4.3.8 problem 46

Let $X\sim \text{NHGeom}(4,48,1)$ be the number of non-aces before the first ace. Then, $\text{E}\left(X\right)=\frac{\mathrm{rb}}{w+1}=\frac{1\ast 48}{5}=9.6$.

Let $Y\sim \text{NHGeom}(4,48,2)$ be the number of non-aces before the second ace is drawn. Then, $\text{E}\left(X\right)=\frac{\mathrm{rb}}{w+1}=\frac{2\ast 48}{5}=19.2$

Let $Z=Y-X$. Notice that $Z$ represents the number of non-aces between the first and the second ace. $\text{E}\left(Z\right)=\text{E}\left(Y\right)-\text{E}\left(X\right)=19.2-9.6=9.6$.