# Why isn't $\frac1{3p} =$ the probability that the strategy of always switching succeeds, given that Monty opens door 2?

1. Consider the Monty Hall problem, except that Monty enjoys opening door 2 more than he enjoys opening door 3, and if he has a choice between opening these two doors, he opens door 2 with probability p, where $$1/2 \le p \le 1$$.

To recap: there are three doors, behind one of which there is a car (which you want), and behind the other two of which there are goats (which you don't want). Initially, all possibilities are equally likely for where the car is. You choose a door, which for concreteness we assume is door 1. Monty Hall then opens a door to reveal a goat, and offers you the option of switching. Assume that Monty Hall knows which door has the car, will always open a goat door and offer the option of switching, and as above assume that if Monty Hall has a choice between opening door 2 and door 3, he chooses door 2 with probability p (with $$1/2 \le p \le 1$$).

(b) Find the probability that the strategy of always switching succeeds, given that Monty opens door 2.

Blitzstein, Introduction to Probability (2019 2 edn), Chapter 2, Exercise 40, p 91.

### My wrong approach

Let $$C_i$$ be the event that car is behind the door $$i$$, $$O_i$$ be the event that Monty opened door $$i$$ and $$X_i$$ be the event that intially I chose door $$i$$. Here $$i=1,2,3$$.
$$P(O_2|C_1, X_1) = p$$
$$P(O_2|C_2, X_1) = 0$$
$$P(O_2|C_3, X_1) = 1$$

Now, we are to find $$P(C_3|O_2, X_1)$$.
$$P(C_3|O_2, X_1) = \dfrac{P(O_2|C_3, X_1) . P(C_3|X_1)}{P(O_2|X_1)}$$ $$= \dfrac{1 . \frac{1}{3}}{p}$$ $$= \dfrac{1}{3p}$$

This is not the correct answer. Can anybody comment on what I am missing in my approach here.

## 1 Answer

Let $C_i$ be the event that car is behind the door $i$, $O_i$ be the event that Monty opened door $i$ and $X_i$ be the event that intially I chose door $i$. Here $i=1,2,3$.
$P(O_2|C_1, X_1) = p$
$P(O_2|C_2, X_1) = 0$
$P(O_2|C_3, X_1) = 1$

Now, we are to find $P(C_3|O_2, X_1)$.
$P(C_3|O_2, X_1) = \frac{P(O_2|C_3, X_1) . P(C_3|X_1)}{P(O_2|X_1)}$
$= \frac{P(O_2|C_3, X_1) . P(C_3|X_1)}{P(O_2|C_3, X_1) . P(C_3|X_1) + P(O_2|C_2, X_1) . P(C_2|X_1) + P(O_2|C_1, X_1) . P(C_1|X_1)}$

$= \frac{1 . \frac{1}{3}}{1.\frac{1}{3} + 0.\frac{1}{3} + p.\frac{1}{3}}$ $= \frac{1}{1+p}$

This is the correct answer.

• $\checkmark$ In short: $P(O_2\mid X_1) \neq p = P(O_2\mid C_1, X_1)$ . Rather, use the law of total probability. Commented Feb 28, 2018 at 23:58
• This also works if $0 \le p \lt \frac12$. And since $\frac{1}{2} \le \frac{1}{1+p} \le 1$ for all possible $p$ with $0 \le p \le 1$, it is still a good idea to always switch, no matter how biased Monty is and whether or not you know how biased he is. Commented Mar 2, 2018 at 16:51