Explaining the Monty Hall problem

Question

Explaining the rules of the game:

There are three doors. Only one contains the prize. You choose one door first, then Monty Hall opens one of the other two doors that doesn't have the prize. If there are two doors to pick from, he'd choose at random. If you choose a door, then Monty Hall opens another door, should you change your choice to the other door?

Let's call the doors are 1, 2 and 3. $C_1$ means the prize is behind door 1 and $O_1$ means door 1 is opened. So $P(C_2|O_3)$ would be the probability that the car is behind door 2 given that door 3 is opened while $P(O_3|C_2)$ would be the probability that Monty hall opens door 3 given that the prize is behind door 2.

We assume we pick door 1 first. Then Monty Hall opens door 3. So we're to find the probability that the prize is behind door 2 given that Monty Hall opens door 3.

Using Bayes theorem: $$P(C_2|O_3) = \frac{P(O_3|C_2)\cdot P(C_2)}{P(O_3)}$$ $$=\frac{P(O_3|C_2)\cdot P(C_2)}{P(O_3|C_1)\cdot P(C_1) + P(O_3|C_2)\cdot P(C_2) + P(O_3|C_3)\cdot P(C_3)}$$

There is an equal chance of the prize being behind every door so $P(C_1) = P(C_2) = P(C_3) = \frac{1}{3}$

$P(O_3|C_1)$: The car is behind door 1 and we've picked door 1. Therefore Monty Hall has doors 2 and 3 to choose from. $P(O_3|C_1) = \frac{1}{2}$

$P(O_3|C_2)$: The car is behind door 2 and we've picked door 1. Therefore Monty Hall only has door 1 to pick. $P(O_3|C_2) = 1$

$P(O_3|C_3)$: The car is behind door 3 so Monty Hall cannot open door 3 so this would be equal to 0.

$$P(C_2|O_3) = \frac{1\cdot \frac{1}{3}}{\frac{1}{2} \cdot \frac{1}{3} + 1 \cdot \frac{1}{3} + 0 \cdot \frac{1}{3}} = \frac{2}{3}$$

There's a higher chance that the car is behind the other door so it's a better choice to switch.

I'm not sure if I'm allowed to do this. Just wanted to answer the question since I didn't find any solutions that fully explained it.

Answer 1

Monty hall is only confusing because n=3. Here's an easy way to conceptualize the solution.

Imagine God picks 1 atom in the entire universe and asks you to guess it. You make your pick, then he tells you that the atom he was thinking of is either the one you picked, or atom #897432789489723. Which atom is more likely to be the one he picked? Obviously #897432789489723 because the chance you initially picked his atom is so infinitesimally small.

Answer 2

I am not sure the solution is correct. It assumes that we are in the universe where a player opened door 1 and there is no car behind the door. At the same time, it uses $P(C_1)=1/3$

Answer 3

Think of it this way: if you switch, the only way you can lose is if you initially pick the door with the prize and then switch away from it. The probability of initially picking the door with the prize is $\frac{1}{3}$. But, you win if you initially pick one of the wrong doors and then switch to the door with the prize after the other wrong door has been revealed. The probability of initially picking a wrong door is $\frac{2}{3}$.