AMC 10 · 2025 · #16

Grade 7 probability

Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count

expected-valueprobability-basiccombinations-basic casework ↑ Prerequisites: expected-valueprobability-basic

📏 Long solution 💡 2 insights

Problem

Three coins are dropped one at a time into three jars. For each coin, the jar is chosen at random, each jar equally likely, and the coins are placed independently. After all three coins land, look at whichever jar holds the most coins and record that count. Find the long-run average of that largest count.

Pick an answer.

(A)

$\frac{4}{3}$

(B)

$\frac{13}{9}$

(C)

$\frac{5}{8}$

(D)

$\frac{17}{9}$

(E)

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

Expected value is a weighted average: to get it we need the probability of each value M can take, and to get those probabilities we must count outcomes. That is exactly Tool #2 (Make a Systematic List) — the largest count M is only ever 1, 2, or 3, so we split the 27 outcomes into those three buckets and count each. Two buckets (M=1 all-different, M=3 all-same) are quick to count directly, so rather than build the messy M=2 bucket by hand we use Tool #16 (Change Focus / Count the Complement): whatever is left after removing the easy two must be the M=2 bucket. Tool #7 (Identify Subproblems) frames the whole solve as 'first the counts, then the probabilities, then the weighted sum,' keeping the arithmetic in tidy stages.

1STEP 1

Size the sample space

Each coin independently picks one of 3 jars, so all 27 placements are equally likely — probabilities become pure counting.

3³ = 27 equally likely outcomes

2STEP 2

Name the possible maxima

Coins must land somewhere and there are only 3 of them, so the largest count M is 1, 2, or 3 — three buckets to fill.

M ∈ {1, 2, 3}

3STEP 3

Count M = 3 (all in one jar)

The maximum is 3 only when all three coins share one jar, and there are just 3 jars to pick from.

#{M = 3} = 3

4STEP 4

Count M = 1 (one coin per jar)

The maximum is 1 only when every jar gets one coin — a one-to-one matching, so 3! = 6 ways.

#{M = 1} = 3! = 6

5STEP 5

Count M = 2 by complement

The three cases cover all 27, so subtract the two already counted: the remaining 18 outcomes have a maximum of exactly 2.

#{M = 2} = 27 - 3 - 6 = 18

6STEP 6

Turn counts into probabilities

Outcomes are equally likely, so each probability is that case's count over 27, and the three add to 1 — the cases miss nothing.

P(M=1)=6/27, P(M=2)=18/27, P(M=3)=3/27; 6/27+18/27+3/27=1

7STEP 7

Weight each value by its probability

The expected value is a weighted average: multiply each possible maximum by its probability, all of which share the denominator 27.

E[M] = 1·6/27 + 2·18/27 + 3·3/27 = 6/27+36/27+9/27

8STEP 8

Add and simplify

Same denominator, so add the numerators to get 51 over 27, then divide top and bottom by 3 to reach $\frac{17}{9}$ — choice (D).

E[M] = (6+36+9)/27 = 51/27 = 17/9 → (D)

Answer

17/9

The answer 17/9 ≈ 1.89 sits between the smallest possible maximum (1) and the largest (3), so it is in range. It must be at least 1 because some jar always holds the most, and it lands just under 2 — sensible, because the perfectly-spread case M=1 (probability 6/27) pulls the average down below the very common M=2. Choice (E) 2 is the trap for anyone who forgets that the all-different outcomes drag the mean down. The counts also check out: 3 + 6 + 18 = 27, all outcomes accounted for.

💡Key takeaway

To find an average outcome, list every case, count how likely each one is, then multiply each value by its chance and add — the biggest pile of three coins averages 17/9, a little under 2.

Size the sample space
Name the possible maxima
Count M = 3 (all in one jar)
Count M = 1 (one coin per jar)
Count M = 2 by complement
Turn counts into probabilities
Weight each value by its probability
Add and simplify