Monty Hall Problem - Strategy that maximizes chances of winning the prize

Question

On a game show, there are three doors, behind one of which is a prize. I choose a door and the host opens one of the other doors that has no prize behind it. I get to switch my door choice if I wish.

Now suppose we have three positive numbers $p_1$, $p_2$, $p_3$ such that $p_1+p_2+p_3=1$ and the prize is behind door $i$ with probability $p_i$. By labeling the doors suitably we can assume $p_1>p_2>p_3$. Assume that you know the probabilities $p_1$, $p_2$, $p_3$ associated to each door. What is the strategy that maximizes my chances of winning the prize?

Answer 1

The usual form of the "Monty Hall" question asks for the probabilities of the car being behind each of the doors, conditioned on the fact that the contestant initially chose a particular door and host opened a particular door. Those probabilities cannot be computed unless we know (or make some assumptions about) the probability that the host will open a particular door when two doors are eligible to be opened.

Since the question asked here does not say that any doors have been opened yet, I take it that the question is only about the strategy that gives the best chance to win under the condition that the game is about to start, and that the strategy can specify both how to pick the door to choose initially and how to react (switch or stay) depending on which door the host opens.

Start by choosing door 3. After the host opens a door, always switch.

Answer 2

The usual interpretation on this problem is that Monty knows where the car is.

You pick a door. The probability that the car is behind it is $\dfrac{1}{3}$.

The probability that it is behind one of the other two is $\dfrac{2}{3}$. Since there are two goats and Monty knows where the car is he will always show you a goat. The probability that the car is behind the remaining door is $\dfrac{2}{3}$. You are twice as likely to get the car if you swap.