The Monty Hall problem in plain English

The following is based on the book ‘Das Ziegenproblem‘ by Gero von Randow.

Setup:

  • There are 3 doors.
  • 1 door has a car behind it.
  • 2 doors have a goat behind it.
  • The content behind the doors is randomly chosen.
  • There are 2 candidates, A and B.
  • There is 1 moderator who knows which door has a car behind it.

Actions:

  1. A and B are asked to choose a door and both choose the same door.
  2. The moderator chooses one door which has a goat behind it.
  3. A and B are asked if they would like to switch their choice and pick the remaining door.
  4. A always stays with his choice, the door that has been initially chosen by both A and B.
  5. B always changes her choice to the remaining third door.

Repeat the actions 999 times:

If you repeat the above list of actions 999 times, given the same setup, what will happen?

Candidate A always stays with his initial choice. Which means that he will on average win 1/3 of all games. He will win 1/3*999, 333 cars.

But who won the remaining 666 cars?

Given the setup of the game, the moderator has to choose a door with a goat behind it. Therefore the moderator does win 0 cars.

Candidate B, who always switched her choice, after the moderator picked a door with a goat behind it, must have won the remaining 666 cars (2/3*999)!

1 candidate and 100 doors:

Alter the above setup of the game in the following way

  • There are 100 doors.
  • 1 door has a car behind it.
  • 99 doors have a goat behind it.
  • There is 1 candidate, A.

Alter the above actions in the following way

  • The moderator opens 98 doors with goats behind them.

Now let’s say the candidate picks door number 8. By rule of the game the moderator now has to open 98 of the remaining 99 doors behind which there is no car.

Afterwards there is only one door left besides door 8 that the candidate has chosen.

You would probably switch your choice to the remaining door now. If so, the same should be the case with only 3 doors!

Further explanation:

Your chance of picking the car with your initial choice is 1/3 but your chance of choosing a door with a goat behind it, at the beginning, is 2/3. Thus on average, 2/3 of times that you are playing this game you’ll pick a goat at first go. That also means that 2/3 of times that you are playing this game, and by definition pick a goat, the moderator will have to pick the only remaining goat. Because given the laws of the game the moderator knows where the car is and is only allowed to open a door with a goat in it.

What does that mean?

On average, at first go, you pick a goat 2/3 of the time and hence the moderator is forced to pick the remaining goat 2/3 of the time. That means 2/3 of the time there is no goat left, only the car is left behind the remaining door. Therefore 2/3 of the time the remaining door has the car. Which makes switching the winning strategy.

Further reading

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