Players are of equal skill, and in a contest the probability is 0.5
that a specified one of the two contestants will be the victor. a
group of 2^n players is paired off against another at random. the
2^(n-1) winners are again paired off randomly, and so on, until a
single winner remains. consider two specified contestants, a and b,
and define the events a(i), i< =n, and e by:
a(i): a plays in exactly i contests
e: a and b ever play each other
a) find p(a( i = 0, n
b) find p(e)
c) let p(n) = p(e). show that:
p(n) = 1/(2^n - 1) + (2^n - 2)/(2^n - 1) (1/4) p(n-1)
d) explain why a total of 2^n-1 games are played. number these games and let b(i) denote the event that a and b play each other in game i, i = 1, 2^n-1
e) what is p(b(
f) use part (e) to find p(e).