2025 AMC 12B Problem 18

The first game produces one outcome. If it was a win (probability $\tfrac{1}{3}$), the expected wait for a loss is $\tfrac{1}{\frac{2}{3}} = \tfrac{3}{2};$ if it was a loss (probability $\tfrac{2}{3}$), the expected wait for a win is $\tfrac{1}{\frac{1}{3}} = 3.$ So the expected total is $1 + \tfrac{1}{3}\cdot\tfrac{3}{2} + \tfrac{2}{3}\cdot 3 = 1 + \tfrac{1}{2} + 2$ $= \tfrac{7}{2}.$ Thus, the correct answer is D.