Question

Alice, Bob, and Carol play a game in which each of them chooses a real number between $0$ and $1.$ The winner of the game is the one whose number is between the numbers chosen by the other two players. Alice announces that she will choose her number uniformly at random from all the numbers between $0$ and $1,$ and Bob announces that he will choose his number uniformly at random from all the numbers between $\tfrac12$ and $\tfrac23.$ Armed with this information, what number should Carol choose to maximize her chance of winning?

$\tfrac12$

$\tfrac{13}{24}$

$\tfrac{7}{12}$

$\tfrac58$

$\tfrac23$

Answer: B

Concepts:geometric probability casework optimization

Difficulty rating: 2520

Small Hint:

Split into cases by where Carol’s number $c$ sits relative to $\tfrac12$ and $\tfrac23;$ for $\tfrac12 \lt c \lt \tfrac23$ she can win in two ways

Big Hint:

In that middle range her win probability is $-12c^2 + 13c - 3;$ maximize this downward quadratic at its vertex $c = \tfrac{-b}{2a}$

Solution:

If $c \le \tfrac12,$ Carol beats Bob automatically, so she wins only if Alice is below $c,$ probability $c \le \tfrac12.$ If $c \ge \tfrac23,$ she wins with probability $1 - c \le \tfrac13.$ Neither case exceeds $\tfrac12.$

For $\tfrac12 \lt c \lt \tfrac23,$ the chance Bob’s number exceeds $c$ is $\frac{\frac{2}{3} - c}{\frac{2}{3} - \frac{1}{2}} = 4 - 6c,$ so the probability Carol is above Alice and below Bob is $c(4 - 6c);$ the reverse ordering has probability $(1 - c)(6c - 3).$ Adding, $\begin{aligned} &c(4 - 6c) + (1 - c)(6c - 3) \\ &= -12c^2 + 13c - 3. \end{aligned}$ This downward parabola is maximized at $c = \tfrac{13}{24},$ which lies in $\left(\tfrac12, \tfrac23\right),$ and its value exceeds $\tfrac12.$

Thus, the correct answer is B.

2018 AMC 12A Problem 24

Answer: B

Difficulty rating: 2520

Small Hint:

Big Hint:

Solution:

Problem 24 in Other Years