# Ask Uncle Colin: The Probability of Winning

Dear Uncle Colin,

Suppose Team 1 beats Team 2 by a score of 10-7, and Team 2 beats Team 3 by a score of 10-4. How would we predict the score of a match between Team 1 and Team 3?

- Make A Team Calculation Happen

Hi, MATCH, and thanks for your message!

I don’t have any information about what sport this is, but I’m going to assume that it’s a first-to-ten affair, and that the points are independent and identically distributed – just depending on the quality of the teams playing.

Our best estimate of the probability of Team 1 winning a given point against Team 2 is $p_{12} = \frac{10}{17}$ – they’ve won 10 of the 17 points played.

Similarly, we can say that $p_{23} = \frac{10}{14}$ (or $\frac{5}{7}$ if you prefer).

## A simple model

My favourite model for this sort of match is to say that the probability of Team 1 winning any given point against Team 3 is $\frac{p_{12}p_{23}}{p_{12}p_{23} + (1-p_{12})(1-p_{23})}$.

This has several benefits (not least of which is that $p_{13} + p_{31} = 1$), and can be justified as the probability of Team 1 beating Team 2 on a random point, and Team 2 beating Team 3 on a random point, given that Team 2 wins one point and loses the other.

So, in this example, $p_{13} = \frac{ \frac{10}{17} \times \frac{10}{14} }{ \frac{10}{17}\times \frac{10}{14} + \frac{7}{17} \times \frac{4}{14}}$. With a bit of work, that turns out to be $\frac{25}{32}$.

So, on average, Team 1 would score 10 points in every 12.8, so the most likely first-to-ten result would be somewhere around 10-3.

## But winning the match?

To find the probability of Team 1 winning the match, we could use a binomial distribution. “But wait!” – I hear the anguished cries – “doesn’t a binomial require a fixed number of trials?” Why yes, yes it does. We shall have 19 of them. If Team 1 wins at least 10 of the first 19 points, they’ve won the game.

So, let $X \sim B\left(19, \frac{25}{32}\right)$. We want $P(X \ge 10)$, which is $1 - P(X \le 9)$.

Turning to our trusty Classwiz, we find that is 0.9968 or so. Team 3 would be massive underdogs, winning about one game in 300.

Hope that helps!

- Uncle Colin