Solution
When the Better Drug Loses
Show the problem again
Drug A succeeds in 80% of mild cases and 30% of severe cases. Drug B succeeds in 70% of mild cases and 20% of severe cases, so A is better in each group. Drug A is given to 20 mild patients and $n$ severe patients; drug B is given to 70 mild and 30 severe patients. What is the smallest integer $n$ for which B's overall success rate is strictly higher than A's?
Worked solution
B's overall rate is $(0.7\cdot70 + 0.2\cdot30)/100 = 55/100$. A's is $(16 + 0.3n)/(20+n)$. Setting this equal to $0.55$ gives $16 + 0.3n = 11 + 0.55n$, so $n = 20$ is exactly a tie. For larger $n$ the severe cases, with their low rate, carry more weight in A's average, so A falls below $0.55$. At $n = 21$ the rate is $22.3/41 \approx 0.544 < 0.55$. The answer is **21**.
Source: Simpson's paradox (Yule 1903, Simpson 1951); numbers original. Statement written for AxiomIQ.