# Calculating possible option of switching pairs in group

I am struggling with this probability/combinatorics problem.

the problem states: "11 people are seated around a round table. At some point, everyone stands up and sits again at a randomly chosen seat. Show that the expected value of the number of pairs which swapped places (A sat at B's place and B sat at A's place) is 1/2."

I struggle with calculating the number of orderings in which pairs switch places. I know we have 10! possible permutations for 11 people sitting in a circle, and there are $$11\choose2$$ possible pairs that could switch places. But I'm not sure how to count for example the number of permutation in which only one couple switches places.

## 2 Answers

Number the people from 1 to 11 and define $$X_{A,B} = \begin{cases} 1 \quad \text{if A and B switch places} \\ 0 \quad \text{otherwise} \end{cases}$$ for $$1 \le A < B \le 11$$.

The probability that $$A$$ ends up in $$B$$'s former position is $$1/11$$, and then the probability that $$B$$ also ends up in $$A$$'s former position is $$1/10$$; so $$P(X_{A,B} = 1) = \frac {1} {11 \cdot 10}$$

The expected number of pairs who switch places is \begin{align} E \left( \sum_{1 \le A < B \le 11} X_{A,B} \right) &= \sum_{1 \le A < B \le 11} E(X_{A,B} ) \tag{1}\\ &= \binom{11}{2} \frac {1} {11 \cdot 10} \tag{2}\\ &= \frac{1}{2} \end{align} At $$(1)$$ we have used linearity of expectation, and at $$(2)$$ we have used the fact that the number of pairs $$(A,B)$$ satisfying $$1 \le A < B \le 11$$ is $$\binom{11}{2}$$, the number of subsets of $$\{1,2,3,\dots,11\}$$ of size $$2$$.

There's a total of $$11!$$ possible permutations. A given pair $$(a,b)$$ gets switched in exactly $$(11-2)! = 9!$$ of these: the elements $$a$$ and $$b$$ get swapped while the rest (the other 9 elements) are free to do whatever (to permute among themselves). There is a total of $$\frac{11*10}{2}$$ pairs $$(a,b)$$, giving a total of $$9!\frac{11*10}{2}$$ switchings for all pairs in all permutations. Dividing by the number of permutations gives an average of $$\frac{1}{2}$$ switchings per permutation.