Help, I can't get any of these parts.
Eight players enter a knock-out singles tennis tournament in which each of the
four first round winners plays one second round game to decide who enters the final.
(a) Assuming that all players are equally likely to win a game, show that the probability
that two particular entrants play each other in the tournament is 1/4.
(b) Also show that if sixteen persons enter the tournament, then the probability that
the two players meet is 1/8 .
(c) Prove that for a similar knock-out tournament for 2^n players, the probability that
two players meet is 2^{1−n}.
for part (a)
my solution is:
prob meeting in first game = 1/[8C2 x 6C2 x 4C2) = 1/2520
prob meeting in second game = (1/2)^2 (1/[{4C2)] = 1/24
prob meeting in third game = (1/2)^2 = 1/4
total = 92/315
this isn't the answer
without part (a), I can't do part (b) or (c)
Thank you