Matching gender expectation 
If there are x men and y women and we pair them randomly (do not consider gender). What are the expected number of the pair man-man, man-woman, woman-woman respectively?

(Assume x,y large so that even or odd in total is neglectible, I just want an approximation)
Note: I am confusing. Why can we compute it just like the probability of drawing two balls out of the box with red balls and white balls inside? Many answers suggested that they are the same, but I think they are completely different, this one is like drawing two balls out of box repeatedly without putting back, and counting the numbers of each type.
 A: We assume that the total number $x+y$ is even, so that everyone gets a partner.  
Take a particular woman, say Alicia. The probability that Alicia is paired with a woman is $\frac{y-1}{x+y-1}$. For the expected number of woman-woman pairs, sum over all women, and divide by $2$. The expected number of woman-woman pairs is
$\frac{y}{2}\cdot\frac{y-1}{x+y-1}$.
We get a similar expression for the expected number of man-man pairs. For the mixed pairs, do a similar analysis, or subtract the expected number of unmixed from $\frac{x+y}{2}$.
Remark: If $x$ and $y$ are large (with still $x+y$ even), we could say that for example the expected number of woman-woman pairs is approximately $\frac{y^2}{2(x+y)}$. But one might as well have an exact answer. 
A: I get slightly different probabilities than the two answers already given.  I would say the pairing is male-male with probability $\dfrac{\binom{x}{2}}{\binom{x+y}{2}} = \dfrac{x(x-1)}{(x+y)(x+y-1)}$
Male-female with probability $\dfrac{\binom{x}{1}\binom{y}{1}}{\binom{x+y}{2}} = \dfrac{2xy}{(x+y)(x+y-1)}$
Female-female with probability $\dfrac{\binom{y}{2}}{\binom{x+y}{2}} = \dfrac{y(y-1)}{(x+y)(x+y-1)}$
A: A random pair is m-m with probability $\frac x{x+y}\cdot \frac x{x+y}$, it is w-w with probability $\frac y{x+y}\cdot \frac y{x+y}$, and mixed with probability $2\cdot\frac x{x+y}\cdot \frac y{x+y}$
A: $$(x+y)^2 = x^2 + 2 x y + y^2$$
Consequently, we expect $\frac{x^2}{(x+y)^2}$ male-male assignments, $\frac{2 x y}{(x+y)^2}$ male-female assignments, and $\frac{y^2}{(x+y)^2}$ female-female assignments.
