**Standard notation:**

probability of A given B is written \( \Pr(A \mid B) \)

probability of B is written \( \Pr(B) \)

**Bayes' Theorem:**

Using the notation above, Bayes' Theorem can be written: \[ \Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)} \]Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty shows us a goat behind Door 2. Now let A be the event that the car is behind Door 1 and B be the event that Monty shows us a goat behind Door 2. Then

\begin{aligned}

\Pr (A \mid B) &= \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)} \\

&= \frac{1/2\times 1/3}{1/3\times 1/2+1/3\times 0+1/3\times 1} \\

&= 1/3.

\end{aligned}The tricky calculation is \( \Pr(B) \). Remember, we are assuming we initially chose Door 1. It follows that if the car is behind Door 1, Monty will show us a goat behind Door 2 half the time. If the car is behind Door 2, Monty never shows us a goat behind Door 2. Finally, if the car is behind Door 3, Monty shows us a goat behind Door 2 every time. Thus, \[ \Pr(B) = 1/3\times 1/2+1/3\times 0+1/3\times 1 = 1/2. \]The car is either behind Door 1 or Door 3, and since the probability that it's behind Door 1 is 1/3 and the sum of the two probabilities must equal 1, the probability the car is behind Door 3 is \( 1-1/3 = 2/3 \). You could also apply Bayes' Theorem directly, but this is simpler.

So Bayes says we should switch, as our probability of winning the car jumps from 1/3 to 2/3.