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There are 2 teams - Team A and Team B. Both teams play 2 matches. In both matches, Team A has a higher pass rate than Team B. Can the overall combined pass rate for Team B be higher than that of Team A?

Pass rate = (number of passes successful) / (number of passes attempted)

Combined Pass rate = (total no. of passes successful in both matches) / (total no. of passes attempted in both matches)

My gut feeling is that it should be possible based on Simpson's Paradox where if there is a huge disparity in the number of passes attempted then it's possible. Not sure if I'm correct or wrong.

Wondering if it's possible, could anyone come up with an example of the same in terms of numbers? I'm unable to think of numbers that could make it possible.

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If Team A achieves 1 of 1 (100%) and 1 of 4 (25%), they have an overall rate of 2 in 5 (40%).

If Team B achieves 3 of 4 (75%) and 0 of 1 (0%), they have an overall rate of 3 in 5 (60%) while doing worse than team A in each individual match.

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I always consider the extremes in these types of problems,

Game 1 Team A: 1 / 1 = 100% Team B: 99 / 100 = 99% (some really extreme case like this)

Game 2 Team A: 1 / 2 = 50% Team B: 1 / 3 = 33%

Overall Team A: 2 / 3 = 66% Team B: 100 / 103 = 97%

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  • $\begingroup$ This doesn't appear to add anything to the accepted answer which was posted a while before. Please do not duplicate answers unless you have something substantial to add. $\endgroup$
    – bobble
    Commented Feb 12 at 21:33

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