2026 AMC 10A Problem 12

Taking socks out one at a time at random is the same as first shuffling all $200$ socks into a random line and then taking them from the front. Pippi stops early, so some socks at the back of the line are never taken. The whole problem is about what the back of the line looks like.

1. A tiny example. Suppose the drawer had only $2$ red (R) and $2$ blue (B) socks. There are $6$ equally likely lines:

In every row, the socks left are exactly the block of same-colored socks at the very end of the line. For example, RRBB ends in BB, with a red sock just before that block.

2. Why that is always true. Pippi stops right after taking the last sock of some color, say the last red. Every sock behind it in the line is blue, since no reds are left. And the sock she just took is red, so the blue block behind it starts right there. So:
$$\text{socks left} = \text{the block of one color at the end of the line.}$$

1. What "exactly 2 left" means. The end block has length exactly $2$ when the last two socks in the line are the same color and the third sock from the end is the other color.

4. Fill the line from the back. A random line read backwards is still a random line, so the last spot is equally likely to hold any of the $200$ socks, the spot before it is equally likely to hold any of the $199$ others, and so on. Take the case "last two red, third from the end blue":
- last sock red: $\frac{100}{200}$;
- second to last red, given that: $99$ reds among $199$ socks, so $\frac{99}{199}$;
- third from the end blue: $100$ blues among $198$ socks, so $\frac{100}{198}$.

Multiply, using $\frac{100}{200} = \frac12$ and $\frac{99}{198} = \frac12$:
$$\frac{100}{200} \cdot \frac{99}{199} \cdot \frac{100}{198} = \frac12 \cdot \frac12 \cdot \frac{100}{199} = \frac{25}{199}.$$

5. The other color. "Last two blue, third from the end red" has the same probability, $\frac{25}{199}$, because the two colors have the same number of socks. The two cases cannot happen together, so add them:
$$\frac{25}{199} + \frac{25}{199} = \frac{50}{199}.$$

1. Check with the tiny example. The same method gives $2 \cdot \frac{2}{4} \cdot \frac{1}{3} \cdot \frac{2}{2} = \frac{1}{3}$, and the table above shows $2$ of the $6$ lines leaving exactly $2$ socks. They agree.

The answer is $\boxed{\textbf{(D)}\ \frac{50}{199}}$.

A process that stops "when something runs out" is easier to study as one complete random order: what is left over is decided entirely by the end of that order. Reading the order from the back replaces a long process (up to $198$ draws) with three simple draws. Whenever a question asks about the last items or the leftovers, try looking at the order from the end.

Count lines instead of multiplying probabilities. Every arrangement of $100$ R's and $100$ B's in a row is equally likely, and there are $\binom{200}{100}$ of them.

A line ends with "B, R, R" when its last three spots are fixed that way and the other $197$ spots hold the remaining $98$ R's and $99$ B's. That is $\binom{197}{98}$ lines. Dividing and cancelling the factorials,
$$\frac{\binom{197}{98}}{\binom{200}{100}} = \frac{100 \cdot 99 \cdot 100}{200 \cdot 199 \cdot 198} = \frac{25}{199},$$
the same as Step 4. Doubling for lines ending "R, B, B" gives $\frac{50}{199}$.