Question

2. (Monty Hall) Suppose you are on a game show and are presented with three closed doors
marked door 1, 2, and 3. Behind one door is a prize and behind the other two are goats.
Suppose the host allows you to select one door, but the following two rules apply:
• Before it is opened the host opens one of the two unselected doors that has a goat behind
it.
• The host then allows you to switch your choice to the remaining door or stay with your
original choice.
Say you select door 1. If the host then opens door 3 to reveal a goat, compute the probability
the prize is behind door 2. To do this, use the following events:
• D1 = The prize is behind door 1
• D2 = The prize is behind door 2
• D3 = The prize is behind door 3
• H1 = the host opens door 1
• H2 = the host opens door 2
• H3 = the host opens door 3
In other words, compute P(D2 | H3). What would you do in this situation, stay or switch?
Hint: Use Bayes’ Theorem and the Law of Total Probability

Answer 1

Solution

Back-up Theory

Bayes’ Theorem and Law of Total Probability

If A and B are two events such that probability of B is influenced by occurrence or otherwise of A, then, conditional Probability of B given A, denoted by P(B/A) = P(B ∩ A)/P(A)……...................................................................................................….(1)

Law of Total Probability

P(B) = {P(B/A) x P(A)} + {P(B/A^C) x P(A^C)}………………….......................................................…………………………….(2)

Baye’s Theorem:

P(A/B) = P(B/A) x {P(A)/P(B)}…….................................................………………………..………...............…………..…….(3)

Now to work out the solution,

Let

D1 = The prize is behind door 1
D2 = The prize is behind door 2
D3 = The prize is behind door 3
H1 = the host opens door 1
H2 = the host opens door 2
H3 = the host opens door 3
Given that the player selected door 1 .................................................................................................................................(4)

And

The host opens one of the two unselected doors that has a goat behind it........................................................................(5)

We want P(D2|H3)

= {P(H3/D2) x P(D2)}/P(H3) [vide (3)] .............................................................................................................................. (6)

Given that the player selected door 1, if the prize is behind door 2, vide (5), the host

would surely open door 3 only. Thus, P(H3/D2) = 1 .......................................................................................................... (7)

Since the prize could be behind any one of the three doors, the unconditional probability

P(D2) = 1/3 ......................................................................................................................................................................... (8)

Now, vide (2),

P(H3) = {P(H3/D2) x P(D2)} + {P(H3/D3) x P(D3)}

= {1 x (1/2)} + {0 x (1/3)}

[vide (7), P(H3/D2) = 1; Since the player selected door 1, the host is left with only two options and so P(D2) = ½. Again, vide (5), host would not open door 3 if it has the prize.

So, P(H3/D3) = 0. And trivially, P(D3) = 1/3]

Thus, P(H3) = ½ .................................................................................................................................................................(9)

(7), (8), (9) and (6) =>

P(D2|H3) = (1/3)/(1/2) = 2/3 ............................................................................................................................................. (10)

Also, once the host opens the door which has got the goat, the prize is behind one of the

other two doors with equal probability.

Since by (10), the probability is 2/3 > ½, the player must switch. Answer

DONE