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You are asked to open one door. There is a large prize behind one of the three doors and the other two doors are losers. After you select a door, Monty Hall opens one of the two doors you did not select that he knows is a losing door, selecting at random if both are losing doors. Monty asks you whether you would like to switch doors. Suppose that the three doors in the puzzle are labeled 1, 2, and 3. Let W be the random variable whose value is the number of the winning door; assume that p(W = k) = 1/3 for k = 1, 2, 3. Let M denote the random variable whose value is the number of the door that Monty opens. Suppose you choose door i.

Solve the following parts:
a) What is the probability that you will win the prize if the game ends without Monty asking you whether you want to change doors?
b) Find p(M = j | W = k) for j = 1, 2, 3 and k = 1, 2, 3.
c) Use Bayes’theorem to find p(W = j | M = k) where i and j and k are distinct values.
d) Explain why the answer to part (c) tells you whether you should change doors when Monty gives you the
chance to do so.

2 Answers

Best answer
5 votes
5 votes
(a) 1/3 as one out of 3 is the favourable case and all are equally likely

(b)
P(M=1|W=1, i = 1) = 0
P(M=2|W=1, i = 1) = 0.5
P(M=3|W=1, i = 1) = 0.5

P(M=1|W=2, i = 1) = 0
P(M=2|W=2, i = 1) = 0
P(M=3|W=2, i = 1) = 1

P(M=1|W=3, i = 1) = 0
P(M=2|W=3, i = 1) = 1
P(M=3|W=3, i = 1) = 0

Similarly we can write for i = 2, and i = 3.

(c)

(i, j, k) are distinct

P(W=1|M=2) = 2/3 (i = 3 implied by distinct values, and i=3 with probability 1/3 and Monty never selects the winning one making the other one's probability to win 2/3)
P(W=1|M=3) = 2/3 (i = 2)
P(W=2|M=1) = 2/3 (i = 3)
P(W=2|M=3) = 2/3 (i = 1)
P(W=3|M=1) = 2/3 (i = 2)
P(W=3|M=2) = 2/3 (i = 1)

(d). The probability of my selection winning is 1/3 and probability of the remaining door winning is 2/3 => so, if I believe in probability I should change the doors.

If you strictly want to use Bayes' theorem

http://faculty.washington.edu/tamre/BayesTheorem.pdf
selected by
1 votes
1 votes
D).yes the door should be changed as the probability of winning after switching the doors is 66.66%.Here it goes how? If the prize is behind Door 1, Monty will open either Door 2 or Door 3 to reveal an empty door. You switch to the other of Door 2 or Door 3, and in either case you switched to a door with nothing behind it (remember, the car is behind Door 1). If the car is behind Door 2, Monty will open Door 3. This is because he always opens a door which has nothing behind it, and he can't open Door 1 because that was your original choice. So the only door you can switch to is Door 2, which is the door with the car behind it.'You win!' If the car is behind Door 3, Monty will open Door 2. This is because he always opens a door which has nothing behind it.and he can't open Door 1 because that was your original choice. So the only door you can switch to is Door 3, which again is the door with the car behind it. Ding! You win!So out of three you win two times so probability is 66.66%.

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