Are Current Odds Reasonable?

If we are given odds for a home win, draw, and away win, we can convert them to probabilities. One way to do this is to take the reciprocals and then normalize them. So if the odds are $r_H$, $r_D$, $r_A$ the inferred probability for a home win is

$$ p_H = \frac{\frac{1}{r_H}}{\frac{1}{r_H} + \frac{1}{r_D} + \frac{1}{r_A}} $$

And we can define $p_D$ and $p_A$ similarly. So say that we have a bunch of matches with odds, and therefore with inferred probabilities. For now, consider only the probability of a home win, and say we have $n$ matches with inferred probabilities $p_1$, $p_2$, $\ldots$, $p_n$. Say that $X_i$ is $1$ if the result of the $i$’th match is a home win and $0$ otherwise. Also assume that $X_1$, $\ldots$, $X_n$ are ordered chronologically and that

$$ E \left ( X_i \mid X_1, \ldots, X_{i-1} \right ) = p_i $$

Let $X$ be given by

$$X = \frac{1}{n}\sum_{i=1}^n X_i\, ,$$

then we can conclude that

$$ E(X) = \frac{p_1 + \ldots + p_n}{n}, \, V(X) = \frac{p_1(1-p_1) + \ldots + p_n(1 – p_n)}{n^2} $$

So what we can do in order to check that the odds are reasonable is to bucket all the matches by the probability and then look at their average expected number of home wins vs the actual number of home wins.

I’ve done this in the figures below, and overall they look reasonable. This doesn’t mean that the odds are perfect, but it implies that they are at least correctly scaled.