2010 AMC 10A Problem 18

Because the digits are always put in descending order, each number comes from exactly one group of three digits. For example, the digits $2, 7, 5$ always give $752$. So we just count groups of three digits.

1. How many choices. Bernardo picks $3$ of $9$ digits: $\dfrac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$ groups. Silvia picks $3$ of $8$: $\dfrac{8 \times 7 \times 6}{6} = 56$ groups.
2. The only difference is the digit 9, which only Bernardo can have. So split into two cases.

Case 1: Bernardo picks a 9.

1. He picks $9$ plus $2$ of the other $8$ digits: $\dfrac{8 \times 7}{2} = 28$ groups. The chance is $\dfrac{28}{84} = \dfrac13$.
2. His number then starts with $9$, and Silvia's starts with $8$ or less. Bernardo wins every time.

Case 2: Bernardo does not pick a 9 (chance $1 - \tfrac13 = \tfrac23$).

5. Now Bernardo picks from $\{1, \ldots, 8\}$, exactly like Silvia. The two players are in the same position.
6. They tie only if they pick the very same group. Whatever Bernardo picks, Silvia matches it with chance $\dfrac{1}{56}$.
7. The rest of the time, $1 - \dfrac{1}{56} = \dfrac{55}{56}$, someone wins, and because the two are in identical positions, each wins half of that:
$$\frac12 \times \frac{55}{56} = \frac{55}{112}.$$

Put the cases together.
$$\frac13 \times 1 + \frac23 \times \frac{55}{112} = \frac13 + \frac{55}{168} = \frac{56}{168} + \frac{55}{168} = \frac{111}{168} = \frac{37}{56}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{37}{56}}$.

Find the one thing that makes the two players different (here, the digit $9$) and split on it. In the case where they are exactly alike, you do not need to compare numbers at all: winning and losing are equally likely, and only the tie needs to be taken out first. Remembering ties is the difference between the right answer and choice (C).