The Monty Hall Problem: You Really Need to Switch

“Marbles….” by kevinjay. is licensed under CC BY-NC-SA 2.0

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Feel free to skip next five paragraphs.

Description of the problem (straight out of Wikipedia)

Let’s reason with 100 marbles (no math!)

A lot of people that come across the Monty Hall problem have a hard time understanding why the chance of winning is not 50%. That’s the first thing they think of. I know I did. To see why it’s not, let’s not think about 3 doors. Let’s do 100 doors. Or actually, let’s think in terms of marbles.

Continuing with 100 marbles (with some math thrown in)

In the larger bag of marbles, any of the 99 remaining marbles could be the red marble. It’s a 99% chance that any of the marbles in the larger bag is red. That’s huuuge! On the other hand, there is only a 1% chance it’s in the smaller bag. If I, the host, take the larger bag and remove 98 marbles that are not red, I am making it easier for you to pick a red marble by removing the 98 that aren’t red.

If this is not a 50–50 chance scenario, what is then?

I will leave this part to you, but I will ask a question to get you started. Consider a scenario where there is initially only two doors, one with a goat and one with a car. You pick your door. I then add a third door, open it, and reveal a goat. Should you switch to the second unopened door? Does it matter?

If still not convinced:

  1. Simulate it yourself with code. Thinking through it with code might be the way you arrive at your own conclusion why the math makes sense.
  2. Play it with a friend! Take three index cards, each representing the three doors and what’s behind them (door number on one side and what is behind the door on the other side). Flip the cards to only show numbers and start playing. Play several times without switching and then several other times with switching. Compare probabilities of winning for the different cases. As you play, you will hopefully see why the math makes sense.

Statistician and data scientist based in Boston, MA

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