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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?

RobPratt
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1 Answers1

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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.

Parcly Taxel
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  • Sorry, I don't understand your first sentence. "After door 3 is shown, the probabilities (:2) drop to zero, and the other probabilities (:3) increase proportionally so as to sum to one". What is Pi? also, I am showing door 2, right? Not door 3? –  Apr 10 '20 at 13:40
  • @Oishika $P_i$ simply means $P_1$, $P_2$ and $P_3$ comvined. – Parcly Taxel Apr 10 '20 at 14:19