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dionysian
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07 Jun 2011, 10:46 am

Natty_Boh wrote:
dionysian wrote:
The monty hall problem is stupid... seems to be pretty clearly misusing maths.


Extend the example out to the 1,000,000-door version makes the point more clearly. The one-in-a-million chance that you picked the right door to start with, vs. the 999,999,998-to-one chance that the other remaining door, left by process of elimination, is correct.

Well, to be fair, it would be 1,000,000 to 1 vs 999,999 to 1. Not 999,998 to 1, because there is still the possibility that you initially selected the right door.


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Thom_Fuleri
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07 Jun 2011, 2:51 pm

dionysian wrote:
Well, to be fair, it would be 1,000,000 to 1 vs 999,999 to 1. Not 999,998 to 1, because there is still the possibility that you initially selected the right door.


This is where it gets messy. See what I mean? Arguments again!!

In the original problem, there are three boxes. You choose one. The host opens one of the others to reveal it is empty, and offers you the chance to swap. Doing so gives you a 50/50 shot of winning rather than the 1/3 chance you get by sticking - and most people will stick with their original choice. The odds are not now "50/50" simply because there are only two boxes - the probability is not even, because the host knows which box has the prize. If you pick the prize box, he can open either of the others - if you don't, he will open the empty box you didn't pick.

It's the same with a million boxes. You choose one - your odds are 1/1000000 that it's the prize. Your host then opens all the other boxes one at a time, except one - there are now two left. Are your odds now 50/50? No sir. Your odds of having chosen the right box in the first place are still 1/1000000. Your odds of the prize being in the other box are damn near approaching 100%.

Incidentally, this approach doesn't work for "Deal or No Deal". The host genuinely has no idea what is in which box in that game.



dionysian
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07 Jun 2011, 3:40 pm

It's not really very complicated. Most people intuitively understand that the host showing you an empty door doesn't change what's behind the door you already selected. Attempts at showing how it does are kind of ridiculous.. meaningless parlor tricks.


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CrinklyCrustacean
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10 Jun 2011, 3:39 am

Thom_Fuleri wrote:
In the original problem, there are three boxes. You choose one. The host opens one of the others to reveal it is empty, and offers you the chance to swap. Doing so gives you a 50/50 shot of winning rather than the 1/3 chance you get by sticking.

No, swapping doesn't give you a 50/50 shot of winning, it gives you a 2/3 chance of winning, which is a significantly greater probability. You could think of it like this: with 2 empty boxes and 1 prize box, you are twice as likely to pick an empty box as a prize box the first time around. When the game host opens the empty box and asks you whether you want to stick or swap, you are still twice as likely to have picked an empty box the first time around. Therefore, you are twice as likely to win a prize by swapping than you would be by sticking. The odds of winning by swapping aren't 50%, but 66%.



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10 Jun 2011, 11:56 am

CrinklyCrustacean wrote:
No, swapping doesn't give you a 50/50 shot of winning, it gives you a 2/3 chance of winning, which is a significantly greater probability.


Sorry, yes. I must stop trying to work out probabilities when half asleep!