Average Shooting Score
Jane Street
Two teams A and B played football twice. Two independent matches. The average shoot of A in match 1 is greater than B. The average shoot of A in match 2 is also greater than B. So can we say the average shoot of A is greater than B in two matches together? If so, why?
Answer
No, we cannot conclude this - it is still possible for B to have a higher overall average. This is an example of Simpson’s Paradox. Even though Player A has a higher average than Player B in both matches, Player B can still have a higher average overall if the number of attempted shots differs significantly between the matches.
To reason this out in an interview, first recall that average = shots hit / shots taken. The overall average for the season is the total number of shots hit divided by the total number of shots taken across both matches.
The key is that averages are weighted by the number of observations (in this case, shots taken). If Player A only took a few shots with a high average and Player B took more shots with slightly lower averages, B’s total performance could outweigh A’s. For example:
First Match
A: 3/10 = 0.300 -- A is better in the first half.
B: 2/10 = 0.200
Second Match
A: 1/1 = 1.000 -- A is better in the second half.
B: 40/100 = 0.400
Both Matches
A: 4/11 = 0.364
B: 42/110 = 0.382 -- B is better for the season.