# 2023 AMC 12B Problems/Problem 19

The following problem is from both the 2023 AMC 10B #21 and 2023 AMC 12B #19, so both problems redirect to this page.

## Problem

Each of $2023$ balls is randomly placed into one of $3$ bins. Which of the following is closest to the probability that each of the bins will contain an odd number of balls?

$\textbf{(A) } \frac{2}{3} \qquad\textbf{(B) } \frac{3}{10} \qquad\textbf{(C) } \frac{1}{2} \qquad\textbf{(D) } \frac{1}{3} \qquad\textbf{(E) } \frac{1}{4}$

## Important Clarification

Stars and Bars does not provide an exact probability. However, it does provide a good estimate for the approximate answer as on average, the number of arrangements will be almost the same when each container has an odd # of balls or when each container has an even # of balls. (similar binomial distributions)

The above clarification is a good explanation of why the stars and bars argument still yields an expression sufficiently closed to $\frac{1}{4}$. However, I think the content of this clarification should be clearly explained WITHIN any stars and bars solution. Even with this clarification, this page is currently a redundant mess of poorly explained fakesolves:

Solutions 1 and 2 both use stars and bars to arrive at an incorrect probability and fail to explain why their method is a good approximation of the actual probability.

The explanation in solution 4 does not make sense at all.

The nature of this problem makes it so that pretty much any approach will arrive at an answer close to $\frac{1}{4}$. However, I'm concerned that the current state of this page will make most readers come away with a worsened understanding of the structure of the problem, especially since the first two solutions are both fakesolves. This page should be fixed ASAP.

Please don't delete this message without addressing any of my concerns. If you have any questions, feel free to PM me. ~ CT17

(Solution 5 by Dissmo does a very good job at explaining Solution 4.) -Multpi12

## Solution 1

Because each bin will have an odd number, they will have at least one ball. So we can put one ball in each bin prematurely. We then can add groups of 2 balls into each bin, meaning we now just have to spread 1010 pairs over 3 bins. This will force every bin to have an odd number of balls. Using stars and bars, we find that this is equal to $\binom{1012}{2}$. This is equal to $\frac{1012\cdot1011}{2}$. The total amount of ways would also be found using stars and bars. That would be $\binom{2023+3-1}{3-1} = \binom{2025}{2}$. Dividing our two quantities, we get $\frac{1012 \cdot 1011 \cdot 2}{2 \cdot 2025 \cdot 2024}$. We can roughly cancel $\frac{1012 \cdot 1011}{2025 \cdot 2024}$ to get $\frac{1}{4}$. The 2 in the numerator and denominator also cancels out, so we're left with $\boxed{\frac{1}{4}}$.

Note: My solution does not provide an exact probability, but is a good estimate, which is how this problem was designed. Most of the solutions on this page will give a decent estimate. This is due to the fact that the binomial expansions are symmetric for even vs odd. Try experimenting to determine why this is true.

## Solution 2 (Systematic Algebraic Way)

Suppose the numbers are $a_1$, $a_2$, and $a_3$. First, we try to calculate the amount of ways for all three balls to be placed in a bin so the number of balls in each bin is odd. $a_1+a_2+a_3=2023$ and each bin has at least one ball because they are positive odd numbers. Changing the equation, we see that $\frac{[(a_1)+1]}{2}+\frac{[(a_2)+1]}{2}+\frac{[(a_3)+1]}{2}=\frac{(2023+3)}{2}$. Let $a_1+1=2b_1$, $a_2+1=2b_2$, and $a_3+1=2b_3$. Thus $b_1+b_2+b_3=1013$. We can also see that $b_1$, $b_2$, and $b_3$ are all positive. Using the positive version of stars and bars, we get ${1013-1 \choose 3-1}$ = ${1012 \choose 2}$ choices.

Now, we want to find the total amount of cases. Using the non-negative version of stars and bars, we find that the total is ${2023+3-1 \choose 3-1}$=${2025 \choose 2}$.

Now we need to calculate ${1012 \choose 2}$/${2025 \choose 2}$, which is just $\frac{1012 \cdot 1011 \cdot 2}{2 \cdot 2025 \cdot 2024}$. Cancelling the twos, we get $\frac{1012 \cdot 1011}{2025 \cdot 2024}$. This is roughly equal to $\frac{1}{4}$. The answer is $\boxed{\textbf{(E)} \frac{1}{4}}$.

~Aopsthedude

## Solution 3

Having 2 bins with an odd number of balls means the 3rd bin also has an odd number. The probability of the first bin having an odd number of balls is $\frac{1}{2}$, since even and odd have roughly the same probability. The probability of the second bin having an odd number of balls is also $\frac{1}{2}$ for the same reason. If both of these bins have an odd number of balls, the number of balls remaining for the third bin is also odd. Therefore the probability is $\frac{1}{2} \cdot \frac{1}{2} = \boxed{\textbf{(E)} \frac{1}{4}}$.

~Yash C

## Solution 4

We first examine the possible arrangements for parity of number of balls in each box for $2022$ balls.

If a $E$ denotes an even number and a $O$ denotes an odd number, then the distribution of balls for $2022$ balls could be $EEE,EOO,OEO,$ or $OOE$. With the insanely overpowered magic of cheese, we assume that each case is about equally likely.

From $EEE$, it is not possible to get to all odd by adding one ball; we could either get $OEE,EOE,$ or $EEO$. For the other $3$ cases, though, if we add a ball to the exact right place, then it'll work.

For each of the working cases, we have $1$ possible slot the ball can go into (for $OEO$, for example, the new ball must go in the center slot to make $OOO$) out of the $3$ slots, so there's a $\dfrac13$ chance. We have a $\dfrac34$ chance of getting one of these working cases, so our answer is $\dfrac34\cdot\dfrac13=\boxed{\textbf{(E) }\dfrac14.}$

~pengf ~Technodoggo

## Solution 5

2023 is an arbitrary large number. So, we proceed assuming that an arbitrarily large number of balls have been placed.

For an odd-numbered amount of balls case, the 3 bins can only be one of these 2 combinations:

$OEE$ ($OEE$,$EOE$,$EEO$)

$OOO$ ($OOO$)

Let the probability of achieving the $OOO$ case to be $P(OOO) = p$ and any of the $OEE$ permutations to be $P(OEE) = 1-p$.

Because the amount of balls is arbitrarily large, $P(OOO) = p$ even after another two balls are be placed.

There are two cases for which placing another two balls results in $OOO$:

$OOO$: The two balls are placed in the same bin ($OOO\to OOE\to OOO$)

$OEE$: The two balls are placed in the two even bins ($OEE\to OOE \to OOO$)

So,

$P(OOO) = P(OOO) * \frac{1}{3} + P(OEE) * \frac{2}{3} * \frac{1}{3}$

$p = p * \frac{1}{3} + (1-p) * \frac{2}{3} * \frac{1}{3}$

$\frac{8}{9}p = \frac{2}{9}$

$p = \boxed {\textbf {(E)} \frac {1}{4}}$

~Dissmo

## Solution 6

We use the generating functions approach to solve this problem. Define $\Delta = \left\{ \left( a, b, c \right) \in \Bbb Z_+: a+b+c = 2023 \right\}$.

We have $$\left( x + y + z \right)^{2023} = \sum_{(a,b,c) \in \Delta} \binom{2023}{a,b,c} x^a y^b z^c .$$

First, we set $x \leftarrow 1$, $y \leftarrow 1$, $z \leftarrow 1$. We get $$3^{2023} = \sum_{(a,b,c) \in \Delta} \binom{2023}{a,b,c} 1 . \hspace{1cm} (1)$$

Second, we set $x \leftarrow 1$, $y \leftarrow -1$, $z \leftarrow 1$. We get $$1 = \sum_{(a,b,c) \in \Delta} \binom{2023}{a,b,c} (-1)^b . \hspace{1cm} (2)$$

Third, we set $x \leftarrow 1$, $y \leftarrow 1$, $z \leftarrow -1$. We get $$1 = \sum_{(a,b,c) \in \Delta} \binom{2023}{a,b,c} (-1)^c . \hspace{1cm} (3)$$

Fourth, we set $x \leftarrow 1$, $y \leftarrow -1$, $z \leftarrow -1$. We get $$-1 = \sum_{(a,b,c) \in \Delta} \binom{2023}{a,b,c} (-1)^{b+c} . \hspace{1cm} (4)$$

Taking $\frac{(1)-(2) - (3)+(4)}{4}$, we get \begin{align*} \frac{3^{2023} - 1 - 1 + (-1)}{4} & = \frac{1}{4} \sum_{(a,b,c) \in \Delta} \binom{2023}{a,b,c} \left( 1 - (-1)^b - (-1)^c + (-1)^{b+c} \right) \\ & = \frac{1}{4} \sum_{(a,b,c) \in \Delta} \binom{2023}{a,b,c} \left( 1 - (-1)^b \right) \left( 1 - (-1)^c \right) \\ & = \sum_{\substack{(a,b,c) \in \Delta \\ a, b, c \mbox{ are odds}}} \binom{2023}{a,b,c} . \end{align*}

The last expression above is the number of ways to get all three bins with odd numbers of balls. Therefore, this happens with probability \begin{align*} \frac{\frac{3^{2023} - 1 - 1 + (-1)}{4}}{3^{2023}} & \approx \boxed{\textbf{(E) } \frac{1}{4}}. \end{align*}

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

## Solution 7

Four even-odd splittings divides $2023$ in to three, namely $(O,O,O)$, $(E,E,O)$, $(E,O,E)$, and $(O,E,E)$. Here if we define a "move" as relocated one ball, then we will notice in each case, that a random "move" will be evenly likely to be one of the other three splittings. Hence by Group theory (or by intuition), we will find the structure of this splitting is $V_4$ group, and it's symmetric for all four elements in this Group.

Thus, no matter what is the initial starting point, four cases will be evenly likely to appear when repeated many times. The answer is $\boxed{\textbf{(E)} \frac{1}{4}}$.

~Prof. Joker

## Solution 8 (Exact Probability)

Let $p(n)$ be the probability of all 3 bins having odd number of balls when we randomly put $2n+1$ balls into these bins. We need find out $p(1011)$. Clearly $p(0)=0$. For $n>1$, there are only two scenarios for the first $2n-1$ balls: Case 1 all 3 bins are odd; Case 2 only 1 bin is odd and the other 2 bins are even.

For case 1 which has a probability $p(n-1)$, in order to get all 3 bins to be odd, we need put the last 2 balls in the same bin. This has a probability of $\frac{1}{3}$.

For case 2 which has a probability of $1-p(n-1)$, in order to get all 3 bins to be odd, we need to put the last 2 balls into the 2 even bins. The first one has a probability of $\frac{2}{3}$ to be put in an even bin, and the second has a probability of $\frac{1}{3}$ to be put the in the last even bin, so the probability is $\frac{2}{9}$ to get all 3 bins to be odd in this case.

So we have $p(n)=\frac{1}{3} p(n-1) + \frac{2}{9} (1-p(n-1)) = \frac{1}{9} p(n-1) + \frac{2}{9}$.

Solving for the fixed point: $x=\frac{1}{9} x + \frac{2}{9}$ yields $x=\frac{1}{4}$, and this helps us the rewrite the above as $p(n) - \frac{1}{4} = \frac{1}{9} (p(n-1) - \frac{1}{4} )$.

Therefore $p(n)-\frac{1}{4} = \frac{1}{9^n} (p(0) - \frac{1}{4})$. Using $p(0)=0$, we have $p(n)=\frac{1}{4} (1 - \frac{1}{9^n})$.

The probability for 2023 balls is $\frac{1}{4} (1 - \frac{1}{9^{1011}})$, and this is closest to $\boxed{\textbf{(E)} \frac{1}{4}}$.

~Qing

## Solution 9 (Recursion, rigorous)

We can use recursion to figure out an expression that determines the probability of all the bins being odd.

Define $P_n$ to be the probability that with $2n-1$ balls, randomly inserting them makes all 3 bins have an odd count. We would first want to find $P_{n+1}=f(P_n)$ for some function $f$.

Now, we can say that if we have an O-O-O combination, where O denotes odd and E denotes even, then to add 2 more balls to keep it O-O-O, we need to put them in the same bin. This would result in a $\frac{1}{3}$ probability of staying O-O-O.

If we have an E-E-O combination, then we need to add 2 more balls, one in each of the even bins. This would result in a $\frac{2}{9}$ probability of turning to O-O-O. Now, we can write a function of $P_{n+1}$ in terms of $P_n$.

$P_{n+1}=\frac{1}{3}P_n+\frac{2}{9}-\frac{2}{9}P_n=\frac{1}{9}P_n+\frac{2}{9}$

Now, we want to express $P_{n+1}=g(n)$ for some function $g$. Here, the $\frac{2}{9}$ is a problem because now we can't just put it as a geometric sequence. However, we can rewrite it as followed:

$P_{n+1}=\frac{1}{9}P_n+\frac{2}{9}$

$P_{n+1}-\frac{1}{4}=\frac{1}{9}(P_n-\frac{1}{4})$

Now, we can let $Q_n=P_n-\frac{1}{4}$. We can rewrite that as $Q_{n+1}=\frac{1}{9}Q_n$.

We can note that $Q_1=P_1-\frac{1}{4}$. Since $P_1=0$ (you only have one ball, so two of the bins have to be even), $Q_1=-\frac{1}{4}$. Therefore, $Q_n=Q_1(\frac{1}{9})^{n-1}=-\frac{1}{4} \cdot (\frac{1}{9})^{n-1}$.

Substituting, we can find that $P_n=-\frac{1}{4} \cdot (\frac{1}{9})^{n-1}+\frac{1}{4}$.

2023 is the 1012th odd number, so we can substitute $n=1012$ into the equation.

$-\frac{1}{4} \cdot (\frac{1}{9})^{1011}+\frac{1}{4} \approx 0+\frac{1}{4}=\frac{1}{4}$.

Therefore, the answer is $\boxed{\textbf{(A) }\frac{1}{4}}$

~ethanzhang1001

## Solution 10 (Generating Functions)

Define \begin{align*} [x^{2023}]G(x)&=(x^1 + x^3 + ... + x^{2023} + ...)^3 \\ [x^{2023}]G(x)&= x^3(1+x^2+x^4+...+x^{2022}+...)^3 \\ [x^{2020}]G(x)&= (\frac{1}{1-x^2})^3 \\ [x^{2020}]G(x)&= (1-x^2)^{-3} \\ [x^{2020}]G(x)&= (1 + \frac{3 \cdot 2}{2 \cdot 1}(x^2)^1 + ... \frac{1012 \cdot 1011}{2 \cdot 1}(x^2)^{1010}) \\ [x^{2020}] &= \frac{1012 \cdot 1011}{2 \cdot 1} \end{align*}

Similarly we can compute the total number of cases which is: $\frac{2025\cdot 2022}{2 \cdot 1}$

So, the probability is: \begin{align*} &= \frac {\frac{1012 \cdot 1011}{2 \cdot 1}}{\frac{2025\cdot 2024}{2 \cdot 1}} \\ &= \frac{1012 \cdot 1011}{2025 \cdot 2024} \\ &\approx \frac{1}{4} \\ \end{align*}

Therefore, the answer is $\boxed{\textbf{(E) }\frac{1}{4}}$

## Solution 11(Close and Simple)

By Stars and Bars, the total number of ways to place the 2023 balls into 3 distinct bins is 2025C2.

Let's do some case work now.

If Bin A has 1 ball, Bin 2 and Bin 3 have how 2022 balls total.

If Bin 2 has 1 ball, bin 3 has 2021.

If Bin 2 has 3 balls, Bin 3 has 2019 balls.

So, in total, there are $\frac{2021-1}{2}+1=1011$ ways for case 1.


By symmatry, if Bin A now has 3 balls, there are 1010 ways. Continuing down to if Bin A has 2021 balls, we have the total amount of successful ways is 1011+1010+1009+....+1. So the probability is closest to $\frac{\frac{1011*1012}{2}}{2025C2}$, which gives us $\boxed{\textbf{(E) }\frac{1}{4}}$.

~MC