The Monty Hall problem

Suppose you're on a game show, and you're given the choice of three doors: $a$, $b$, and $c$. Behind one door is a car; behind the others, goats. You pick a door, say $a$, and the host, who knows what's behind the doors, opens another door, say $b$, which has a goat. He then says to you, "Do you want to pick door $c$?" Is it to your advantage to switch your choice?

What were the odds he would choose that door to open?

• If the car is in box $a$ then he could open either $b$ or $c$ to reveal a goat. The probability is $1/2$.
• If the car were in box $b$ there's no way he would open $b$. The probability is $0$.
• If the car is in box $c$ then Monty would definitely open box $b$ to reveal a goat. The probability is $1$.

So what's the probability that the car is in box $c$ given that he opened $b$?

Bayes' Theorem

P(AB)=P(BA)P(A)P(B)P(A|B)={P(B|A)P(A) \over P(B)}P(AB)=P(B)P(BA)P(A)

There are three possibilities for the denominator P(Openb)P(Open_b)P(Openb):

• Cara(OpenbCara)=1312=16Car_a (Open_b|Car_a)=\frac 1 3 \cdot \frac 1 2=\frac 1 6Cara(OpenbCara)=3121=61
• Carb(OpenbCarb)=130=0Car_b(Open_b|Car_b)=\frac 1 3 \cdot 0=0Carb(OpenbCarb)=310=0
• Carc(OpenbCarc)=131=13Car_c(Open_b|Car_c)=\frac 1 3 \cdot 1=\frac 1 3Carc(OpenbCarc)=311=31

so we'll add them up and plug them into our formula:

giving us a $2/3$ probability that the car is in the other box and therefore you should switch.