2017 AMC 10A Problem 15

Scale the intervals to $[0, 1]$ for Chloé and $[0, 2]$ for Laurent; the probability is unchanged.

Split on Laurent's number $L$.

With probability $\tfrac12$, $L \in [1, 2]$. Then $L$ is at least $1$, which is at least Chloé's number, so Laurent wins for sure.

With probability $\tfrac12$, $L \in [0, 1]$. Now both numbers are independent and uniform on the same interval, so by symmetry each is larger with probability $\tfrac12$ (a tie has probability $0$).

Combining:
$$\frac12 \cdot 1 + \frac12 \cdot \frac12 = \frac34.$$

The answer is $\boxed{\textbf{(C)}\ \frac{3}{4}}$.

Two independent uniform choices form a uniform point in a rectangle, so probability is area. The rectangle here is $1 \times 2$; the losing region $L \le C$ is a right triangle with legs $1$ and $1$, area $\tfrac12$, out of total area $2$, giving a losing probability $\tfrac14$. Conditioning on which half Laurent lands in is the same computation organized so that symmetry does the work.

Draw the $1 \times 2$ rectangle with $C$ on one axis and $L$ on the other. The line $L = C$ cuts off the triangle $L \le C$, of area $\tfrac12 \cdot 1 \cdot 1 = \tfrac12$. So the probability that $L > C$ is $\frac{2 - \frac12}{2} = \frac34$.

• Treating the two numbers as symmetric and answering 1/2, forgetting that Laurent's interval is twice as long.
• Averaging the two probabilities as $\frac{1 + \frac12}{2}$ but then also multiplying by an extra $\tfrac12$, or forgetting the sure-win half entirely.