Three Door Game Show Problem And Solution
Problem : Suppose you are on a game show and you are given the choice of three doors : Behind
one door is a car; behind the other two doors are goats. You pick a door, say # 1, and the host, who
knows what’s behind the doors, opens another door which has a goat. The host then asks you if you
would like to switch to the door you did not pick (and the host did not open). Is it to your advantage
to switch?
Intuitive Solution : There are 3 possible arrangments
Case 1
Case 2
Case 3
Door 1
Car
Goat
Goat
Door 2
Goat
Car
Goat
Door 3
Goat
Goat
Car
Case 1 : You pick door 1 and the host picks either doors 2 or 3. If you switch you lose. If you don’t
switch you win.
Case 2 : You pick door 1 and the host picks door 3 (the host will NOT pick door 2). If you switch
you win. If you don’t cwitch you lose.
Case 3 : You pick door 1 and the host picks door 2 (the host will NOT pick door 3). If you switch
you win. If you don’t cwitch you lose.
Clearly if you switch, your chances of winning are 2 out of 3 (winning probability is 2/3) and if you
do not switch, your chances of winning are 1 out of 3 (winning probability is 1/3). So yes it is to your
advantage to switch.
Law of Total Probability Solution : You always pick door 1. (Note that the solution does not
change if you pick always pick door 2. So this is general enough). Let W be the event of winning,
D1 the event that the car is behind door 1, D2 the event that the car is behind door 2, and D3 the
event that the car is behind door 3. There are two policies : policy 1, you always switch and policy 2,
you never switch. Under each of the two policies find P (W ). Remember that in both cases you know
P (W |D1), P (W |D2), P (W |D3), P (D1), P (D2), and, P (D3). When you do the calculations (yes I
have skipped the calculations on purpose, I want you to try it on your own) you will see that P (W )
equals 2/3 with policy 1 and 1/3 with policy 2.