A tournament among five teams is to be played as a single-round round-robin (each team plays each other team once).
Teams A, B< C and D are equivalent; each game among them has a 12\frac12 chance of each team winning. Team E is better -- they have a probability p>12p > \frac12 of winning each game. But in order to be happy with their results, team E has to win the tournament outright, that is, has to end with more wins than any other team.
How large does pp have to be form team E to have a 20%20\% chance of a happy outcome?
(Hint – if p=12p=\frac12 the chance of an outright win for one team in an all-teams-equal tournament is quite a bit below 20%20\%)
Yes, you can solve this by enumerting the 64 cases for E winning 3 out of four games. I was wondering if anybody has a feel for how much of an edge, in a round-robin tournament, a superiority of ϵ\epsilon gives you, and how likely is a tie for first.
BTW the answer is very close, but I believe the 60% team does have more than a 20% of a chance of outright winning — but a 59% team does not.