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Imagine we have 2 teams playing against each other and that we know each of their overall win rates (wr). As an example lets say team A has a wr of 60% whilst team B has a wr of 55%. Is it possible to calculate the probability of team A beating team B in a game?

My initial intuition is just to normalize it i.e. $p_{ab} = r_a / (r_a + r_b)$ where $p_{ij}$ is the probability of team $i$ beating team $j$ and $r_i$ is the win rate of team $i$.

However its easy to show this doesn't work by example:

Imagine we have 3 teams A,B,C and that they all play each other the same amount. Now lets say that team A has a 100% wr whilst B and C are both equally matched and therefore each have a 25% wr (as they win 50% of their games against each other but lose all of their games against team A).

Under the above formula we would get $p_{ab} = 1 / (1 + 0.25) = 0.8$ which doesn't seem right as we know team A win all of their games.

So yer any help on how to handle this would be greatly appreciated.

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There is no ‘correct formula’. The chances that team A beat team B in particular match cannot be deduced from the long-run win rates of teams A and B.

Just think of all the factors that influence any sporting contest, and the different predictions and expectations made by fans and pundits. The information that we have in real life about two teams - which is a lot more than their win rates - isn’t enough to bring us all into agreement about the relative plausibilities of different outcomes.

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Wrong formula

The formula for $p_{a,b}$ you give does not answer your question. I'll first show you the "meaningful" formula which is closest to the one you gave.

Note that $r_i$ is equal to $P(\mathrm{Win}\mid \mathrm{Team}\;i)$. If you have access to the number of match each team has played $n_i$, then you can compute $P(\mathrm{Team}\;i) = n_i/n = p(i)$ where $n=\sum_i n_i$ is the total number of match played.

Once you have this you can compute $P(\mathrm{Win}) = \sum_i p(i)\,r_i$ and deduce using bayes' rule the probability $$ P(\mathrm{Team}\;i \mid \mathrm{Win}) = \dfrac{P(\mathrm{Team}\;i)P(\mathrm{Win}\mid \mathrm{Team}\;i)}{P(\mathrm{Win})} = \dfrac{p(i)r_i}{\sum_i p(i)\,r_i}$$ This is similar to your formula $r_i / \sum_i r_i$ but you weight each $r_i$ by the probability that team $i$ plays.

This formula answers the question:

Some team just won, what is the probability that team number $i$ was the winner ?

However this does not answer your question, which is rather:

Given teams $i$ and $j$ play each other, what is the probability that each team wins ?

Correct formula

You'll need more data. You need to break down $r_i$ in terms of $r_{ij}$, a quantity that tells you the winrate of $i$ when playing against $j$. If you have this data your question is directly answered because $$r_{ij} = P(\mathrm{Team}\;i\;\mathrm{wins}\mid i,j)$$

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I only see an easy solution if you assume that these winning probabilities are independent (which is only reasonable for a non-interactive game, where the two teams do not use their skills to interfere with one another).

If this is the case, you can think of each of the two teams as two orthogonal axis, X and Y, with overall win rates x and y, respectively.

Then, in this graph, you can plot two vertical lines at X = x and X = 1: the first interval gives you the probability of team X winning and the second interval the probability of team X losing. You can also plot two horizontal lines at Y = y and Y = 1: the two intervals will again give you the probability of team Y winning and losing, respectively.

This will result in 4 different areas in your graph (top-down, left-right): X wins & Y loses, X wins & Y wins, X loses & Y loses, X loses & Y wins.

From this visualization, you can see the winning probability for team X as P(X wins & Y loses)/(P(X wins & Y loses) + P(X loses & Y wins))

Symbolically: P(X wins & Y loses) = x*(1-y)/(x+y-2xy)

PS: Sorry for the bad formatting, it is my first time posting an answer here...

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  • $\begingroup$ Welcome to CV! This is a clever approach. I am having a hard time conceiving of the outcome of such a game as being conditional, though, because that means you are entertaining the possibility that both players win or both players lose and are viewing the outcome as conditional on that. But since obviously these events have zero probability, it seems like we are forced to conclude that this independence model is mathematically impossible--and so any formula is logically valid. How would you address that concern? $\endgroup$
    – whuber ♦
    Commented May 18, 2022 at 15:44
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Where a and b are the win% of each team:

P(a) = a* (1 - b) / (a * (1 - b) + b* (1 - a))

This does not factor in home court advantage which varies greatly from sport to sport. If you want it to be a little more accurate you can first do this league adjustment on each teams win%

aCorrected = a - ((a - .5) / (N - 1))

bCorrected = b - ((b - .5) / (N - 1))

Where N is the number of teams in the league. This will account for the overall strength of the rest of the league.

Obviously this does not account for injuries or back to backs which especially in the nba make a big difference.

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