# monty hall

## 1. assumptions

- you have a 1/3 chance at picking the winning door immediately
- among the losing doors, the host will select uniformly at random

## 2. intuition one:

- Consider the never-switch strategy. This has a 1/3 chance of succeeding.
- Now consider the strategy where you wait for the revealed goat and then randomly select among the two remaining doors – that has a 1/2 chance of succeeding
- now consider the strategy where you always switch. 1/3 of the time you would have been on the correct door and made the switch incorrectly. 2/3 of the time you would have been on the incorrect door and made the switch correctly

## 3. intuition two:

\[\begin{align} P(win(1) \mid reveal(2), choose(1)) & = \frac{P(reveal(2)\mid win(1),choose(1))P(win(1) \mid choose(1))}{P(reveal(2) \mid choose(1))}\\ & = \frac{P(reveal(2)\mid win(1),choose(1))P(win(1) \mid choose(1))}{P(reveal(2)\mid win(1),choose(1))P(win(1) \mid choose(1)) + P(reveal(2)\mid \neg win(1),choose(1))P(\neg win(1) \mid choose(1))}\\ & = \frac{\frac{1}{2} \cdot \frac{1}{3}}{\frac{1}{2}\cdot \frac{1}{3} + 1\cdot\frac{2}{3}}\\ & = \frac{1}{3} \end{align} \] where

- \(win(1)\) is the event of the car being behind door 1
- \(reveal(2)\) is the event of door 2 being revealed
- \(choose(1)\) is the event of me choosing door 1
- \(P(reveal(2)\mid win(1),choose(1)) = \frac{1}{2}\) because if the winning door is 1 and I chose 1, the host picks uniformly at random between the remaining doors
- \(P(reveal(2)\mid \neg win(1),choose(1)) = \frac{1}{2}\) because if the winning door is not 1, then 1/2 the time the winning door will be door 2 and 1/2 the time, the winning door will be door 3
- And of course \(P(win(2) \mid reveal(2)) = 0\), so \(P(win(3) \mid reveal(2), choose(1)) = \frac{2}{3}\)