How Many Outcomes Are Possible in a Three-Team Round Robin?

If three teams each play the other two teams once, the schedule contains three games. If each game can end in either team winning or in a draw, there are 27 possible combinations of game results.

The count treats the games as distinct: a result for Team A versus Team B is recorded separately from a result for Team A versus Team C. It counts complete records of the three games, not just the possible tournament winners or final rankings.

Count the games and outcomes

Call the teams A, B, and C. In a round robin where each pair meets once, the matchups are A–B, A–C, and B–C. The number of matchups is the number of ways to choose two teams from three:

$$\binom{3}{2}=3.$$

For any one matchup, there are three possible results: the first team wins, the second team wins, or the game is drawn. The multiplication rule says to multiply the number of choices for each distinct game:

$$3 \times 3 \times 3=3^3=27.$$

So there are 27 possible outcome records for the round robin.

A worked example

Fix the game order as A–B, A–C, and B–C. One possible record is:

• A beats B;
• C beats A;
• B and C draw.

Another record could have A and B draw, A beat C, and C beat B. These are different records because at least one specific matchup has a different result.

You can describe any complete record as a three-entry list, with one entry for each fixed matchup. Each entry has three choices, so there are $3^3$ such lists. The order in which you write the entries is just a way to label the games; it does not create extra outcomes once the matchups are fixed.

Check by counting the number of draws

A useful check is to group the records by how many games are drawn. For each game that is not drawn, either of its two teams can win.

Number of draws	Count of records	Reason
0	8	Each of 3 games has 2 possible winners: $2^3$
1	12	Choose the drawn game in 3 ways; the other games have $2^2$ winner choices
2	6	Choose the 2 drawn games in 3 ways; the remaining game has 2 winners
3	1	All three games are drawn

Adding the groups gives $8+12+6+1=27$, agreeing with the multiplication-rule calculation.

When does this count give probabilities?

The number 27 is a count of possibilities. It does not mean every record is equally likely in real competition. If you assume that each game independently has three equally likely results, then each complete record has probability $1/27$. In actual games, the teams may have different strengths, and draws may be more or less likely than wins, so equal probabilities are often an unrealistic model.

For example, under that equal-and-independent assumption, the probability that all three games are drawn is $(1/3)^3=1/27$. Without the assumption, the number of possible records is still 27, but calculating event probabilities requires a joint probability model; if games are assumed independent, the probabilities for individual game results suffice.

Common mistakes

• Counting only two results per game. If draws are allowed, each matchup has three results, not just a win for one team or the other.
• Counting only two games. There are three pairings among three teams: A–B, A–C, and B–C.
• Confusing outcome records with standings. Different game records can lead to the same points totals or ranking. The count 27 is not a count of distinct standings or champions.
• Treating the games as indistinguishable. The result of A–B is attached to that matchup. Swapping it with the result of A–C generally gives a different record.

If the tournament uses a different schedule, count the number of games first. With $g$ specified games and three possible outcomes in each, there are $3^g$ outcome records. If some games have a different number of possible outcomes, multiply the number of choices for each game instead.