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?

(In reply to

Answer by K Sengupta)

Let the average number of doors that the individual will have to traverse be x.

Let A = probability of suceess in each attempt.

Then, it follows that:

x = (1-A)*x + 1

Now, the individual is faced with 5 doors, and so: A = 1/5

Thus,

x = 4x/5 + 1

or, x/5 = 1

or, x= 5

Consequently, the individual will have to make precisely 5 attempts on the average before making it out.

*Edited on ***September 17, 2008, 1:38 pm**