Probability

Question

if you have 3 doors and a player has chosen the wrong door, the emcee opens tone of the remaining door that has no prize behind it. when the player has chosen the correct door, the emcee always opens the left-most of the 2 remaining doors. (if the 2 remaining doors are A and B, he selects A; A and C -> A; B and C-> B

 

Given this policy by the emcee, what should the player do if he chose A and the emcee chose B?

Answer 1

You have chosen A.  There are 3 possibilities for the prize, A, B, and C; we assume they are equally likely

If the prize is behind A, the emcee reveals B, the left-most of the 2 remaining doors. (p = 1/3)

If the prize is behind B, the emcee reveals C. (p = 1/3)

If the prize is behind C, the emcee reveals B. (p = 1/3_

Thus, the emcee reveals B in 2 cases, the prize is behind A and the prize is behind C.  The total probability is 1/3+1/3=2/3

Thus, the probability of the prize being behind A if the emcee opens B = 1/3/(2/3) = 1/2, and the probability of the prize being behind C if the emcee opens B = 1/3/(2/3) = 1/2

The prize is equally likely to be behind A or C, so there is no advantage to making either choice.

 

Answer 2

I have a feeling this problem is meant to be a statement of the Monte Hall problem:

  https://en.wikipedia.org/wiki/Monty_Hall_problem

It's tricky ;-)

 

Answer 3

This is known as the Monty Hall Problem, after the host of Let's Make a Deal.

The counter-intuitive answer is that the person should switch since the chances are 2/3 that the prize is there.

For a detailed explanation:

https://en.wikipedia.org/wiki/Monty_Hall_problem

 

Once again the incongruity of the "voting" system is manifest.

My response simultaneously is "Not Approved," and gets one "Thumbs Up" and one "Thumbs Down."