2021 AMC Fall 10B Problem 16

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 $\boxed{\textbf{(D)}\ 2.2}$.

Linearity of expectation lets you ignore the dependence between balls: compute one ball's probability of ending home and multiply by five. Symmetry (every ball sees the same situation) makes that one probability enough. This "indicator per object" trick is the standard route for any "expected number of fixed points" question.

Enumerate the $25$ outcomes directly: $5$ have Silva undo Chris (all $5$ balls fixed), $10$ have the pairs overlapping in one ball (a $3$-cycle; $2$ fixed), and $10$ have disjoint pairs ($1$ fixed). Total $25 + 20 + 10 = 55$, average $55/25 = 2.2$.

• Counting only the case where neither swap touches the ball (9/25) and forgetting the swap-and-swap-back case, giving 1.8.
• Believing Silva can also return $X$ by swapping the other pair containing $X$'s new position; that moves $X$ two spots away, not home.