# Probability of a 5-way tie in a 6-team round robin

Rocky is staging a tiddlywinks tournament at which six evenly matched teams will play a round robin, each team playing each other team once. He will award the winner with the Bullwinkle memorial cup.

Boris points out that he'd better have 5 such cups made, in case there is a five-way tie for first place. Rocky says "How likely is that to happen"?

If our aeronautinc rodent hero comes prepared with only four cups, how likely is he to be embarrassed by a 5-way tie for first place?

(Assuming, of course, that Boris and Natasha do not rig the results.)

We approach via direct counting.

Our first observation (previously a comment above) is that in order for there to be a five-way tie it must be the case that one of the six players won zero games while each of the other five won three games.

This can be seen by noting that there are a total of $\binom{6}{2}=15$ games played, thus a total of fifteen wins need to be distributed among the players. Letting the winners have a total of $W$ wins and the loser has a total of $w$ wins, this implies that $5W+w=15$ and $5\geq W>w\geq 0$. The only integer solution to which is $W=3$ and $w=0$.

• Pick who the loser is: $6$ options.

From among the remaining five players, take the player who appears first in some predefined arbitrary order (e.g. earliest alphabetically, shortest, name first appearing on signup list,... I will refer to the person as being "smallest" from here out). We will call the smallest winner $x$.

• Pick which of the remaining four winners $x$ had beaten versus lost: $\binom{4}{2}=6$ options.

We now make the observation that among the two winners who beat $x$, one of them beat the other. By symmetry, we can see that there are exactly as many occurrences of the smaller of the two beating the larger of the two as vice versa.

• From the two winners who beat $x$, pick whether the larger beat the smaller or vice versa. Let the loser of that match be called $y$: $2$ options.

Now, we make the observation that as soon as we define one more outcome, everything else is strictly determined.

• Pick who the third person that $y$ had beaten was: $2$ options.

There are then $6\cdot 6\cdot 2\cdot 2 = 144$ total tournament outcomes which result in a five-way tie.

As there are $2^{\binom{6}{2}} = 2^{15}$ different equally likely tournament outcomes this implies that the probability of a five-way tie in a six-player roundrobin tournament where all teams are equally likely to win each match is:

$$\frac{144}{2^{15}}=\frac{9}{2048}\approx 0.0043945$$

• Looks right to me. And nice pretty pictures! – Mark Fischler Aug 24 '16 at 14:16