AMC 10 · 2025 · #16
Grade 7 probabilityThere are three jars. Each of three coins is placed in one of the three jars, chosen at random and independently of the placement of the other coins. What is the expected number of coins in a jar with the most coins?
Pick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: There are $3$ jars and $3$ coins; Each coin independently goes into one of the $3$ jars, each jar with probability $\frac{1}{3}$; $M$ = the number of coins in the jar that has the most coins; Answer choices: (A) $\frac{4}{3}$, (B) $\frac{13}{9}$, (C) $\frac{5}{8}$, (D) $\frac{17}{9}$, (E) $2$
Unknowns: The expected value $\mathbb{E}[M]$ — the average, over all equally likely placements, of the largest number of coins in a single jar
Understand
Restated: 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.
Givens: There are $3$ jars and $3$ coins; Each coin independently goes into one of the $3$ jars, each jar with probability $\frac{1}{3}$; $M$ = the number of coins in the jar that has the most coins; Answer choices: (A) $\frac{4}{3}$, (B) $\frac{13}{9}$, (C) $\frac{5}{8}$, (D) $\frac{17}{9}$, (E) $2$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #16 Change Focus / Count the Complement, #7 Identify Subproblems
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.
Execute — Answer: D
7.SP.C.8 Step 1 Size the sample space
- Each coin independently picks one of $3$ jars, so the number of equally likely ways to place all three coins is $3 \times 3 \times 3$.
- Every one of these outcomes has the same chance, which lets us find probabilities just by counting.
💡 Independent choices multiply: three coins each with three options give $27$ equally likely placements.
7.SP.C.8 Step 2 Name the possible maxima
- The largest count in a jar cannot be $0$ (the coins have to land somewhere) and cannot exceed $3$ (there are only $3$ coins).
- So $M$ is $1$, $2$, or $3$.
- We will count how many of the $27$ outcomes fall under each value and treat those three cases as our whole list.
💡 With only three coins the biggest pile is at least one coin and at most all three.
7.SP.C.8 Step 3 Count M = 3 (all in one jar)
- The maximum is $3$ exactly when all three coins land in the same jar.
- Pick which single jar holds all three — there are $3$ jars to choose from, and each choice is one outcome.
💡 All three coins together means you only choose the one jar they share.
7.SP.C.8 Step 4 Count M = 1 (one coin per jar)
- The maximum is $1$ exactly when every jar gets exactly one coin — no jar is doubled up.
- That is a one-to-one matching of the $3$ coins to the $3$ jars, so the count is $3!$.
💡 Perfectly spread out means the coins are just rearranged one-per-jar, which is $3! = 6$ orders.
7.SP.C.8 Step 5 Count M = 2 by complement
- Rather than build the $M=2$ configurations directly, use that the three cases exhaust all $27$ outcomes.
- Subtract the two cases already counted from the total; whatever remains must be the outcomes with a maximum of exactly $2$.
💡 Everything that is not all-same and not all-different has to be the one-pair-plus-a-single shape, so counting the leftovers is fastest.
7.SP.C.7 Step 6 Turn counts into probabilities
- Each outcome is equally likely, so the probability of each value of $M$ is its count divided by $27$.
- This is the probability model for $M$; the three probabilities add to $1$, confirming the cases partition every outcome.
💡 With equally likely outcomes, probability is just favorable count over $27$.
5.NF.B.4 Step 7 Weight each value by its probability
- The expected value is the average of $M$ weighted by how likely each value is: multiply each possible maximum by its probability.
- Because every probability shares the denominator $27$, multiplying a whole-number value by that fraction just scales the numerator.
💡 A value that happens often should count more, so you scale each value by its probability before adding.
4.NF.B.3 Step 8 Add and simplify
- Add the three fractions — they already share the denominator $27$, so add the numerators.
- Then simplify by dividing top and bottom by $3$.
- The result matches choice (D).
💡 Same denominator means just add the tops, then reduce the fraction to lowest terms.
7.SP.C.8 Each coin independently picks one of $3$ jars, so the number of equally likely w 7.SP.C.8 The largest count in a jar cannot be $0$ (the coins have to land somewhere) and 7.SP.C.8 The maximum is $3$ exactly when all three coins land in the same jar. Pick which 7.SP.C.8 The maximum is $1$ exactly when every jar gets exactly one coin — no jar is doub 7.SP.C.8 Rather than build the $M=2$ configurations directly, use that the three cases ex 7.SP.C.7 Each outcome is equally likely, so the probability of each value of $M$ is its c 5.NF.B.4 The expected value is the average of $M$ weighted by how likely each value is: m 4.NF.B.3 Add the three fractions — they already share the denominator $27$, so add the nu Review
Reasonableness: The answer $\frac{17}{9} \approx 1.89$ 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 $\frac{6}{27}$) 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.
Alternative: Tool #16 again, via the tail-sum identity for a positive whole-number variable: $\mathbb{E}[M] = P(M\ge 1) + P(M\ge 2) + P(M\ge 3)$. Here $P(M\ge 1)=1$, $P(M\ge 2) = 1 - P(M{=}1) = 1 - \frac{6}{27} = \frac{21}{27}$, and $P(M\ge 3) = P(M{=}3) = \frac{3}{27}$. Adding gives $1 + \frac{21}{27} + \frac{3}{27} = 1 + \frac{24}{27} = 1 + \frac{8}{9} = \frac{17}{9}$, matching (D).
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting the $3^3 = 27$ equally likely placements and sorting them into the three cases $M=1$ ($6$), $M=2$ ($18$), and $M=3$ ($3$).)7.SP.C.7Develop probability models and use them to find probabilities of events (Turning the case counts into the probability model $P(M{=}1)=\frac{6}{27}$, $P(M{=}2)=\frac{18}{27}$, $P(M{=}3)=\frac{3}{27}$, and checking they sum to $1$.)5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a fraction (Weighting each possible maximum by its probability — multiplying the whole-number values $1$, $2$, $3$ by their fractional probabilities in the expected-value sum.)4.NF.B.3Understand a fraction with numerator greater than one as sum of unit fractions (Adding the like-denominator fractions $\frac{6}{27}+\frac{36}{27}+\frac{9}{27}=\frac{51}{27}$ and reducing to $\frac{17}{9}$.)
⭐ 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 $\frac{17}{9}$, a little under $2$.
⭐ 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 $\frac{17}{9}$, a little under $2$.
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