Why Averaging Averages Gives the Wrong Answer

A trading desk has two teams. Team A trades 90 contracts a day and averages $10 P&L per contract. Team B trades 10 contracts a day and averages $100 P&L per contract. What's the desk's average P&L per contract? "Average the two averages" gives $\dfrac{10+100}{2}=55$, i.e. $55 — but that's wrong, and not by a rounding error. The desk actually made $90(10) + 10(100) = 900+1000=1900$ dollars across $100$ contracts, an average of $19 per contract, less than half the naive answer. Averaging averages silently assumes every group is the same size, and here they aren't even close.

The idea: weight by group size, not by group count

The true overall average of $n_1$ items averaging $\bar x_1$ and $n_2$ items averaging $\bar x_2$ is the weighted average:

$$\bar x = \frac{n_1 \bar x_1 + n_2 \bar x_2}{n_1 + n_2} .$$

In plain English: you're re-deriving the total (each group's average times its own size) and dividing by the true total count — not treating the two group averages as equally-weighted data points. Simply averaging $\bar x_1$ and $\bar x_2$ is only correct in the special case $n_1 = n_2$, where the weights happen to cancel out to $\frac12$ each. The moment group sizes differ, "average of averages" quietly overweights the smaller group and underweights the larger one.

Worked example 1: the desk P&L, done correctly

Using the weighted formula: $\bar x = \dfrac{90(10) + 10(100)}{90+10} = \dfrac{900+1000}{100} = \dfrac{1900}{100} = 19$, i.e. $19. Team A, despite the lower per-contract average, dominates the true blended number because it trades nine times the volume — the naive $55 answer treated Team B's ten contracts as if they carried the same weight as Team A's ninety, which overstates Team B's influence by 9×.

Worked example 2: the classic "average speed" version

A car drives from A to B at 30 mph and returns from B to A at 60 mph, the same distance each way. What's the average speed for the round trip? Averaging the two speeds gives $\frac{30+60}{2}=45$ mph — wrong, because the car spends more time at the slower speed, not more distance. For a distance $d$ each way, the outbound leg takes $d/30$ hours and the return leg takes $d/60$ hours, so average speed $= \dfrac{\text{total distance}}{\text{total time}} = \dfrac{2d}{d/30+d/60} = \dfrac{2d}{d/20} = 40$ mph — the harmonic mean of the two speeds, not the arithmetic mean, because time (not distance) is what's being equally split here.