2025 AMC 12B Problem 18

Question

Awnik repeatedly plays a game that has a probability of winning of

\dfrac{1}{3}.

The outcomes of the games are independent. What is the expected value of the number of games he will play until he has both won and lost at least once?

\dfrac{5}{2}

3

\dfrac{16}{5}

\dfrac{7}{2}

\dfrac{15}{4}

Answer: D

Concepts:expected value geometric distribution

Difficulty rating: 1770

Small hint:

After the first game you already have one outcome; you then wait only for the opposite outcome

Big hint:

The expected number of games to get an outcome of probability

p

is

\dfrac{1}{p};

condition on whether the first game is a win or a loss

Solution:

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.