Question:
Eight teams compete in a tournament. Each pair of teams plays exactly
one game against each other. There are no ties. If the two possible
outcomes of each game are equally likely, what is the probability that
every team loses at least one game and wins at least one game?
I found this question on 2017 Fermat Competition of Waterloo University. The original solution seems a bit wordy and redundant. I wonder if there is a way to solve the probability in more simplified approaches? Original Solution (Page 8)
Best Answer
The probability that team${_k}$ loses its seven matches is $2^{-7}={1\over128}$. The probability that one of the eight teams loses its seven matches therefore is $8\cdot{1\over128}={1\over16}$, and similarly, the probability that one of the eight teams wins its seven matches is ${1\over16}$. The probability that at least one of these two events occurs is ${1\over16}+{1\over16}$ minus the probability that both occur at the same time. The latter is ${8\over128}\cdot{7\over64}={7\over1024}$: We can choose in $8$ ways the team${_k}$ that loses $7$ times and then in $7$ ways the team that wins against team${_k}$ and $6$ other teams.
It follows that the probability $p$ for having nothing of these is given by $$p=1-\left({1\over16}+{1\over16}-{7\over1024}\right)={903\over1024}\ .$$