The Monty Hall problem - a variation

[Hofzinser]

I'm sure most of you are familiar with the Monty Hall problem (http://en.wikipedia.org/wiki/Monty_Hall_problem).

Some friends have been arguing over a subtle variation on this which is as follows: suppose Monty picks one of the other two boxes purely at random, and just happens to pick one with a goat behind it.

Does this affect whether or not you should switch? One of my friends thinks you should still switch as you chances remain 2/3 if you do so. Another friend thinks that it now makes no difference whether you switch or not - the fact that Monty's choice was random means that there is now a 50-50 chance that your original choice was correct.

I think the second friend is correct but I'm not sure how to prove it. Can anyone provide a definitive correct answer and, ideally, a proof?

[bobman0330]

So, there's a possibility that Monty would pick the door with the real prize behind it?

[Hofzinser]

Yeah, exactly.

[bobman0330]

Then it doesn't affect your chances of winning if you switch. You'll win just as often overall with this variant though. 1/3 of the time Monty will show the car and you can pick it. 1/2 of the remaining 2/3 of the time, you'll win the car blind. = 2/3.

[TomCollins]

[ QUOTE ]
I'm sure most of you are familiar with the Monty Hall problem (http://en.wikipedia.org/wiki/Monty_Hall_problem).

Some friends have been arguing over a subtle variation on this which is as follows: suppose Monty picks one of the other two boxes purely at random, and just happens to pick one with a goat behind it.

Does this affect whether or not you should switch? One of my friends thinks you should still switch as you chances remain 2/3 if you do so. Another friend thinks that it now makes no difference whether you switch or not - the fact that Monty's choice was random means that there is now a 50-50 chance that your original choice was correct.

I think the second friend is correct but I'm not sure how to prove it. Can anyone provide a definitive correct answer and, ideally, a proof?

[/ QUOTE ]

Suppose you picked the right box. This happens 1/3 of the time.

In this case, he will always show you an empty box. Switching will be wrong, so if you switch, you lose.

If you picked the wrong box 2/3 of the time, 50% of the time Monty shows you the prize, and 50% he shows you an empty box. Switching will win.

So once he shows us an empty box, we know one of two things happened.

1) We picked right (33% of the time overall)
2) We picked wrong, and he picked wrong (66% * 50% = 33%)

Since each of these events is equally likely to happen, the probability we are right remains the same, 50% of the time.

So this is different than the Monty Hall Problem. Switching is no more of an advantage.

^^ good response

[Hofzinser]

Thanks Tom, that's perfect.