Prob. of finishing in 3: 16807/32768 = (7/8)^5
Expected number of rounds: 2470/651
To calculate this, first we need the probability of fininshing in n rounds.
First, we find the following, similar to the other part of the problem:
P(1) = (1/2)^5
P(2) + P(1) = (3/4)^5
P(3) + P(2) + P(1) = (7/8)^5
P(4) + P(3) + P(2) + P(1) = (15/16)^5
P(5) + P(4) + P(3) + P(2) + P(1) = (31/32)^5
P(n) + P(n-1) + . . . + P(2) + P(1) = (1 - 1/2^n)^5
From that:
P(1) = (1/2)^5
P(2) = (3/4)^5 - (1/2)^5
P(3) = (7/8)^5 - (3/4)^5
P(4) = (15/16)^5 - (7/8)^5
P(5) = (31/32)^5 - (15/16)^5
P(n) = (1 - 1/2^n)^5 - (1 - 1/2^(n-1))^5
The expected number of turns can be calculated from
E = P(1) + 2*P(2) + 3*P(3) + 4*P(4) + 5*P(5) + . . .
The expression for E approaches 2470/651 as a limit. |