Why is there a 50 percent chance of finding the key in half the attempts in the brute force of the attack on the symmetric algorithm?

Question

In any cryptographic text, it is mentioned that with a brute force attack on a symmetric algorithm, there is a 50 percent chance of finding the key after half the attempt.

eg. A DES with a 56-bit key would have a 50% chance of finding the key after the first ^2,555 attempts.

Why is there a 50% chance of finding a key after half the attempts in a brutal attack against any symmetric encryption algorithm? What is this mathematical proof for?

Answer 1

N , , , .

( -: , , , .)

. 1/N, - . (N/2), (1/N) * (N/2), 1/2, 50%.

Answer 2

. , 100 , :

• : 1/100.
• : 2/100.
• ...
• : 50/100 (50%)