# Probability of a round-robin tournament being tied

Consider a round-robin tournament played by 2n teams:

• each team plays each other team once. (total of n(2n-1) games)
• each game has a result (win/loss, no ties/draws possible)
• all the teams are perfectly equal, so each team has a 50/50 chance of winning each game
• the teams are ranked by number of wins, and the top n teams go through to the next stage.

What is the probability that the teams in ‘n’th and ‘n+1’th place are tied, such that some second metric is needed? (For instance, in cricket this might be net run rate, or in soccer, net goal difference.)

When n=1, a tie is impossible.

When n=2, there are 6 games played, and hence 2^6 = 64 different possible outcomes. There are two ways the 2nd and 3rd placed teams could have the same number of wins:

Win distribution 2-2-2-0. There are 8 outcomes that result in this distribution (4 teams which could place last, times 2 different ways for the other three teams to win one and lose one against each other).

Win distribution 3-1-1-1. By analogous reasoning, there are 8 outcomes.

Total: 16 out of 64 → ¼ chance

When n=3, the situation is considerably more complicated. The following win-distributions involve a 3-win tie between 3rd and 4th placed teams: 5-3-3-3-1-0, 4-4-3-3-1-0, 4-3-3-3-2-0, 4-3-3-3-1-1, 3-3-3-3-2-1, 3-3-3-3-3-0, Each of these has an analogous situation which results in a 2-win tie (simply change each of the 15 games’ outcomes). However, counting up the number of different ways to achieve each of these win distributions is daunting. All the more so for n>3.

So the question is: what is the general trend as n increases? Is there a good probabilistic approximation?

NB It seems to me that the probability of a tie should increase as n rises.

• Do you mean the probability of two particular teams being tied ? – true blue anil Apr 22 '18 at 9:24
• No, I mean any two teams being tied across the nth and n+1th place. – Thomas Delaney Apr 22 '18 at 9:59