Q:

Expected value for the number of goals in a game

I'm trying to use odds data from bookmakers to estimate the expected number of goals in a game. We have these known facts:

P(o4.5) = 0.573
P(o5.5) = 0.458
P(o6.5) = 0.279

P(o4.5) is the probability that there will be more than 4.5, P(o5.5) more than 5.5 and so on. The probability that there will be less than 4.5 is the inverse: 1 - P(o.45). 
Can you from these facts estimate the expected number of goals in the game? Something like
P(oX) = 0.5

What is the value of the X variable? Is the information given above enough to figure it out? Is it enough if you make some assumptions like on the shape of the goal distribution and so on?

A:

There's no natural solution to this problem. In a sense, this is because you've specified both too much and too little. If you'd specified the probability that there will be more than $n$ goals for all non-negative integers $n$, then it would be straightforward to calculate the probability that there will be exactly $n$ goals for all $n$. On the other hand, if you'd specified only one or two probabilities, one might use a model and use the specified value(s) to determine its parameter(s). For instance, a natural model for the number of goals would lead to the Poisson distribution. This has a single parameter (usually denoted by $\lambda$), which you could fix by specifying a single probability. One might consider other distributions, such as a binomial distribution or a normal distribution truncated at $0$. However, all the obvious candidates have at most $2$ parameters, so they would be overdetermined by the three values you specify. You'd have to make some specific and probably to some degree arbitrary assumptions to come up with a model that has $3$ parameters to fit.
You could consider performing a non-linear least squares fit of a Poisson distribution to the probabilities you've specified. The data fit that model reasonably well; here's a table of values for $\lambda=5.25$ that gets within $0.03$ of your probabilities; this corresponds to an expected number of $5.25$ goals. (Here $\Gamma(n,\lambda)/\Gamma(n)$ is the probability that there will be less than $n$ goals.)

