Humans And Probabilities - the Three-Goats Problem

Imagine that you are a candidate in a TV show. You have the chance to win a car. The moderator offers you three different doors. Behind one of these doors there is the car. You point to one of the doors, but the moderator ignores that and opens one of the two remaining doors. There is no car behind the opened door.

Now the moderator offers you the possibility to choose one of the remaining doors. What is the best strategy now?

Note that you have three different possibilities:

  1. You can stay with your original choice
  2. You can throw a dice
  3. You can change your mind and choose the other one

Is the decision going to have any impact on the final result at all? If so, why?

When I was asked about this, I thought in first place that there should be no difference between these choices: two doors, one chance, 50%, that’s it.

But I was told that

  1. persisting on your first choice means that you have a 1:3 or 33% chance only.
  2. throwing a dice increases your chance to 1:2 or 50%.
  3. choosing the other door increases your chance to 2:1 or 66%!

I could not believe this. Can you?

>> The Solution

Author: KaiJaeger


Probabilities/3goats/Task (last edited 2010-03-27 13:04:44 by KaiJaeger)