2017 AMC 10A Problem 15

Only the sizes of the ranges matter, so shrink the numbers: Chloé picks from $0$ to $1$ and Laurent picks from $0$ to $2$. Laurent's range is still twice as long, so the probability is the same.

Split on where Laurent's number lands.

1. Laurent lands between $1$ and $2$. This happens half the time, because it is half of his range. Chloé's number is at most $1$, so Laurent wins every time.
2. Laurent lands between $0$ and $1$. This happens the other half of the time. Now both numbers come from the same range, $0$ to $1$, picked the same way. Neither player has an advantage, so each wins half the time. (An exact tie has probability $0$.)
3. Combine the two halves:
$$\tfrac12 \times 1 + \tfrac12 \times \tfrac12 = \tfrac12 + \tfrac14 = \tfrac34.$$

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

When two numbers are picked at random from ranges, split the problem into pieces where the answer is easy: one piece where the result is certain and one piece where the two players are in the same position, so each wins half. This "split, then use fairness" idea works well for many probability questions.

Draw a picture. Put Chloé's number across ($0$ to $1$) and Laurent's number up ($0$ to $2$). Every pair of choices is a point in a $1$ by $2$ rectangle, with area $2$, and each point is equally likely. Laurent loses when his number is at most Chloé's, which is the triangle under the line from $(0, 0)$ to $(1, 1)$. That triangle has area $\tfrac12 \times 1 \times 1 = \tfrac12$. So Laurent wins with probability
$$\frac{2 - \tfrac12}{2} = \frac{1.5}{2} = \frac34.$$