Consider the Monty Hall problem. In the case that the door you initially chose hid the valuable prize, suppose that Monty opens one of the other two doors for you not uniformly at random, but rather opens the second door with probability p and the third door with probability 1 − p, for some p ∈ [0, 1]. How does this affect your chances of getting the prize when switching?

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A: n(S) =300!4!296!n(P) =298!2!296!

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A: Given, P(Stay | Bribe) = 1 P(stay | Not bribed) = 1/3 P(bribe) = 1/4 P(not bribe) = 3/4

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