Ok, that was a long time to think about the problem from May 5th. You haven't read it yet? Stop reading and go back to the post and make up your mind before reading the solution.
Funny enough, the chances of winning the game are 2/3 overall. That said, it's obvious that you have to switch doors to arrive at this high probability. Because obviously, if you chose one of the three doors, and stay with the choice even after the elimination of an empty door, your chances is and always will be only 1/3.
For myself, I came up with the following explanation, which I then afterwards also found on the web: imagine you play the game three times. Everyone would agree that if you don't switch, chances are that you win the game 1 time, at least on average (or say you play it 3 million times, the number of winning games will be pretty close to 1 million).
Put differently, if your are right 1 time out of three, you are wrong 2 times out of three. This means on average your switch will make win in two times out of three and lose only one time. Straightforward, isnt' it?
What I liked about this problem is that it makes you think quite a while, at least me, even after coming up with a reasonable solution and argumentation. If you want to read more about it check out the Ziegenproblem on Wikipedia Germany or the Monty Hall Problem on Wikipedia international.
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