A Monty Hall Probability Simulation

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There are three doors. And hidden behind them are two goats and a car. Your objective is to win the car. Here’s what you do:

  • Pick a door.
  • The host opens one of the doors you didn’t pick that has a goat behind it.
  • Now there are just two doors to choose from.
  • Do you stay with your original choice or switch to the other door?
  • What’s the probability you get the car if you stay?
  • What’s the probability you get the car if you switch?

It’s not a 50/50 choice. I won’t digress into the math behind it, but instead let you play with the simulator below. The game will tally up how many times you win and lose based on your choice.

What’s going on here? Marilyn vos Savant wrote the solution to this game in 1990. You can read vos Savant’s explanations and some of the ignorant responses. But in short, because the door that’s opened is not opened randomly, the host gives you additional information about the set of doors you didn’t choose. Effectively, if you switch, you are select all the other doors. If you choose to stay, you are select just one door.

In her answer, she suggests:

Here’s a good way to visualize what happened. Suppose there are a million doors, and you pick door #1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?

To illustrate that in the simulation, you can increase number of number of doors in the simulator. It becomes pretty clear that switch is the correct choice.

Finally, here’s some Kevin Spacey:

@seandolinar