Any cryptography text mentions that in brute force attack on Symmetric Algorithm there is 50 percent chance of finding the key after half of the attempt.
For e.g. DES with 56 bit key would have 50 percent chance of finding the key after first 2 to the power 55 attempts.
Why in a brute force attack against any symmetric encryption algorithm there is a 50 percent chance of finding the key after half of the attempts? What is the mathematical proof for it.
If there are N boxes in front of you, one of which contains a prize, then on average you only have to look in half the boxes before you find it.
(Look at it another way: you'd be spectacularly unlucky if there were a lot of boxes and the prize was in the last box you tried.)
Proof: The chance of the prize being in any particular box is 1/N, and the prize is in one and only one box. If you look in half the boxes (N/2), your chance of finding it is (1/N) * (N/2) which is 1/2, or 50%.