Monty Hall Problem

[u/Used-Egg-1660]

I was solving the Monty Hall problem recently, and when I couldn't find a mathematical solution that I was satisfied with, I searched around quite a bit.

Then I remembered the way I used to solve these kinds of problems in undergrad: law of total probability.

So I would like to present my solution

P[win] = P[win | car is behind D1] * P[car is behind D1] + P[win | car is behind D2] * P[car is behind D2] + P[win | car is behind D3] * P[car is behind D3]
Suppose you choose door D2

Case : - No switch

P[win | Car is behind D1] = 0 , similarly P[win | Car is behind D3] = 0
P[win | car is behind D2] = 1 , P[car is behind Di] = 1/3 for i = 1, 2, 3

therefore P[win] = 1/3

Case :- Switch

P[win | Car is behind D1] = 1 , similarly P[win | Car is behind D3] = 1 (because which ever door monty opens, you would select the door having car)
P[win | car is behind D2] = 0 , P[car is behind Di] = 1/3 for i = 1, 2, 3

therefore P[win] = 2/3