2021 AMC 12B Problems/Problem 23
Problem 23
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 
is 
for 
More than one ball is allowed in each bin. The probability that the balls end up evenly spaced in distinct bins is 
where 
and 
are relatively prime positive integers. (For example, the balls are evenly spaced if they are tossed into bins 
and 
) What is 
Solution
"Evenly spaced" just means the bins form an arithmetic sequence.
Suppose the middle bin in the sequence is 
. There are 
different possibilities for the first bin, and these two bins uniquely determine the final bin. Now, the probability that these 
bins are chosen is 
, so the probability is the middle bin is 
. 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*}](http://latex.artofproblemsolving.com/b/5/3/b53c2ce3c14c76f8cd9336f940fddc695f69f106.png)
The answer is 
Solution 2
Let us first consider the cases where the balls are in bins 
and . The probability for the first ball is 
, the probability for the second ball is 
and the probability for the third ball is 
. There are 
ways to orient which ball is in which bin since there are 3 colors. Multiply all of them to get 
. Next consider the case where the balls are in bins 
and 5
\frac{3}{256}
1
8
1
\frac{3}{32} * (1 + \frac{1}{8} + \frac{1}{8^2} + …) = \frac{3}{28}
2, 3, and 4
\frac {3}{256}
2\frac{3}{256} * (1 + \frac{1}{8} + \frac{1}{8^2} + …) = \frac{3}{224}$.
