I recently came across the Monty Hall Problem. And while I get the gist of it, what I am having trouble understanding is this specific part of it.
Say I pick door 1. Then the probabilities for the following are as follows: $$P(P_1:S_2)=\frac16$$ $$P(P_1:S_3)=\frac16$$ $$P(P_2:S_3)=\frac13$$ $$P(P_3:S_2)=\frac13$$ Where $P_3:S_2$ is the probability that the prize is behind door number 3, AND Monty opens door 2. I don’t understand these numbers. If he opens door 3, and the prize is behind door 2, hasn’t our probability increased from being a third to a half?
You have to go by the hard probabilities. After door $3$ is shown, the probabilities $P(P_i:S_3)$ drop to zero, and the other probabilities $P(P_i:S_2)$ increase proportionally so as to sum to one. They do not equalise as you (and indeed many others who are shown the problem for the first time) think. So $P(P_3:S_2)$ is now $\frac23$.
I like to think about it this way: $\frac23$ of the time, your door contains a goat, and Monty's hand is forced. He has to open the door with the other goat, and switching wins you the prize. The other $\frac13$ of the time your door contains the prize, and switching will win you only a goat.