Probability - selecting n boys and n girls There are $n$ boys and $n$ girls. One at a time, one of the boys is selected at random. When chosen, the boy selects a girl of his choice. Santiago, one of the boys, wants to choose Mildred. If every boy other than Santiago is equally likely to pick any of the remaining girls when his turn is taken, what is the probability that Santiago gets to select Mildred?
 A: It is equally likely that Santiago is first to choose, second, third, and so on. We assume the choosing is done without replacement. 
If S is first to choose, the probability of success is $1$.  It is convenient to call this $\frac{n}{n}$.
If S is second to choose, the probability of success is $\frac{n-1}{n}$. 
If S is third to choose, the probability of success is $\frac{n-2}{n}$.
And so on. If poor S is last to choose, the probability his desired choice is still unchosen is $\frac{1}{n}$.
Thus our probability is 
$$\frac{1}{n}\left(\frac{n}{n}+\frac{n-1}{n}+\frac{n-2}{n}+\cdots+\frac{1}{n}       \right).\tag{1}$$
To simplify (1), use the fact that $1+2+\cdots+n=\frac{n(n+1)}{2}$. The probability Santiago will be happy is $\frac{n+1}{2n}$. 
A: Santiago gets his girl half the time.
Let's take each of the boys and scramble them. Then each of the girls and scramble them. Each boy takes the girl at the top of the list. Except Santiago, who picks Mildred when he can. So the question is what is the average number of picks before Santiago gets to pick before Mildred is picked? Since each both are equally distributed over $1-n$, there's an equal chance he picks before she's picked as there is being picked after.
