Locked in a dungeon, you are faced with five doors. One of them leads to freedom. The other four will lead you back to the starting room disoriented and confused, so that you will not remember which of the doors you have already tried and have to start again.
How many attempts do you expect to make on the average (statistically) before making it out?
Let's say that the average number of doors you have to go through is N. Since you will always end up in front of the same five doors after a failed attempt, your chance of making it out in each round are equal to 1/5th.
That is to say that there's a 0.8 chance that you'll be right back where you started, with one more attempt wasted, and still (on the average) N of them to go. Therefore
N = 1 + 0.8*N
0.2*N = 1
N = 5
You will average 5 attempts before making it out.
(This solution was submitted by Nick Reed. See comments for a more general solution.)
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