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?

Solution 1 (Case Work)

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

Step 1: Count total outcomes

Each coin has choices, so the total number of equally likely placements is .

Step 2: Casework on the maximum number of coins

Case 1: . This occurs when each jar has exactly one coin. There are assignments of coins to jars. Hence, .

Case 2: . This occurs when all three coins fall into the same jar. There are jars to choose from, so .

Case 3: . This occurs when one jar has coins, another jar has coin, and the last jar has coins. We can choose which jar gets coins in ways, which jar gets coin in ways, and which coins out of the go into the jar with two coins, so we multiply by , which is just (note we don't have to do this for the earlier cases because for case , all coins go into one jar, and for case , the factorial already accounts for that). Therefore, there are outcomes. Thus, .

Step 3: Compute the expected value The expected value of is . Converting everything to ninths, we have .

Hence, the expected number of coins in the jar with the most coins is .

=Probability of

=Expected value

~Boywithnuke(Goal to 10 followers)

~aldzandrtc(Fixed mistake, original = "Case 3: . 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 outcomes. Thus, ." After we choose which jar gets 2 coins, which jar gets one, we don't need to multiply by 3 to see which jar gets 0, because once the first 2 are picked, the last jar is set. However, the number is coincidentally correct, because you need assign which 2 coins go into the jar with 2 coins, so you multiply by 3 choose 2, which is 3. Feel free to fix it up, because I am not experienced with latex)

~ Minor edits by SixthGradeBookWorm927 ~ Minor edit by AlgeBruh16

~ Minor edit by jeffyang2025

Another way of finding cases of M = 2

As described in the solution, there are ways of distributing the coins into the jars. Because there are ways for M=1 and ways for M=3, there are ways for .

Solution 2 (Fast)

Assuming all jars and coins are distinct, there are total outcomes. of them distribute exactly, and thus a max of, per cup, ways to choose a jar to put all coins in for a max of . This leaves for a max of , so the expected value is

~megaboy6679

Solution 3 (Circular Counting)

In circular counting, expected value is determined as Where \( \mathbb{E}_C n(A) \) is the expected value, \( C n(A) \) is the circular gap theorem, \( \mathbb{T} \) is the total number of cases, and \( n(A) \) is the invalid cases.

Arrange the three jars and three coins into a circle of 6 elements. We want to count all the ways in which no 2 or no 3 coins are next to one another. This can easily be done using the circular gap theorem, which states Where \( n \) is the number of elements in our circle and \( k \) is the number of non adjacencies, in which we wish to count \( n = 6 \) and \( k \in \{1,2,3\} \) (The reason we have only 1, 2, and 3 is because if we have \( n=6 \) and \( k \ge 4 \), we see that the combinatorial part of \( C n(A) \) becomes invalid).

So we begin counting.

For \( k=1 \), using the formula, we have \( \frac{6}{1} \binom{4}{0} \) which is 6.

For \( k=2 \), using the formula, we have \( \frac{6}{2} \binom{3}{1} \), which is 9.

For \( k=3 \), using the formula, we have \( \frac{6}{3} \binom{2}{2} \) which is 2.

Thus we have \( C_6(k \in \{1,2,3\}) = 6+9+2 = 17 \).

Now we calculate the denominator of \( \mathbb{E}_C n(A) \). We see the total number of cases is \( 3^3 = 27 \). So we can find \( \mathbb{T} - C n(A) \). We check for \( n(A) \), in which through common sense, we can see that if the triplet represents the number of coins given to each jar, the triplet doesn't satisfy the condition, as all jars have an equal number of coins. Therefore \( n(A)=1 \).

Our answer is

~Pinotation

Solution 4 (Weighted Probabilities)

WLOG label the jars as and the coins as . The probability for a given coin to land in a given jar is The number of ways for all three coins to land in the same jar is , either all in jar , all in jar , or all in jar . Next, the number of ways to have two coins in one jar and one in another in another is . Finally, since there are ways, the last weight is Thus Thus we get

~Aeioujyot

~Minor edit by jeffyang2025

Solution 5

Denote the expected number of coins in a jar with the most coins as

   Process
   └── The 1st coin must add 1 to . Situation: (1, 0, 0)
       ├── The 2nd coin has  chance to add 1 to . Situation: (2, 0, 0)
       │   └── The 3rd coin has  chance to add 1 to . Situation: (3, 0, 0)
       └── The 2nd coin has  chance to leave  unchanged. Situation: (1, 1, 0)
           └── The 3rd coin has  change to add 1 to . Situation: (2, 1, 0)

Hence,

~reda_mandymath

Solution 6

Create a tree diagram and add up the probabilities.

Note: (It does not matter which jar you pick at the head of the diagram so long a you stay consistent) ~Rames_Jong

Chinese Video Solution

https://www.bilibili.com/video/BV1nhkUByEzM/

~metrixgo

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

Video Solution by grogg007

https://youtu.be/CCYoHk2Af34?start=1520

Video Solution by Daily Dose of Math

https://youtu.be/WZmTGRl4O84

~Thesmartgreekmathdude

See Also

• AMC 10
• AMC 10 Problems and Solutions
• Mathematics competitions
• Mathematics competition resources

These problems are copyrighted © by the Mathematical Association of America.