Question

e. Application of Bayes: Using Bayes theorem driven in item d, show that in the following problem switching is the better strategy for the player and makes the probability of wining as 2/3, while not switching gives the probability of wining 1/3. Monty Hall: There are 3 doors 1, 2, 3. Randomly, and equally likely, behind one of the door there is a prize, and other two are empty. The player choose one door. He does not open the door first. The showman; Monty hall (who knows the place of prize) shows the player one of the other two doors that is empty. He can always find a door that is not empty even if the door initially chosen by player is empty and the prize is behind other two doors. If both of them are empty he chose one to be shown with the probability 1/2. The player have the right to switch his door from first choice to the remained door that is not shown or stay with his initial choice Without loss of generality let player always choose door 1. Denote A1, A2, A3 as 3 events of prize being behind door 1, 2 and 3. These three events are mutually exclusive, equally likely with probability 1/3 and exhaustive .Denote the event that Monty hall discloses that the door 2 is empty with "B". First with law of total probability that is in item "c" show that P(B)-1/2 . Then by Bayes theorem that is in item "d" show that P(A3 |B)-2/3 (this is probability of winning while switching after disclosure of an empty door like door 2.) . Then show similarly show P(A1 B)-1/3 (this is probability of winning while not switching after disclosure of an empty door like door 2.) . Conclude by symmetry if we define B the event that Monty shows the door 3 being empty the abovee answers are same and switching is the optimal strategy to win with twice as much chance . Write Q.E.D. (which is "Quod Erat Demonstrandum" which means "that which was to be demonstrated")

Answer 1

a) Consider the following cases: Case I I the prize is behind Door1 then Monty can select door 2 with probability 1/2 ie. P(BA, ) = 1/2. Hence P(BnAa) = P(BAs)P(Aa) = 1 /2 x 1/3-1/6 Case III the prize is behind door 2 then Monty will not open that door Hence P(BnA2)-0 Case III: Observe that if the prize is in door 3 then Monty will obviously choose door 2 to show that it is empty. Hence P(BnAs) 1/3. Therefore by law of total probability -1/6 + 0 + 1/3-1/2 b) By Bayes' Theorem, P(B) =2/3 c) By Bayes" Theorem, P(B) d) Observe that as the player has already chosen door 1, hence there are only two doors for Monty Hal to open. Now either one of door that Monty shows should be empty. Hence f C is the probability of Monty showing that door3 is empty, then observe that the doors 2 and 3 can be swapped which does not alter P(B) and P(C). Hence P(B) P(C)