Yesterday, I bookmarked a Bayesian solution to the Monty Hall problem. The main drawback to that solution is that it requires you to go through some rather complex mechanics and recastings to get how it works. I think there's a simpler, plainly intuitive solution to be had.

But first, in a nutshell, the Monty Hall problem is this:

- You're given a chance to choose one out of three doors, behind one of which is a prize.
- After you make your choice, Monty reveals that one of the two doors you did not choose is incorrect.
- Monty gives you the option of sticking with your original choice or switching to the remaining unknown door that you did not originally choose.

The key intuition to getting the correct solution is this:

Your original choice was one out of three. Therefore, you had a two out of three chance of getting it wrong. Nothing that happens later changes this fact. Your original choice of door continues to have a two out of three chance of having been incorrect throughout the duration of the problem.

Now, when Monty tells you one of the doors you did not choose is incorrect, the door you originally chose still has a two out of three chance of having been incorrect. Given that there is now only one other door remaining, it must have a two out of three chance of being correct. You should switch when Monty gives you the chance.

I must confess that this is not how I would have originally tried to solve the problem. Like most people, I originally assumed that I was going from a one out of three to a one out of two equiprobable decision problem, meaning that both remaining doors had an equal chance of being correct. In other words, there was no reason to switch. The better intuition is to think that Monty has effectively combined the two doors I did not choose into one.

Recasting the problem into the following equivalent form makes this clear:

You may stick with the single door you originally chose or take both of the remaining two doors you did not.

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