Thanks for that extra 33.3% Monty. I will switch ….
I read and heard different explanations of Monty Hall problem. I liked Udacity‘s explanation using Bayes theorem. This is my notes for it.
So, what is Monty Hall problem? There are three doors…. I will let Kevin Spacey explain it.
The prior probability of door1, door2 and door3 having a car is one-third. Now we choose Door1 and Monty open Door3, and you see a goat.
Now the question is do I need to switch. Let us calculate the probability of switching to door2 given that Monty has opened the door3 and importantly “I have chosen door1” using Bayes theorem.
The likelihood that Monty would open door3 given the car is in door2 is 1. Monty cannot open door1 because we already choose it. Given the evidence that Monty has chosen door3 the likelihood door3 is open given the car is in door2 is 1.
We already know the prior probability the car is door2 is 1/3. Let us see how to calculate the marginal.
It is straightforward if we remember that even though Monty opens door2 still the car can be in door2 or door1.
Let us plug it into Bayes theorem
Yes, the final posterior probability is 2/3. So it is always good to switch to improve your chances. Thanks to Monty now the probability for switching has increased by 1/3.
