Fall 2011 Strategic Practice 3: Section 1 (Continuing with Conditioning) - Question 1
Consider the Monty Hall problem, except that Monty enjoys opening Door 2 more than he enjoys opening Door 3, and if he has a choice between opening these two doors, he opens Door 2 with probability p, where $\frac{1}{2}\leq p\leq 1$. To recap: there are three doors, behind one of which there is a car (which you want), and behind the other two of which there are goats (which you don't want). Initially, all possibilities are equally likely for where the car is. You choose a door, which for concreteness we assume is Door 1. Monty Hall then opens a door to reveal a goat, and offers you the option of switching. Assume that Monty Hall knows which door has the car, will always open a goat door and offer the option of switching, and as above assume that if Monty Hall has a choice between opening Door 2 and Door 3, he chooses Door 2 with probability p (with $\frac{1}{2}\leq p\leq 1$).
(a) Find the unconditional probability that the strategy of always switching succeeds (unconditional in the sense that we do not condition on which of Doors 2, 3 Monty opens).
(b) Find the probability that the strategy of always switching succeeds, given that Monty opens Door 2.
(c) Find the probability that the strategy of always switching succeeds, given that Monty opens Door 3.
Solution: (a.) Using the law of total probability, we can find the unconditional probability of winning in the same way as in class and we get 2/3. (b.) A tree method or Bayes' Rule with law of total probability will give both give 1/(1+p). (c.) The structure of the problem is the same as part b (except for the condition that p is greater than or equal to 1/2, which was no needed above). By the previous part, we get 1/(2-p).
"Mathematics is the logic of certainty, but statistics is the logic of uncertainty."