And now I’m doing a post about it.
Yesterday, I tweeted a poll:
First of all, I’m impressed that only 17% of people said they didn’t want to do math today. Secondly, many of you did get the answer perfectly correct, with the same formulas I’m writing below and/or some fun simulation results. You guys are a credit to yourselves and your AP Stats teachers. Thirdly, I tweeted this question so I could get on my soapbox about 2 things as they relate to sports stats, namely:
- looking at the right outcome variable
- dealing in distributions rather than averages
Before I go any further, the takeaway here is that Team A wins more often. Read on to see how to arrive at that answer.
So, how do we begin to approach this? A simple expected value framework would state that Team A’s expected goals = 1, whereas Team’s B’s expected goals = # of shots * P(goal) = 20 * 0.055 = 1.1. 1.1 goals > 1 goal, so Team B is better off.
But that’s not really what we care about! We’d like to know the probability that each team wins the game. [Note: Some of you wanted to get pedantic on me and say that you’d cheer for Team B regardless of the math because it’s more exciting. So I will return the pedantry and say that I did not specify how Team A is scoring that goal, and it could very well be the most exciting hockey you’ve ever seen.]
For Team A to win, Team B has to score 0 goals on 20 shots. The probability of that happening, assuming the shots are independent, is P(no goal) multiplied by itself 20 times, or (1 – 0.055)^20. Therefore, P(Team A wins) = 0.32258, or about 32%.
For the teams to tie, Team B has to score exactly 1 goal. This is where it’s helpful to look up how the binomial distribution works, but I’ll power through this explanation assuming you haven’t.
Let’s say Team B only scores on their first shot, and then misses all the others. The probability of this exact sequence happening is P(goal) * P(no goal)^19 = 0.055 * 0.945^19 = 0.018774.
O X X X X X X X X X X X X X X X X X X X X
This is equivalent to the probability of only scoring on your 2nd, or 3rd, or 4th shot.
X O X X X X X X X X X X X X X X X X X X X
X X O X X X X X X X X X X X X X X X X X X
X X X O X X X X X X X X X X X X X X X X X
And there are 20 possible ways for you to only score on your nth shot, so P(tie) = 20 * 0.055 * 0.945^19 = 0.37549.
Finally, P(Team B wins) is just 1 – P(Team A wins) – P(tie) since Team B wins if they score 2+ goals and there are no other outcomes. Which means P(Team B wins) = 0.30193.
So it’s very close, but Team A wins slightly more often (32% vs. 30%), even though Team B is expected to score more on average (by 0.1 goals per game). Obviously, that isn’t a huge discrepancy, but the point I wanted to make is this: try to look at distributions whenever you can, because averages may not tell the whole story.
Friend of the site Danny Page wrote about some similar ideas in soccer, so be sure to check that out as well!