# 2021 AMC 12B Problems/Problem 23

## Problem

Three balls are randomly and independantly tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $i$ is $2^{-i}$ for $i=1,2,3,....$ More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is $\frac pq,$ where $p$ and $q$ are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins $3,17,$ and $10.$) What is $p+q?$ $\textbf{(A) }55 \qquad \textbf{(B) }56 \qquad \textbf{(C) }57\qquad \textbf{(D) }58 \qquad \textbf{(E) }59$

## Solution

"Evenly spaced" just means the bins form an arithmetic sequence.

Suppose the middle bin in the sequence is $x$. There are $x-1$ different possibilities for the first bin, and these two bins uniquely determine the final bin. Now, the probability that these $3$ bins are chosen is $6\cdot 2^{-3x} = 6\cdot \frac{1}{8^x}$, so the probability $x$ is the middle bin is $6\cdot\frac{x-1}{8^x}$. Then, we want the sum \begin{align*} 6\sum_{x=2}^{\infty}\frac{x-1}{8^x} &= \frac{6}{8}\left[\frac{1}{8} + \frac{2}{8^2} + \frac{3}{8^3}\cdots\right]\\ &= \frac34\left[\left(\frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3}\right) + \left(\frac{1}{8^2} + \frac{1}{8^3} + \frac{1}{8^4}\right) + \cdots\right]\\ &= \frac34\left[\frac17\cdot \left(1 + \frac{1}{8} + \frac{1}{8^2} + \frac{1}{8^3}\right)\right]\\ &= \frac34\cdot \frac{8}{49}\\ &= \frac{6}{49} \end{align*} The answer is $6+49=\boxed{\textbf{(A) }55}.$

## Solution 2

As in solution 1, note that "evenly spaced" means the bins are in arithmetic sequence. We let the first bin be $a$ and the common difference be $d$. Further note that each $(a, d)$ pair uniquely determines a set of 3 bins.

We have $1 \leq a \leq \infty$ because the leftmost bin in the sequence can be any bin, and $1 \leq d \leq \infty$, because the bins must be distinct.

This gives us the following sum for the probability: \begin{align*} 6 \sum_{a=1}^{\infty} \sum_{d=1}^{\infty} 2^{-3a-3d} &= 6 \sum_{a=1}^{\infty} \sum_{d=1}^{\infty} 2^{-3a} \cdot 2^{-3d} \\ &= 6 \left( \sum_{a=1}^{\infty} 2^{-3a} \right) \left( \sum_{d=1}^{\infty} 2^{-3d} \right) \\ &= 6 \left( \sum_{a=1}^{\infty} 8^{-a} \right) \left( \sum_{d=1}^{\infty} 8^{-d} \right) \\ &= 6 \left( \frac{1}{7} \right) \left( \frac{1}{7} \right) \\ &= \frac{6}{49} .\end{align*} Therefore the answer is $6 + 49 = 55$, which is choice (A).

-Darren Yao

## Solution 3

This is a slightly messier variant of solution 2. If the first ball is in bin $i$ and the second ball is in bin $j>i$, then the third ball is in bin $2j-i$. Thus the probability is \begin{align*} 6\sum_{i=1}^{\infty}\sum_{j=i+1}^\infty2^{-i}2^{-j}2^{-2j+i}&=6\sum_{i=1}^{\infty}\sum_{j=i+1}^\infty2^{-3j}\\ &=6\sum_{i=1}^{\infty}\left(\frac{2^{-3(i+1)}}{1-\tfrac{1}{8}}\right)\\ &=6\sum_{i=1}^\infty\frac{8}{7}\cdot2^{-3}\cdot2^{-3i}\\ &=\frac{6}{7}\sum_{i=1}^\infty2^{-3i}\\ &=\frac{6}{7}\frac{2^{-3}}{1-\tfrac18} = \frac{6}{49}. \end{align*}

## Solution 4 (Table: Detailed Explanation of Solution 2) $$\begin{array}{c|c|c|c} & & & \\ [-1.5ex] \textbf{Exactly }\boldsymbol{n}\textbf{ Spaces Apart} & \textbf{Bin \#s} & \textbf{Expression} & \textbf{Prob. of One Such Perm.} \\ [1ex] \hline\hline & & & \\ [-1.5ex] n=1 & 1,2,3 & 2^{-1}\cdot2^{-2}\cdot2^{-3} & 2^{-6} \\ [1ex] & 2,3,4 & 2^{-2}\cdot2^{-3}\cdot2^{-4} & 2^{-9} \\ [1ex] & 3,4,5 & 2^{-3}\cdot2^{-4}\cdot2^{-5} & 2^{-12} \\ [1ex] & 4,5,6 & 2^{-4}\cdot2^{-5}\cdot2^{-6} & 2^{-15} \\ [1ex] & \cdots & \cdots & \cdots \\ [1ex] \hline & & & \\ [-1.5ex] n=2 & 1,3,5 & 2^{-1}\cdot2^{-3}\cdot2^{-5} & 2^{-9} \\ [1ex] & 2,4,6 & 2^{-2}\cdot2^{-4}\cdot2^{-6} & 2^{-12} \\ [1ex] & 3,5,7 & 2^{-3}\cdot2^{-5}\cdot2^{-7} & 2^{-15} \\ [1ex] & 4,6,8 & 2^{-4}\cdot2^{-6}\cdot2^{-8} & 2^{-18} \\ [1ex] & \cdots & \cdots & \cdots \\ [1ex] \hline & & & \\ [-1.5ex] n=3 & 1,4,7 & 2^{-1}\cdot2^{-4}\cdot2^{-7} & 2^{-12} \\ [1ex] & 2,5,8 & 2^{-2}\cdot2^{-5}\cdot2^{-8} & 2^{-15} \\ [1ex] & 3,6,9 & 2^{-3}\cdot2^{-6}\cdot2^{-9} & 2^{-18} \\ [1ex] & 4,7,10 & 2^{-4}\cdot2^{-7}\cdot2^{-10} & 2^{-21} \\ [1ex] & \cdots & \cdots & \cdots \\ [1ex] \hline & & & \\ [-1.5ex] \cdots & \cdots & \cdots & \cdots \\ [1ex] \end{array}$$

Since three balls have $3!=6$ permutations, the requested probability is \begin{align*} 6\left[\sum_{k=0}^{\infty}2^{-6-3k}+\sum_{k=0}^{\infty}2^{-9-3k}+\sum_{k=0}^{\infty}2^{-12-3k}+\cdots\right]&=6\left[2^{-6}\cdot\sum_{k=0}^{\infty}2^{-3k}+2^{-9}\cdot\sum_{k=0}^{\infty}2^{-3k}+2^{-12}\cdot\sum_{k=0}^{\infty}2^{-3k}+\cdots\right] \\ &=6\left[\left(2^{-6}\cdot\sum_{k=0}^{\infty}2^{-3k}\right)\cdot\left(1+2^{-3}+2^{-6}+\cdots\right)\right] \\ &=6\left[\left(2^{-6}\cdot\sum_{k=0}^{\infty}2^{-3k}\right)\cdot\sum_{k=0}^{\infty}2^{-3k}\right] \\ &=6\left[\left(2^{-6}\cdot\frac{1}{1-2^{-3}}\right)\cdot\frac{1}{1-2^{-3}}\right] \\ &=\frac{6}{49} \end{align*} by infinite geometric series, from which the answer is $6+49=\boxed{\textbf{(A) }55}.$

~MRENTHUSIASM

## Video Solution Using infinite Geometric Series

https://youtu.be/3B-3_nOTIu4 ~hippopotamus1

## See Also

 2021 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 24 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 12 Problems and Solutions

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