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

  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 $1/2 \le p \le 1$).

(a) Find the unconditional probability that the strategy of always switching succeeds (unconditional in the sense that we do not condition on which of doors 2 or 3 Monty opens).

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

Solution:

(a) Let $C_j$ be the event that the car is hidden behind door j and let W be the event that we win using the switching strategy. Using the law of total probability, we can find the unconditional probability of winning: $P(W) = P(W \mid C_1)P(C_1) + P(W \mid C_2)P(C_2) + P(W \mid C_3)P(C_3) = 0 \cdot 1/3 + 1 \cdot 1/3 + 1 \cdot 1/3 = 2/3.$

(b) A tree method works well here (delete the paths which are no longer relevant after the conditioning, and reweight the remaining values by dividing by their sum), or we can use Bayes' rule and the law of total probability (as below).

Let $D_i$ be the event that Monty opens Door i. Note that we are looking for $P(W|D_2)$, which is the same as $P(C_3 \mid D_2)$ as we first choose Door 1 and then switch to Door 3. By Bayes' rule and the law of total probability,

$P(C3 \mid D2) = \dfrac{P(D_2 \mid C_3)P(C_3)}{P(D_2)} = \dfrac{P(D_2 \mid C_3)P(C_3)}{P(D_2 \mid C_1)P(C_1) + P(D_2 \mid C_2)P(C_2) + P(D_2 \mid C_3)P(C_3)} = \dfrac{1 \cdot 1/3}{p \cdot 1/3 + 0 \cdot 1/3 + 1 \cdot 1/3} = \dfrac{1}{ 1 + p}.$

I am trying to pinpoint where my thinking and solution went wrong. Let's define for notation the following events:

$D_i$ is the event that the car is behind door $i$.

$O_i$ is the event that Monty opens door $i$ after the first choice.

$S$ denotes the event of success i.e. correctly pick a car.

I arrived at the fact that the conditional probability of success $P(S|O_2) = P(D_3|O_2)$.

I understand this as being because $P(S|O_2,D_1)=0$, $P(O_2|D_2)=0$ and $P(S|O_2,D_3)=1$. So this is what you end up with when expand over possible doors using the law of total probability.

However, I'm not clear on why $P(D_3|O_2) \neq \frac{1}{2}$. In the provided solutions to the problem it shows that the correct thing to do is to expand this out using Bayes theorem, and this makes sense to me. What doesn't make sense to me is why the answer doesn't come out as .5. Also, I'm unsure about whether the entire calculation be conditioned on the first choice?

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If $O_2$ happens, then either $D_1$ or $D_3$. $D_1$ and $D_3$ are equally likely but in $D_1$, Monty had a choice, he could have picked $O_2$ or $O_3$ but did pick $O_2$. In $D_3$ Monty didn't have a choice, he always has to pick $O_2$.

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  • $\begingroup$ ah yes thank you! $\endgroup$ Aug 16, 2019 at 14:24

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