Win Distributions

What is a Win Distribution?

A win distribution looks at every possible schedules and counts the records from each schedule. (eg. In how many schedules did Team 1 go 0-14. In how many schedules did Team 1 go 1-13, etc.)

With over $10^{56}$ possible schedules in most league formats, the counts are extremely large, so the counts are converted to a percentage by dividing by the total number of possible schedules.

Example Win Distribution (Full Season)

	0-14	1-13	2-12	3-11	4-10	5-9	6-8	7-7	8-6	9-5	10-4	11-3	12-2	13-1	14-0
Win Percentage (%)	0.030	0.560	3.510	11.05	20.75	25.25	20.79	11.82	4.67	1.27	0.230	0.020	0.010	0.000	0.000

In this example win distribution, in $11.05\%$ of schedules, the team's record was 3-11

Creating a Win Distribution

One Week

The simplest win distribution is one week. For a given team, $T_1$, and week, $W_1$, count how many teams they would have beat if they played them, and how many teams they would have lost to.

For example, in a 6 team league, if $T_1$ would have won against 3 teams and lost against the other 2 teams, then the percentage of schedules that $T_1$ was 1-0 is $\frac{3}{5} = 60 \%$ and the percentage of schedules that $T_1$ was 0-1 is $\frac{2}{5} = 40 \%$.

	0-1	1-0
Win Percentage (%)	40	60

tip

In a 6 team league, there are 5 other teams to play against, since you cannot play against yourself.

Adding Another Week

To add another week, $W_2$, follow the same process as above to determine how teams $T_1$ would have won or lost against in $W_2$.

tip

When adding another week, the number of schedules increases exponentially. In this example, there is 25 schedules that we need to determine the outcome of.

For example, let's say $T_1$ would have won 4 games and lost 1 games in $W_2$. To find the percentage of schedules that $T_1$ went 2-0, we start by looking at the $\frac{3}{5}$ schedules where $T_1$ won $W_1$. Each of the 3 winning schedules from $W_1$ will have 4 schedules where $T_1$ also won in $W_2$.

$$\frac{3*4}{5*5} = \frac{12}{25} = 48 \%$$

The percentage of schedules where $T_1$ was 1-1 is the percentage of schedules where $T_1$ won $W_1$ and lost $W_2$ plus the percentage of schedules where $T_1$ lost $W_1$ and won $W_2$.

$$\frac{2*4}{5*5} + \frac{3*1}{5*5} = \frac{8}{25} + \frac{3}{25} = \frac{11}{25} = 44 \%$$

Finally, calculating the percentage of schedules where $T_1$ went 0-2,

$$\frac{2*1}{5*5} = \frac{2}{25} = 8 \%$$

	0-2	1-1	2-0
Win Percentage (%)	8	44	48

To add more weeks, continue this process with additional weeks!

tip

Try creating scores for each of the 6 teams for 2 weeks. Then create all 25 possible matchups for one team count the number of 0-2, 1-1, 2-0 records.

Using Percentages Instead of Counts

Since the counts can get massive (over $10^{50}$) there can be problems with integer overflow. To avoid this problem, percentages are used instead.

Looking at the calulation for the percentage of schedules that $T_1$ went 1-1, the fractions can be split up into percentages.

$$\frac{2*4}{5*5} + \frac{3*1}{5*5} = (\frac{2}{5})(\frac{4}{5}) + (\frac{3}{5})(\frac{1}{5}) = (0.4)(0.8) + (0.6)(0.2) = 44 \%$$

This can be be generalized. For:

• $N$ = Number of wins
• $K$ = Number of weeks
• $R_{N}^{K}$ = Percentage of records with $N$ wins after $K$ weeks
• $P_{K}'$ = Percentage of wins in week $K$
• $P_{K}$ = Percentage of losses in week $K$

$$R_N^K = (R_{N - 1}^{K - 1})(P_{K}') + (R_{N}^{K - 1})(P_{K})$$

Edge Cases

Ties

In most scoring formats, ties are extremely rare. Due to their rarity and to help simplify the math, winners of a tied game is chosen with a coin flip.

Losses

If ties are ignored, then losses can be inferred by the number of wins a team has and how many weeks have been played in the season.

$$Losses = WeeksPlayed - Wins$$

Divisions and Playing Teams Equally

Our win distributions look at every possible schedule, including schedules where a Team 1 plays Team 2 every single week of the season. The math is significantly more complex to account for divisions / playing teams an equal amount of times and simulating schedules with code would take a very long time.

Rounding and Floating Point Math

Due to rounding and/or very small accuracy issues from floating point arithmetic, all numbers in a win distribution may not perfectly sum up to $100\%$.