# What's the probability that a bookmaker is mispricing odds on soccer games?

An English soccer team plays a series of matches against different opponents of varying ability. A bookmaker offers odds for each match as to whether it will be a home win, away win, or draw. Part-way through the season, the team has played $n$ matches and has drawn $k$ of them, which is more than might be expected from the odds.

What is the probability that the bookmaker is mis-pricing the odds on these matches, rather than just being unlucky? If the bookmaker continues to price the team's remaining matches in a similar way, and I bet $\$1$that each one will be a draw, what is my expected return? ## 2 Answers The answer to your question depends intricately on what information and assumptions you are going to use. This is because the result of a game is an extraordinarily complicated process. It can become arbitrarily complicated depending on what information you have about: 1. Players in the particular team - perhaps even particular combinations of players may be relevant. 2. Players in other teams 3. Past history of the league 4. How stable the team's players are - do players keep getting selected and dropped, or is it the same 11. 5. The time that you place your bet (during the game? before? how much before? what info is lost from betting before to betting on the day?) 6. some other relevant feature of soccer which I have omitted. The odds that a book-maker gives are not a reflection of the book-makers odds. which is impossible if they are probabilities. A book-maker will adjust the odds down when someone bets on a draw, and adjust them up when someone bets on a non-draw. Thus, the odds are a reflection of the gamblers (who use that book-maker) odds as a whole. So it is not the bookmaker who is miss-pricing per se, it is the gambling collective - or the "average gambler". Now if you are willing to assume that whatever "causal mechanism" is resulting in a draw remains constant across the season (reasonable? probably not...), then a simple mathematical problem is obtained (but note there is no reason for this to be "more right" than some other simplifying assumption). To remind us that this is the assumption being used, an$A$will be put on the conditioning side of the probabilities. Under this assumption the binomial distribution applies: $$P(\text{k Draws in n matches}|\theta,A)={n \choose k}\theta^{k}(1-\theta)^{n-k}$$ And we want to calculate the following $$P(\text{next match is a draw}|\text{k Draws in n matches},A)$$ $$=\int_{0}^{1}P(\text{next match is a draw}|\theta,A)P(\theta|\text{k Draws in n matches},A)d\theta$$ where $$P(\theta|\text{k Draws in n matches},A)=P(\theta|A)\frac{P(\text{k Draws in n matches}|\theta,A)}{P(\text{k Draws in n matches}|A)}$$ is the posterior for$\theta$. Now in this case, it is fairly obvious that it is possible for a draw to happen, and also possible for it to not happen, so a uniform prior is appropriate (unless there is extra information we wish to include beyond the results of the season) and we set$P(\theta|A)=1$. The posterior is then given by a beta distribution (where$B(\alpha,\beta)$is the beta function) $$P(\theta|\text{k Draws in n matches},A)=\frac{\theta^{k}(1-\theta)^{n-k}}{B(k+1,n-k+1)}$$ Given$\theta$and$A$the probability that the next match is a draw is just$\theta$so the integral becomes: $$\int_{0}^{1}\theta \frac{\theta^{k}(1-\theta)^{n-k}}{B(k+1,n-k+1)}d\theta=\frac{B(k+2,n-k+1)}{B(k+1,n-k+1)}=\frac{k+1}{n+2}$$ and hence the probability is just: $$P(\text{next match is a draw}|\text{k Draws in n matches},A)=\frac{k+1}{n+2}$$ But note that it depends on$A$- the assumptions that were made. Call the "priced odds" a probability conditional on some other unknown complex information, say$B$. So if the published odds are different to the the above fraction, then this says that$A$and$B$lead to different conclusions, so both can't be right about the "true outcome" (but both can be right conditional on the assumptions each made). THE KILLER BLOW This example showed that the answer to your question boiled down to deciding if$A$was "more accurate" than$B$in describing the mechanics of the soccer game. This will happen regardless of what the proposition$A$happens to be. We will always boil down to the question of asking "whose assumptions are right, the gambling collective's or mine?" This last question is a basically unanswerable question until you know exactly what the proposition$B$consists of (or at least some key features of it). For how can you compare something that is known with something that is not? UPDATE: An actual answer :) As @whuber has cheekily pointed out, I haven't actually given an expected value here - so this part simply completes that part of my answer. If one was to assume that$A$is true with priced odds of$Q$, then you would expect, in the next game to receive $$Q\times P(\text{next match is a draw}|\text{k Draws in n matches},A)-1$$ $$=Q\times \frac{k+1}{n+2}-1=\frac{Q(k+1)-n-2}{n+2}$$ Now if you assume that the value of$Q$is based on the same model as yours then we can predict exactly how$Q$will change into the future. Suppose$Q$was based on a different prior to the uniform one, say$Beta(\alpha_Q,\beta_Q)$, then the corresponding probability is $$P(\text{next match is a draw}|\text{k Draws in n matches},A_Q)=\frac{k+\alpha_Q}{n+\alpha_Q+\beta_Q}$$ with expected return of $$\frac{Q(k+\alpha_Q)-n-\alpha_Q-\beta_Q}{n+\alpha_Q+\beta_Q}$$ Now if we make the "prior weight"$\alpha_Q+\beta_Q = \frac{N}{2}$where$N$is the length of the season (this will allow the "miss-pricing" to continue into the remainder of the season) and set the expected return to zero we get: $$\alpha_Q=\frac{2n+N}{2Q}-k$$ (NOTE: unless this is the actual model,$\alpha_Q$will depend on when this calculation was done, as it depends on$n,k,Q$which will vary over time). Now we are able to predict how$Q$will be adjusted into the future, it will add$1$to the denominator for each match, and$1\$ to the numerator if the match was a draw. So the expected odds after the first match are:

$$(1+\frac{n+\beta_Q-k+1}{k+\alpha_Q})\frac{n-k+\beta_Q}{n+\alpha_Q+\beta_Q}+(1+\frac{n+\beta_Q-k}{k+\alpha_Q+1})\frac{k+\alpha_Q}{n+\alpha_Q+\beta_Q}$$ $$=1+\frac{n+\beta_Q-k}{k+\alpha_Q}\left(1+\frac{2}{(2n+N)(k+\alpha_Q+1)}\right)\approx 1+\frac{n+\beta_Q-k}{k+\alpha_Q}$$

That is the odds won't change much over the season. Using this approximation, we get the expected return over the remainder of the season as:

$$(N-n)\frac{Q(k+1)-n-2}{n+2}$$

But remember that this is based on the overly simplistic model of a draw (note: this does not necessarily mean that it will be a "crap" predictor). There can be no unique answer to your question, because there has been no specified model, and no specified prior information (e.g. how many people use this bookie? what is the bookie's turnover? how will my bets influence the odds they price?). The only thing which has been specified is the data from one season, and that for "some unspecified model" the probabilities are inconsistent with those implied by the odds pricing.

Bookmakers use an overround so they don't actually care what the result is because they win whatever. That is why you never meet a poor bookie. If a bookmaker is mispricing draws your ability to make profit would depend on the odds the bookmaker was offering and whether the profits generated would cover the times you lose.

• This may be true, but mostly irrelevant, because the question asks for the expected return of the gambler, not the expected return of the bookie Apr 10, 2011 at 2:26
• @probability So what is the gambler's expected return? I couldn't find it in your reply :-).
– whuber
Apr 10, 2011 at 17:06