The Monty Hall Problem, Enumerated

The Monty Hall problem is as follows:

You are a contestant in a game show. There are three doors. Behind one of the doors is a car; the other doors, each, a goat. The host asks you to choose a door. After you do so, the host will open one of the other two doors to reveal a goat, and allow you to stick to your choice, or to switch to the other unopened door. Which gives the higher probability to win the car?

It seems that there are a few ways to arrive at the answer, but none of them makes sense to me. In the end I got it by enumerating the only two winning possibilities:

  • Winning possibility #1 (WP1): I choose the car; the host shows me one of the two goats; I stick to my choice.
  • Winning possibility #2 (WP2): I choose a goat; the host shows me the remaining goat; I switch to the car.

For WP1, it is very clear that I have a 1/3 chance of winning the car by sticking to my first choice.

For WP2, I need to choose a goat, and then switch. The probability of choosing a goat is 2/3. Therefore, it is better for me to switch, because I have a higher chance of initially choosing a goat than the car. In other words, to ensure that switching is the winning move, I need to have initially chosen one of the goats; I have a 2/3 chance of initially choosing a goat; therefore I’m better off assuming that I had chosen a goat—which has a higher chance of being true—and then switch to what could only be the car.

Goat it?

15 July 2013 | Uncategorized | Comments

One Response to “The Monty Hall Problem, Enumerated”

  1. 1 Wild guess 13 September 2013 @ 10:02 pm

    Its 50:50 mah. So u cud switch or not n it will still be 50:50.