In the context of a football ("soccer") match, if I have the following for a single game:
How do I derive the following?
Let's start with some notation to make our lives easier.
Define $X_A$ to be number of goals scored by team A and $X_B$ to be number of goals scored by team B.
Define $P(A)$ as the probability team A wins the game which is $P(X_A > X_B)$. Likewise, define $P(B)$ as the probability team B wins which is $P(X_B > X_A)$, and $P(D) = 1 - P(A) - P(B) = P(X_A = X_B)$ as the probability of a draw.
Currently we know $P(A)$, $P(B)$, $P(D)$, and $E(X_A + X_B) = E(X_A) + E(X_B)$. The question is to "derive" $E(X_A)$ and $E(X_B)$ given that information. For simplicity I'll define $E(X_A) = \mu_A$ and $E(X_B) = \mu_B$.
Unfortunately, this is an impossible task without either collecting more data or making strict assumptions. Here is one approach:
Assume the number of goals scored in a game follow a Poisson process. In practice, they don't, but it could be a decent approximation. You will also need to assume that each team scores goals independently of each other, which also isn't really true.
Thus $(X_A + X_B) \sim Poisson(\mu_A + \mu_B)$, $X_A \sim Poisson(\mu_A)$ and $X_B \sim Poisson(\mu_B)$.
What I would do is set $\mu_A = 0$ and $\mu_B = \mu_A + \mu_B$ (remember you already know $\mu_A + \mu_B$). Under these assumptions you can either calculate $P(A)$ and $P(B)$ using the PMFs or estimate it with simulation. Then I would increase $\mu_A$ by some interval $\epsilon$ and decrease $\mu_B$ by $\epsilon$ and repeat until $\mu_A = \mu_A + \mu_B$ and $\mu_B = 0$.
At this point you will have a set of win/draw probabilities for each combination of $\mu_A$ and $\mu_B$ that you tried so you can pick the one that closest fits your actuals of $P(A)$ and $P(B)$. You can also "zoom in" and use a smaller $\epsilon$ over a certain window.
There are ways to fine tune this, for example you can find a notional distribution for xG and assume each shot follows this distribution. Then instead of varying the xG for each team, you vary the number of shots for each team. For example, if each shot has an xG of .1, and the total xG of the game was 2.5, you would assume there were 25 shots in the game. Then you could calculate $P(A)$ given A had 0 shots and B had 25, then if A had 1 shot and B had 24, etc.
You need to translate this into probability statements like this, where A is goals for team A and B is goals for team B
$P(A > B) = p1, P(A < B) = p2, P(A = B) = 1 - p1 - p2, E(A + B) = x$
Let's use an example, p1 = .5, p2 = .3 and x = 4. We then have P(A < 2) = .3, P(A = 2) = .2 and P(A > 2) = .5 etc. You can then derive the expected goals for E(A) = .3 * .5 + .2 * 2 + .5 * 3.5 = 2.3 (assuming equal weight between 0 and 1 goals for A) etc.
Note that a high P(A = B) and odd x are not very consistent.
ischmidt20 + Dirk N - Thank you so much for your insight on this topic.
I largely followed ischmidt20's approach and built a function in python to calculate the expected goals from the home & away sides.
If it's of any interest, the code is placed here -> https://github.com/philbywalsh/Betting/tree/main
Sample results
INPUTS
Total expected goals: 2.633995
Probability of home team winning = 0.17543859649122806
Probability of away team winning= 0.5988023952095809
OUTPUTS
Expected goals for Home Team (A): 0.839999981087032
Expected goals for Away Team (B): 1.7939611795736825
Sum of home & away expected goals: 2.6339611606607145
