Iv'e been asked the following:
10 pairs of balls are distributed randomly around a circle, what is the expected number of pairs that end up one next to each other?
My analysis so far:
There are 20! ways of ordering 20 balls around a circle, 20 rotational symmetries turn this into $\frac{20!}{20}=19!$ and since we do not care about directionality (Clockwise or anti clockwise) we can cut this in half to $\frac{1}{2}\cdot19!$. If we require at least one pair to stick together we may view this pair as a single object and arrange 19 balls in a circle $\frac{1}{2}\cdot18!$, multiplying by the number of inner arrangements (2) we have 18! arrangements for $X\geq1$. Likewise, for $X\geq2$ we have 17! arrangements and so on.
$\displaystyle \mathbb{E}(X)=\sum_{n=1}^{10}\mathbb{P}(X\geq n)=\frac{2\cdot18!}{19!}+\frac{2\cdot17!}{19!}+\cdots+\frac{2\cdot10!}{19!}=\frac{2}{19}+\frac{2}{19\cdot18}+\cdots+\frac{2}{19\cdot18\cdot\ldots\cdot10}=0.111\ldots$
I can't find a fault in the calculation. However, this number seems suspiciously low...
Linearity of expectation: the expected number is ten times the probability that one particular pair end up together. That probability is $2/19$, so the expected number is $20/19$.
The first term in your sum is correct; the remaining nine terms aren't, as they should all equal the first.
Your second term is not the probability the second pair end up together (which is what you need), it is the conditional probability the second pair end up together conditional on the first pair ending up together, etc.