Choosing pairs from a set of n distinct elements, easy double counting confusion Consider we have elements $ \{1,2,...,n \} $. Suppose we choose a pair of elements from the set- there are $\binom {n}{2}$ ways of possible distinct pairs. Probability that the chosen pair contains the element $1$ is $\frac{(n-1)}{\binom {n}{2}}=\frac{2}{n}$. However, we must be double counting here then then probability of the pair containing the element $1 $or $2$ or .... or $n$ is $2$ (not $1!$). The answer should be that the probability that the chosen pair contains the element $1$ is $\frac{1}{n}$, by symmetry, not  $\frac{2}{n}$.
Could someone please explain where I am double counting?
 A: To expand Jean-Clauds comment into an answer: 
The probability that a subset of $\{1,2,\ldots,n\}$ contains the element $1$ is $\frac{2}{n}$ just as you have calculated. So you are not double counting here. However when you are checking your the answer, you are double counting:
The probability that a pair contain $1$ is $\frac{2}{n}$ and the probability that a pair contain $2$ is $\frac{2}{n}$, however, the probability that a pair contain $1$ or $2$ is not $$\frac{2}{n}+\frac{2}{n}$$
Here you are double counting the pairs $(1,2)$ and $(2,1)$ since these pairs belong both to pairs containing 1 and pairs containing 2. Instead to calculate the probability that a pair contain $1$ or $2$ we calculate the probability that a pair contain $1$ and then calculate the probability that a pair contain $2$ but does not contain 1 i.e. the probaility equals
$$\frac{(n-1)}{\binom {n}{2}}+\frac{(n-2)}{\binom {n}{2}}$$
If we want to, in this way calculate the probability that a pair contain $1$ or $2$ or $3$ or... or $n$ we then get 
$$\frac{(n-1)}{\binom {n}{2}}+\frac{(n-2)}{\binom {n}{2}}+\ldots +\frac{1}{\binom {n}{2}} $$
Which we may calculate using arithmeticmetic sum to be
$$=\frac{n(n-1)/2}{\binom {n}{2}}=1$$
Thus probability have not broken down, your answer is correct, but you are doubleconting when checking your answer.
