WNBA: Three Road Victors with 88 pts! What are the odds?
In each of the 3 WNBA games yesterday, the visiting team came out on top. The weird part? All 3 visiting teams scored exactly 88 points.
We’ve been asked on Twitter to break down exactly how likely this event was. Using data from the last two seasons plus this year’s games (excluding last night), here’s the breakdown:
First off, according to our 2016–2018 data, the probability of winning on the road during the regular season is 42.8% (191 of 446 games).
But, we’re interested in the probability of 3 road victories in one night. We will make the simplifying assumptions that 1) only 3 games are played on the night of interest (as was the case last night), and 2) that the result of each game is independent of the other 2 games.
Given these assumptions, the probability of our event (all 3 games won on the road) is given by the formula:
So, what is the probability of 3 teams winning on the road in one night (assuming there are only 3 games)?
As a baseline, the odds of 3 teams winning on the road in the same day are about 8 in 100. With the added condition of score, we expect the odds to be much, much lower.
Now for the real problem: How likely is it for all 3 of those road victors to score exactly 88 points?
First, we have to account for the changing number of points scored per game in different seasons due to talent level, rule changes, pace of the game, etc. — the equivalent of accounting for inflation into our calculations.
To do this, we normalize the points scored in 2016 and 2017 in terms of 2018 points. For example, scoring 86 points in 2017 is actually the equivalent of scoring 88 points in 2018.
In the past 3 seasons (excluding last night) only 10 teams have won on the road scoring 88 points. Before last night, this had happened only once this season.
So, that’s 10 games out of 446 where our desired result occurred. That means the probability one team winning on the road with exactly 88 points is 2.2%, based our data from the last 3 seasons.
Again, we treat each game as an independent event, so the probability of 3 of these games in one night in which exactly 3 games are played is:
Our final answer? 0.001%. Therefore, out of 100,000 nights where exactly 3 WNBA games are played, we would expect all 3 visitors put up 88 points and win on only 1 of those 100,000 nights.
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