I'm struggling to work out this answer. Say Team A is estimated to score $1.6$ goals in a match and Team B is estimated to score $1.1$ goals. Team A's striker is expected to score 40% of his team's goals in any given match.
My question is, how do I figure out the probability of him scoring at least one goal, two or more goals in a match, three or more goals, etc. ? I have a spreadsheet in Excel of all the probabilities of different scores so if it is a case of brute force summation I can do it, I just happen to have hit a brick wall!
Best Answer
You have to assume something about the distribution of goals. The average of $1.6$ goals could come from scoring one goal $40\%$ of the time and two goals $60\%$ of the time. It could also come from scoring $0$ goals $95\%$ of the time and $32$ goals $5\%$ of the time. Your calculation says the expected number of goals scored by the striker is $0.64$, but again this could come from many distributions. In our first, the striker will score more than one goal $0.6\cdot 0.4^2=9.6\%$ of the time under the assumption that which $40\%$ the striker scores are uniformly distributed. In the second, in a game where the team does score a goal (and therefore scores $32$ goals) the striker is essentially certain to score more than one, so he will score multiple goals $5\%$ of the time.
Again, assuming the probability a goal came from the striker is uniform, and given that the team scores $n$ goals, the chance the striker doesn't score any is $0.6^n$ and the chance that he scores $1$ is ${n \choose 1}0.4\cdot 0.6^{n-1}$ (do you see why?). The chance he scores more than one is obtained by subtracting the sum of those from $1$.