Balls. Once you know that there is no prize behind door B, prob(B) becomes 0. Then, since the prize must be behind A or C, and you don't know anything about them, prob(A) = prob(C) = 1/2.

Any of the doors is OK.

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Balls. Once you know that there is no prize behind door B, prob(B) becomes 0. Then, since the prize must be behind A or C, and you don't know anything about them, prob(A) = prob(C) = 1/2.

Any of the doors is OK.

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Sorry Filippo (F?) but u r wrong on this one. U may see also it in this way: in the game You ll never be told that your initial choice is wrong, the information added is only about the other two doors. This breaks the symmetry, the probability that your initial choice stays at 1/3 ( It is secluded by the bit of more information added ) , but now the prob of b goes to zero and because all the probabilities must add up to 1, the prob(c) becomes 2/3. I understand it is not intuitive, but not all the math is :)

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Sorry guys. none of the above is correct.

You forgot the check on the sentence, as Interviewer may be lying or not (50% probability for each of the two).

If he is NOT lying than there is NO price behind B --> you have 50% it is behind A and 50% it is behind C.

If he IS lying than price is behind door B (100% sure)

If you draw the brances tree and put it all together, you have that:

Probab. = 0,5*[(0.5*A+0.5*C] + 0.5*1*B

Ossia:

Probab = 0,25*A + 0.25*C + 0.5B

which means that both A and C have 25% probabiity of having it right, while B has 50% to be right.

Therefore answer is: "you leave door A and choose door B ". Not because this IS behind door B, but because you have a higher probability this is there.

:-)

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my vote is with Rob. this is covered in every basic probability and statistics course. I think Andrea is bringing unneeded complexity to the question

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Filippo's answer would be correct if he did actually open the door when he chose door A, but because it doesn't say that. So assuming that he telling the truth and that door B does not have a prize i.e probability of a zero, door C must have a probability of 2/3 . Therefore Rob's answer is correct.

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This is a the Monty Hall problem. Switching yields a win 2/3 of the time, and is preferred. I ran a simulation to confirm :-)

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

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Change it. The probability for door A is 1/3, the probability for the set Door C + Door B is 2/3. The interview adds information on the set stating that Door B prob is 0, so the probability for Door C is 2/3 while Door A stays at 1/3