2025 AMC 10A Problem 16

Call the coins $a$, $b$, $c$ and the jars $1$, $2$, $3$.

Step 1: count the outcomes. Each coin has $3$ jars to choose from, so there are $3 \cdot 3 \cdot 3 = 27$ placements, all equally likely. For example, "$a$ in jar $1$, $b$ in jar $1$, $c$ in jar $3$" is one placement.

Step 2: sort the placements by the size of the fullest jar. The fullest jar can hold $3$, $2$ or $1$ coins.

1. Fullest jar has $3$. All coins go in the same jar. There are $3$ jars to choose, so $3$ placements.
2. Fullest jar has $1$. Each jar gets exactly one coin. Coin $a$ has $3$ jars to choose, then $b$ has $2$, then $c$ has $1$: that is $3 \cdot 2 \cdot 1 = 6$ placements.
3. Fullest jar has $2$. This is every other placement: $27 - 3 - 6 = 18$.

Check it directly: pick which coin is alone ($3$ ways), pick the jar for the other two ($3$ ways), pick a different jar for the lonely coin ($2$ ways). That gives $3 \cdot 3 \cdot 2 = 18$. It matches.

Step 3: take the average. The expected value is the average of the fullest-jar count over all $27$ placements:
$$\frac{3 \cdot 3 + 2 \cdot 18 + 1 \cdot 6}{27} = \frac{9 + 36 + 6}{27} = \frac{51}{27} = \frac{17}{9}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{17}{9}}$.

The $27$ placements are equally likely, but the patterns "all together", "two and one" and "one each" are not: "two and one" happens $18$ times out of $27$. So count placements, not patterns. Counting the two easy cases and getting the middle one by subtraction saves work, and checking that the counts add to $27$ catches mistakes. Also notice that the answer $\frac{17}{9}$ is not a whole number, even though every jar holds a whole number of coins. An expected value is an average, so that is fine.

Instead of the table, ask three yes-or-no questions about each placement and count the "yes" answers.

• "Does the fullest jar have at least $1$ coin?" Always yes: $27$ placements.
• "At least $2$ coins?" Yes unless the coins are all in different jars: $27 - 6 = 21$ placements.
• "At least $3$ coins?" Yes only when all are together: $3$ placements.

A fullest jar with $3$ coins answers yes three times, one with $2$ coins answers yes twice, and one with $1$ coin answers yes once. So the number of yes answers equals the fullest-jar count, and the total over all placements is $27 + 21 + 3 = 51$. The average is $\frac{51}{27} = \frac{17}{9}$.