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

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?

• When Monty has a choice between two doors to open, how does he decide which one to open? The answer to your question depends on the answer to mine. – Gerry Myerson Feb 7 '19 at 10:13
• I assume he won't open the door with the prize. – Teta K Feb 7 '19 at 10:29
• I think, the best strategy depends on the concrete probabilities. If $p_3$ is very small, for instance, a very good chance is to choose that door and switch. If $p_1$ is very large, a good idea (not sure whether always the best idea) is to choose number $2$ and to switch only if $3$ is opened. – Peter Feb 7 '19 at 10:45
• This question is well known and the solution while counter intuitive to most has been covered here on this site: What's wrong with this equal probability solution for Monty Hall Problem? and many other places on the internet. – Warren Hill Feb 7 '19 at 12:33
• Possible duplicate of What's wrong with this equal probability solution for Monty Hall Problem? – Warren Hill Feb 7 '19 at 12:33

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.