Probability of winning based on probability of individual scoring opportunities Let's take a sport like Hockey, and say we have "Team A" playing "Team B". Assume that I have information on all the scoring opportunities that each team has had, and somehow I also know the probability for each of them ending up as a goal. For example, I can assign each opportunity a score between 0 and 1, with 1 meaning it would have gone in every single time if they tried it again, and 0 meaning it was impossible. 
If I don't know the outcome of the game or each opportunity, how can I use these two sets of numbers to determine which team had the higher probability of winning? One simple way would be to simply sum up the values for each opportunity, but I feel like there might be a better way. A more sophisticated model could also differentiate between the winning probability for a game where a team has 4 opportunities with 0.6 scoring likelihood versus one where a team has 6 opportunities with 0.4 likelihood, if there really is a difference. 
 A: The expected number of goals scored is the sum of the probabilities of the opportunities.  So the team with the higher expected score is more likely to win.
But if the expected scores are very close you might want to test that the difference is statistically significant. I see no better way to treat the problem unless you have additional information for comparing the teams. 
A: To add some more detail to the other excellent answer, lets assume the number of scoring opportunities during a game is a random variable $N$, with distribution on the non-negative integers.  Then lets assume that given a scoring opportunity $i$ with result $X_i$ (one if scored, zero if not) with scoring probability $p$.  Then the total number of scores is 
$$
   S  = \sum_{i=1}^N X_i
$$
and to get its expectation (which is the expected number of scores) we can use the double expectation theorem (see https://en.wikipedia.org/wiki/Law_of_total_expectation) to get
$$ \DeclareMathOperator{\E}{\mathbb{E}}
   \E S = \E \E S \mid N= p \E N
$$
that is, the scoring probability (one for each scoring opportunity) times the expected number of opportunities. 
