Suppose Sir Joseph had scored \(r \) runs in \( n \) [completed] innings before this score, and call the before-and-after averages \(A_0\) and \( A_1 \). Then: * \( r = A_0 n \) * \( r + 50 = A_1 (n+1) \) So \( 50 = n(A_1 - A_0) + A_1 \) and \( n = \frac{50 - A_1}{A_0 - A_1} \) The RHS is all numbers we know, and we get \( n = \frac{0.01587\dots}{0.0000835\dots} \approx 189 \) and \( r = A_0 n \approx 9453 \).