2025 AMC 10A Problem 16

Each coin picks one of three jars, so there are $3^3 = 27$ equally likely placements. The largest jar count is $3$, $2$, or $1$; count the placements of each kind.

Largest count $3$. All three coins land in the same jar: $3$ placements, one per jar.

Largest count $1$. The coins are spread $1, 1, 1$, which means the three coins go to three different jars: $3! = 6$ placements.

Largest count $2$. Everything else: $27 - 3 - 6 = 18$ placements. (Directly: choose the jar holding two coins, $3$ ways; the jar holding one, $2$ ways; and which coin is alone, $3$ ways, giving $3 \cdot 2 \cdot 3 = 18$.)

The expected value is therefore
$$\frac{3 \cdot 3 + 2 \cdot 18 + 1 \cdot 6}{27} = \frac{9 + 36 + 6}{27} = \frac{51}{27} = \frac{17}{9}.$$

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

The coins are distinguishable and each of the $27$ placements is equally likely, so the safest route is to count placements rather than to reason about "shapes", which are not equally likely at all: the $2$-$1$-$0$ shape accounts for two thirds of the outcomes while $3$-$0$-$0$ accounts for only $3$ of $27$. Counting the two extreme shapes and taking the rest by subtraction keeps the arithmetic short and also self-checks, since the three counts must total $27$. Note also that the expectation $\frac{17}{9}$ is not attainable as an outcome; an expected value need not be one of the possible values.

Use the tail formula $E[M] = \sum_{m\ge1} P(M \ge m)$ for the nonnegative integer maximum $M$. Here $P(M \ge 1) = 1$, $P(M \ge 2) = 1 - P(\text{all different}) = 1 - \frac{6}{27} = \frac{21}{27}$, and $P(M \ge 3) = \frac{3}{27}$. Adding, $E[M] = 1 + \frac{21}{27} + \frac{3}{27} = \frac{27 + 24}{27} = \frac{17}{9}$.

• Treating the three shapes 3-0-0, 2-1-0 and 1-1-1 as equally likely and averaging 3, 2, 1 to get 2, choice (E).
• Counting the $1$-$1$-$1$ case as a single placement instead of $3! = 6$, which leaves the three counts failing to sum to $27$.
• Computing the expected number of coins in one fixed jar, which is $3 \cdot \frac13 = 1$, rather than in the fullest jar.