Bayes Theorem: WTF

[spanshcastlemagc]

I feel like I heard this wrong. But I was watching a show on odds, and it mentioned Bayes Theorem. The example it used was simple. There are three doors, behind one of which is a prize. If you chose door A or B, and then someone reveals to you that the prize is not behind door C, you should change your choice in light of the new information, increasing your chances of guessing the right door. I thought that is how they explained it, but that doesn't make any sense to me. Does anyone know the correct way of explaining it, or if this is the correct way how does changing your decision increase your chances of making the correct one?

[traz]

It makes sense because if you choose any door...the chance of getting it right is 1/3. Once you are told that a 3rd door DOES NOT contain the prize, then there are 2 doors left...the one you chose, and another door. Now, the door you chose has a probability of 1/3, we already established that. So the other door must have a probability of 2/3, and thus is more likely to win.

[spanshcastlemagc]

so lets say i chose door A, and i know that nothing is behind door C, doesnt that make door A's probabilty 1/2, the same as door B? im missing something here

[BruceZ]

[ QUOTE ]
so lets say i chose door A, and i know that nothing is behind door C, doesnt that make door A's probabilty 1/2, the same as door B? im missing something here

[/ QUOTE ]

What you're missing is that if the prize is behind B, then the host HAD TO show you that C was empty. He had no choice, since he couldn't open door B and show you the prize, and he couldn't open your door. These are the rules that the host must operate under, and they were not clearly stated, but they are the key to the problem. Now if the prize were behind your door A, the host had a choice of opening either door B or door C, from which we assume that he would choose C only half the time. So the fact that he revealed door C makes it twice as likely that the prize is behind door B.

Now if the game were played so that the host could choose any of the 3 doors at random to show you, including possibly opening the prize door or your door, then it would indeed be equally likely that the prize is behind door A or door B.

This is called the Monty Hall problem, and it has been discussed to death on this forum. Do get acquainted with the search function and the recent archives.

[spanshcastlemagc]

ok now i see, the host is not allowed to open the door that you choose, now that makes sense, thanks

[The_student]

I still have my doubts about this theory... since there is nothing to loose by staying in the game until the end, and since no betting is added in this theoretically middle phase, the game should be viewed as one single phase, no? And as such, by choosing any of the three doors, you would have a 50% chance of winning since one of the loosing doors are eliminated for you. The game should be no different from a hypothetical game were the host sais "choose one of these two doors" or "choose one of these three doors, but not this one becouse it doesnt have anything behind it"

Ofcourse i may be lacking the insight to understand this theory fully, but i would like to know if anyone have made computer simulations which tests this theory, and if so, what the result was?

[Howling]

[ QUOTE ]
I still have my doubts about this theory... since there is nothing to loose by staying in the game until the end, and since no betting is added in this theoretically middle phase, the game should be viewed as one single phase, no? And as such, by choosing any of the three doors, you would have a 50% chance of winning since one of the loosing doors are eliminated for you. The game should be no different from a hypothetical game were the host sais "choose one of these two doors" or "choose one of these three doors, but not this one becouse it doesnt have anything behind it"

Ofcourse i may be lacking the insight to understand this theory fully, but i would like to know if anyone have made computer simulations which tests this theory, and if so, what the result was?

[/ QUOTE ]

This has been answered already, but I shall attempt to break it down.

You pick A. If you keep A, you win 33% of the time.

But new info has been added, if the prize is behind door B or C, the host will open whichever door does not have the prize. It looks like 50/50, but here is how it works.

THE OPENING OF B OR C IS NOT RANDOM.

If you missed with guess A, and the prize is behind B, the host is FORCED TO OPEN C, (or the reverse, prize C, open B).

Whenever you pick A (or any door), and switch, you will win WHENEVER THE PRIZE WAS BEHIND B OR C.

If it was behind B, C is opened, or if it was behind C, B is opened.

The only way you lose if if you picked the winner, then of course both the other doors are empty, and you will switch and lose.

Soooo. Pick A and stay, win whenever prize was behind A.

Pick A and switch, win whenever prize is behind B or C.


Note: I am not pretending that I have 'solved' this puzzle, it has been around a long time. I was just attempting to break it down for the poster I quoted.

[RustyBrooks]

The funny thing is, when you attempt to simulate this on a computer, the answer is immediately, obviously true, because while you are putting in the rules for each branch, it's obvious what all the possible branches are, and how often they'll happen. At any rate, I bet simulation would show either the correct answer or the commonly believed one, depending on whether the programmer made the proper interpretation, i.e. the host is constrained in which window he can open and which he can not. In other words, if you don't "get" that part, your simulation won't either.

[Philo]

[ QUOTE ]
Of course i may be lacking the insight to understand this theory fully, but i would like to know if anyone have made computer simulations which tests this theory, and if so, what the result was?

[/ QUOTE ]

Try it yourself:

http://people.hofstra.edu/staff/stev...tyHallSim.html

[Trylan]

for one of my C++ classes we had to write a program demonstrating what the outcome would be.

Over 1 000 000 iterations it was a 66.67% chance you'd be right by switching.