Simpson's Paradox

No. Interpret “average accuracy” as goals divided by shots. A team’s combined accuracy is then a weighted average of its two match accuracies. Each team’s weights are the fractions of its own shots taken in each match. If team $A$ takes almost all its shots in the match where everyone’s accuracy is low, and team $B$ in the match where it is high, the combined comparison can reverse.

Write $p_{X,i}$ for team $X$’s accuracy in match $i$, and $w_X$ for the fraction of team $X$’s total shots taken in match $1$. Then

The inequalities $p_{A,1}>p_{B,1}$ and $p_{A,2}>p_{B,2}$ say nothing about $p_A$ versus $p_B$ once $w_A\neq w_B$.

Counter-example.

Team $A$ is more accurate in match 1 ($1\% > 0\%$) and in match 2 ($100\% > 99\%$). Yet over both matches $A$ scores $2$ of $101$ shots and $B$ scores $99$ of $101$. So $B$’s combined accuracy is far higher.

The reason is the weights: $w_A = 100/101$ while $w_B = 1/101$. Almost all of $A$’s shots come in the low-accuracy match, and almost all of $B$’s come in the high-accuracy match. This reversal is Simpson’s paradox.

Generalisation. The conclusion does hold if both teams split their shots between the matches in the same proportion, i.e. $w_A=w_B=w$. In that case

Key idea. Pooled rates are weighted averages with group-specific weights, so a comparison that holds within every group can reverse in the aggregate.