Before reading this node, be sure you have read
Monty Hall Problem first...
Answer:
Yes! It improves your chances of getting the car if you switch! In fact, it doubles your chances. Why?
In conceptual terms:
There are three possible arrangements of the prizes:
Case 1: Door1: Goat, Door2: Goat, Door3: Car
Case 2: Door1: Goat, Door2: Car, Door3: Goat
Case 3: Door1: Car, Door2: Goat, Door3: Goat
Say you pick door 1. There is a 1/3 chance that you picked the door hiding the car. Since Monty knows which door the car is behind, he opens a door with a goat.
In cases one and two, when you did not guess the car, he opens the second and third door, respectively.
Only in case three would it be advantageous to not switch.
So, 2/3 of the time, you will win the car if you switch, where only 1/3 of the time, you will win the car if you don't switch, because one of the choices has been removed from the picture.
If you still don't buy it, imagine that there are 10,000 doors. 9,999 of the doors hide goats, and one of the doors hide a new car.
Say you pick door #1. Monty then proceeds to open every door except door #4,654, revealing goats behind each one. Should you switch to door #4,654?
Since your first guess had only a 1/10,000 chance of picking the car, isn't is much more likely that the one door Monty didn't show you contains the car than your 1 in 10,000 guess?
Mathematical Solution
The probability that the prize is behind door X, P(X) = 1/3
The probability that Monty opens door B if the prize were behind A,
P(Monty opens B|A) = 1/2
The probability that Monty opens door B if the prize were behind B,
P(Monty opens B|B) = 0
The probability that Monty Hall opens door B if the prize were behind C,
P(Monty opens B|C) = 1
The probability that Monty opens door B is then
p(Monty opens B) = p(A)*p(M.o. B|A) + p(B)*p(M.o. B|B) + p(C)*p(M.o. B|C)
= 1/6 + 0 + 1/3 = 1/2
Then, by Bayes' Theorem,
P(A|Monty opens B) = p(A)*p(Monty opens B|A)/p(Monty opens B)
= (1/6)/(1/2)
= 1/3
and
P(C|Monty opens B) = p(C)*p(Monty opens B|C)/p(Monty opens B)
= (1/3)/(1/2)
= 2/3
Thus, your odds increase to 2/3 when you switch doors.
However...
A cool side-note to this problem exploits a
loophole in the problem. Check out the
Problem with the Monty Hall Problem.