2025 AMC 10A Problems/Problem 16

Problem

There 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?

$\textbf{(A) } \frac{4}{3}\qquad\textbf{(B) } \frac{13}{9}\qquad\textbf{(C) } \frac{5}{8}\qquad\textbf{(D) } \frac{17}{9}\qquad\textbf{(E) } 2$

Solution 1

We have three coins and three jars. Each coin is placed independently and randomly into one of the jars. Let $M$ be the maximum number of coins in any jar. We want to compute the expected value of $M$ .

Step 1: Count total outcomes

Each coin has 3 choices, so the total number of equally likely placements is $3^3 = 27$ .

Step 2: Casework on the maximum number of coins

Case 1: $M = 1$ . This occurs when each jar has exactly one coin. There are $3! = 6$ assignments of coins to jars. Hence, $\Pr(M=1) = \frac{6}{27} = \frac{2}{9}$ .

Case 2: $M = 3$ . This occurs when all three coins fall into the same jar. There are 3 jars to choose from, so $\Pr(M=3) = \frac{3}{27} = \frac{1}{9}$ .

Case 3: $M = 2$ . This occurs when one jar has 2 coins, another jar has 1 coin, and the last jar has 0 coins. We can choose which jar gets 2 coins in 3 ways, which jar gets 1 coin in 2 ways, and which jar is alone in 3 ways. Therefore, there are $3 \cdot 2 \cdot 3 = 18$ outcomes. Thus, $\Pr(M=2) = \frac{18}{27} = \frac{2}{3}$ .

Step 3: Compute the expected value The expected value of $M$ is $\mathbb{E}[M] = 1 \cdot \frac{2}{9} + 2 \cdot \frac{2}{3} + 3 \cdot \frac{1}{9}$ . Converting everything to ninths, we have $\mathbb{E}[M] = \frac{2}{9} + \frac{12}{9} + \frac{3}{9} = \frac{17}{9}$ .

Hence, the expected number of coins in the jar with the most coins is $\boxed{\text{(D) }\frac{17}{9}}$ .

$Pr$ =Probability of

$\mathbb{E}$ =Expected value

-Boywithnuke(Goal to 10 followers)

~ Minor edits by SixthGradeBookWorm927 ~ Minor edit by AlgeBruh16

Another way of finding cases of M = 2

As described in the solution, there are $3^3=27$ ways of distributing the coins into the $3$ jars. Because there are $6$ ways for M=1 and $3$ ways for M=3, there are $27 - 6 - 3 = 18$ ways for M=2.

Solution 2 (Fast)

Assuming all jars and coins are distinct, there are $27$ total outcomes. $3!=6$ of them distribute exactly, and thus a max of, $1$ per cup, $3$ ways to choose a jar to put all $3$ coins in for a max of $3$ . This leaves $27-6-3=18$ for a max of $2$ , so the expected value is $\frac{6 \cdot 1 + 18 \cdot 2 + 3 \cdot 3}{27} = \boxed{\text{(D) }\frac{17}{9}}$

~megaboy6679

Solution 3 (Weighted Probabilities)

WLOG label the jars 1,2,3 and the coins a,b,c. The probability for a given coin to land in a given jar is $\frac{1}{3}$ The number of ways for all three coins to land in the same jar is 3, either all in jar 1, all in jar 2, or all in jar 3. Next, the number of ways to have two coins in one jar and one in another in another is 18. Finally, since there are 27 ways, the last weight is $27-18-3=6$ Thus $\frac{1}{27} \times 3 \times 3 + \frac{1}{27} \times 2 \times 18 + \frac{1}{27} \times 1 \times 6$ Thus we get $\boxed{\text{(D) }\frac{17}{9}}$

~Aeioujyot

Video Solution (In 1 Min)

https://youtu.be/8iDugBBzei8?si=QXHiwxmA5eqXyEZY ~ Pi Academy

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search