 = 19 / 27 chance of winning.] That's only 70%, for those of us who don't like fractions.

The formula [( 1- (2/3)^n) where n is number of tries] applies only if one assumes that no one les has been checking the shelf for winners. If other people do peek, then the chance of winning drops.

[Dragon]

Originally posted by HarryKeogh


so if there are 900,000 bottles and 300,000 winners are you saying that my odds have changed from 300,000 in 900,000 to 300,000 in 899,999?

Yup.
Also, if each store gets a fair allocation of winning bottles then eliminating a "loser" in any particular store has more of an effect - say 30 in 90 to 30 in 89.
Of course, as Ladewing says, if previous customers have been peeking ...
I suppose the best strategy would be to go to a store which has just had a fresh delivery.

Oh, and we all have too much time on our hands.

[(S)]

The ... well, I wouldn't call it /best/, but the 'optimal' strategy is to peek yourself ;-)

[scribble]

Originally posted by HarryKeogh
[B]I was never good at this so if someone could figure it out for me I'd appreciate it...


I can answer without using math. Your chances are zero, because every other asshat in the universe was smart enough to keep looking and all the winning bottles are long gone. Sorry.

[scribble]

Just for kicks, I'll share my personal winning strategy with you folks. I've discovered a marvelous mathematical proof behind this strategy but it is lengthy and I can't fit it here.

Step 1. Get kazaa-lite or some other decent non-spyware-infested P2P software.
Step 2. Download whichever songs you like.
Step 3. Enjoy a nice Diet Coke, and sit sound in the knowledge that while you may be raping musical artists, at least you're giving Jack Valenti nightmares.

[Yahweh]

Originally posted by HarryKeogh
so if there are 900,000 bottles and 300,000 winners are you saying that my odds have changed from 300,000 in 900,000 to 300,000 in 899,999?
I like those odds!

[sorgoth]

It also gives you a pretty damn good excuse to drink pop. :p

[Jim_MDP]

Originally posted by scribble

Step 3. Enjoy a nice Diet Coke, and sit sound in the knowledge that while you may be raping musical artists, at least you're giving Jack Valenti nightmares.

This gave me a good laugh till I realized…

When Jack has nightmares, hes comes up with idiotic ideas. Such as eliminating 'screener' copies for Academy members. This would mean I could no longer download them from BT.

Hey… now I'M gonna have nightmares.
Nice move, ya mook.

:p

[Deetee]

Originally posted by HarryKeogh

ok, Pepsi is running a promotion where if the bottle cap has a message written under it you win a free song download from Apple. The funny part is if you tilt the bottle just right you could read the winning (or non-winning) message. The odds of getting a winning cap are one in three.


I bet that they lie about the odds.
In the UK a couple of years ago I tried just this same tactic with some softdrink offer which said there were "100 000 winning bottles".
I checked about 30 bottles in each of 3 different supermarkets and found no winners.

I doubt there were more than a few million bottles nationally, so concluded that the company lied about the number of possible winners (or as scribble suggested, someone got there first and took all the winning bottles)

[HarryKeogh]

Originally posted by Deetee


I bet that they lie about the odds.
In the UK a couple of years ago I tried just this same tactic with some softdrink offer which said there were "100 000 winning bottles".
I checked about 30 bottles in each of 3 different supermarkets and found no winners.

I doubt there were more than a few million bottles nationally, so concluded that the company lied about the number of possible winners (or as scribble suggested, someone got there first and took all the winning bottles)

I doubt it. To redeem the free song you have to install Itunes so it's a good way for Apple to get people to put their software on their computer (and a percentage of them will start buying songs regularly)

[(S)]

Originally posted by Deetee
I bet that they lie about the odds.
In the UK a couple of years ago I tried just this same tactic with some softdrink offer which said there were "100 000 winning bottles".
I checked about 30 bottles in each of 3 different supermarkets and found no winners.

I doubt there were more than a few million bottles nationally, so concluded that the company lied about the number of possible winners (or as scribble suggested, someone got there first and took all the winning bottles)

p = chance-of-not-winning-at-all
q = chance-of-not-winning-once
n = number of tries
p = q^n

So, p^(1/n) = q.

So, if you have a 0.5 chance of not winning after 30 tries [you should't be surprised not to win on a 50-50 shot], that indicates that you had a 0.977 chance of not winning on any particular shot. That in turn says you have a 0.023 chance of winning, or 1-in-50. 50 * 100,000 winners gives you 5 million total bottles.
Working the math backwards (or forwards) from 10 million total bottles, you have a 0.01 chance of winning on any particular go, or a 0.99 chance of losing. That gives you 0.74 chance of losing after 30 tries, which makes winning with 30 bottles a 1-in-4 shot.

Long story short, there were probably a 100k winners, and you shouldn't be too surprised at having not won.

[Deetee]

That gives you 0.74 chance of losing after 30 tries, which makes winning with 30 bottles a 1-in-4 shot.

Long story short, there were probably a 100k winners, and you shouldn't be too surprised at having not won.

I guess.
Thanks. Nevertheless I checked about 3 times that number. I can't remember how many exactly, but at the time I felt I was extremely unlucky not to find at least one winner considering the number of "attempts" (even if I was cheating by not purchasing the product):p

[slimshady2357]

Originally posted by (S)


p = chance-of-not-winning-at-all
q = chance-of-not-winning-once
n = number of tries
p = q^n

So, p^(1/n) = q.

So, if you have a 0.5 chance of not winning after 30 tries [you should't be surprised not to win on a 50-50 shot], that indicates that you had a 0.977 chance of not winning on any particular shot. That in turn says you have a 0.023 chance of winning, or 1-in-50. 50 * 100,000 winners gives you 5 million total bottles.
Working the math backwards (or forwards) from 10 million total bottles, you have a 0.01 chance of winning on any particular go, or a 0.99 chance of losing. That gives you 0.74 chance of losing after 30 tries, which makes winning with 30 bottles a 1-in-4 shot.

Long story short, there were probably a 100k winners, and you shouldn't be too surprised at having not won.

Assume winning with 30 bottles is 25% chance, he said he did that three times.

Chances of not winning after checking 90 bottles would be (0.75)^3 = 42%

Therefore he still would have only had about 58% chance of winning after checking 90 bottles :eek:

Crappy contest, I want to WIN :mad:

Adam

[wipeout]

Originally posted by Rolfe
Is this like the two goats and the car all over again?

Because my brain swallowed itself trying to make sense of that one.

Rolfe.

Ah, the infamous "Monty Hall problem"... :D It's that there are only 3 doors which screws everybody up.

I tried to make it more intuitive for myself, so I increased the number of doors and that worked.

With 1000 doors to 999 goats and 1 car, a first choice, 998 doors to goats opened in one go, and then the offer to switch the final choice of the 2 remaining doors, it becomes obvious that the chances are the car was elsewhere to my first-round choice was 999/1000, and there is now only 1 door elsewhere, so switching makes a lot of sense.

In the equivalent with 3 doors to 2 goats and 1 car, the chances are the car was elsewhere to my first round choice was 2/3, but there's now only one door elsewhere, so again switching makes sense.

[Drooper]

Originally posted by HarryKeogh
I was never good at this so if someone could figure it out for me I'd appreciate it...

ok, Pepsi is running a promotion where if the bottle cap has a message written under it you win a free song download from Apple. The funny part is if you tilt the bottle just right you could read the winning (or non-winning) message. The odds of getting a winning cap are one in three.

the question: if I'm at the convenience store and tilt the first bottle and see that it's not a winner and I put it back and grab a second bottle (from among dozens of bottles) and pay for it (without checking the cap) what are the chances the second bottle is a winner? my guess is 1 in 2 but it's pretty much that...a guess. Can someone provide the correct answer along with the math behind it?

The odds are one in three,. because the events are indpendent.

[Drooper]

Originally posted by The Don
[pedant mode] If there are exactly 1/3 of each and you have eliminated one, even if there are millions haven't the odds shifted, albeit in a miniscule fashion, towards 1/2[/pedant mode]

If there are millions of bottles then you have made no MATERIAL difference to your odds of winning.

given the vagiaries of production and distibution, never will exactly 33.33.....% of the bottles in circulation be "winners".