Eric F. Savage

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Monty Hall Problem

A classic brain teaser is the Monty Hall Problem. It fools people with a basic understanding of probability (i.e. most of us), and even after a good amount of consternation seems unintuitive. If complex mathematical formulas don’t convince you, then you’ll likely have to see it actually happen to believe it, as I did.

I came across it again today, and I was unhappy with the idea that this remained a difficult hurdle for my brain to jump, even though at this point I know the answer.  I came up with the following explanation that I think is as simple as it gets, so if it still sticks in your craw as it did mine, perhaps this will help.

If you never switch:

  • 1/3 of time you pick right door, and win.
  • 2/3 of time you pick wrong door, and lose.

(1/3 * 1) + (2/3 * 0) = 1/3

So if you never switch, you will win 1/3 of the time.

If you always switch:

  • 1/3 of time you pick right door, Monty will open either of the empty doors, and then you will open the other empty door, and lose.
  • 2/3 of time you pick an empty door, Monty will open the other empty door, and then you will open the door with the prize and win.

(1/3 * 0) + (2/3 * 1) = 2/3

If you always switch, you will win 2/3 of the time.

Diagrams and formulas aside, the phrase “the other empty door” is what made it come together for me.

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