2021 AMC Fall 10B Problem 16

Each person picks one of the $5$ pairs of neighboring balls, so there are $5 \times 5 = 25$ equally likely outcomes.

Idea. The expected number of balls at home equals the sum, over all five balls, of the chance that each one ends at home. Every ball is in the same situation, so find the chance for one ball and multiply by $5$.

1. Pick one ball, call it $X$. It belongs to $2$ of the $5$ neighbor pairs (the pair with its left neighbor and the pair with its right neighbor).
2. Case 1: nobody touches $X$. Chris avoids $X$ with chance $\frac35$, and Silva avoids $X$ with chance $\frac35$. Chance $\frac35 \times \frac35 = \frac{9}{25}$.
3. Case 2: Chris moves $X$, Silva moves it back. Chris picks a pair containing $X$ with chance $\frac25$. Now $X$ is one spot away. The only way it gets home is if Silva swaps that exact same pair, chance $\frac15$. Chance $\frac25 \times \frac15 = \frac{2}{25}$.
4. Any other outcome leaves $X$ away from home. (If Chris leaves $X$ alone but Silva moves it, it ends one spot away. If Chris moves it and Silva picks a different pair, it either stays moved or gets pushed farther.)
5. Chance $X$ is home: $\frac{9}{25} + \frac{2}{25} = \frac{11}{25}$.
6. Expected number at home:
$$5 \times \frac{11}{25} = \frac{11}{5} = 2.2.$$

The answer is $\boxed{\textbf{(D)}\ 2.2}$.

To find the expected number of things that end up a certain way, add up each thing's chance. This works even though the balls affect each other, which is what makes it so useful. When all the things are in the same position, like five balls around a circle, you only need to work out one chance and multiply.

Sort the $25$ outcomes by how Silva's pair relates to Chris's pair.

Adding up, $25 + 20 + 10 = 55$ balls at home over $25$ outcomes, an average of $\frac{55}{25} = 2.2$.

There are $5$ adjacent pairs, and each person picks one uniformly, so there are $25$ equally likely (Chris, Silva) choices.

By linearity of expectation, the expected number of balls in their original spots is $5$ times the probability that one particular ball, say ball $X$, ends up where it started. Ball $X$ belongs to $2$ of the $5$ adjacent pairs.

• Chris does not touch $X$ (probability $\frac35$) and Silva does not touch $X$ (probability $\frac35$): $X$ never moves. Probability $\frac{9}{25}$.
• Chris swaps a pair containing $X$ (probability $\frac25$), moving $X$ one spot over. $X$ returns only if Silva swaps that same pair (probability $\frac15$); any other choice leaves $X$ displaced or pushes it further. Probability $\frac{2}{25}$.
• Chris does not touch $X$ but Silva does: $X$ ends displaced.

So $P(X \text{ home}) = \frac{9}{25} + \frac{2}{25} = \frac{11}{25}$, and the expected count is
$$5 \cdot \frac{11}{25} = \frac{11}{5} = 2.2.$$

The answer is $\boxed{\textbf{(D)}\ 2.2}$.