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I have been cracking my head trying to get a solution to this problem for awhile, but I can't seem to get it:

"In a soccer match, given team A and team B are both expected to score 1.3 goals each, and that goals follow a poisson distribution, what is the probability that team A wins having gone behind in the match?"

I know quite a bit about the Poisson and have used it to calculate soccer probabilities before, however this one stumps me. My gut feeling is that the whole 90 minutes will have to be modeled to come up with an answer but I just need to know I'm not missing something from conditional probability that can help solve this problem....

clattenburg cake
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1 Answers1

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Hint:

Denote the number of goals scored in the whole match by $S$ and denote the special event by $E$. Then: $$P\left(E\right)=\sum_{n=0}^{\infty}P\left(E\mid S=n\right)P\left(S=n\right)$$ where $S$ is Poisson-$2.6$ distributed.

The probabilities $P\left(E\mid S=n\right)$ can be solved on base of symmetry. If e.g. $n=3$ then there is one possibility ($BAA$) for the event to occur, and symmetry tells you that $P\left(E\mid S=3\right)=2^{-3}$.

I admit that this is only a setup. There is a deeper problem lying behind. I have no hold on that yet.

drhab
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