How does P(Monty opens door 2) = P(Monty opens door 3), and $P(\text{get car}M_2)P(M_2) = P(\text{get car}M_3)P(M_3)$?
"Monty, who knows where the car is, then opens one of the two remaining doors. The door he opens always has a goat behind it (he never reveals the car!)." So Monty must open ONE of the $M_j (j = 2,3$), the one with the goat! But Monty mustn't and won't open the other $M_j$ with the car. So $P(M_2) \neq P(M_3)$!

How then $P(M_2) = P(M_3) = 1/2$ and $P(\text{get car}M_2)P(M_2) = P(\text{get car}M_3)P(M_3)$? Please see the sentences underlined in red and purple.

Please see the sentence underlined in green. How do you deduce the $2/3$ WITHOUT relying on the previous solution using the unconditional probability of success. Up to that sentence, you can deduce only that $P(\text{get car}) = \frac x2 + \frac x2 = x$.
There's a subtlety though, which is that when the contestant chooses whether to switch, she also knows which door Monty opened. We showed that the unconditional probability of success is 2/3 (when following the switching strategy), but let's also show that the conditional probability of success for switching, given the information that Monty provides, is also 2/3.
Let $M_j$ be the event that Monty opens door j, for j = 2; 3. Then
$P(\text{get car}) = P(\text{get car}M_2)P(M_2) + P(\text{get car}M_3)P(M_3)$;
Blitzstein. Introduction to Probability (2019 2 ed). pp 6970.
0 comment threads