Probability of a player scoring multiple goals in a match 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!    
 A: 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 $1$.
A: The most common assumption in this setting (and one must make some assumption to solve this) is that the times when striker S scores a goal are the events of a homogenous Poisson process (of course, this is a huge assumption). Since S scores $1.6\times40\%=0.64$ goals by match, the rate (by match) of this Poisson process should be $\lambda=0.64$. 
Then, in any given match, S will score exactly $n$ goals with probability $\mathrm e^{-\lambda}\lambda^n/n!$. For example, S will score no goal during one whole match with probability $\mathrm e^{-\lambda}\approx52.7\%$. Likewise, the mean number of matches between S's successive goals is $1/\lambda\approx1.56$ matches (that is, if we are talking about soccer matches, which are 90 minutes long, about 140 minutes), S's hattricks (3 or more goals by the same striker during the same match) happen with probability $1-\mathrm e^{-\lambda}(1+\lambda+\lambda^2/2)\approx2.7\%$, that is, roughly every 37th match, and so on.
