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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.

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  • $\begingroup$ Let's say we pair all people and by some strange luck, men are only paired with men and women are only paired with women. If you look at the probability that men are paired with women, for this particular pairing, there is a zero percent chance. Let's say we repeat this experiment, and this time, no men are paired with men and no women are paired with women. Now, there is a 100% chance that men are paired with women. $\endgroup$ – SlipEternal Jun 13 '14 at 22:00
  • $\begingroup$ It doesn't matter what the particular pairings are. If you are looking for the overall probability for a single pair (among all possible pairings), then you are not looking at drawing two balls out of a box repeatedly without putting back and counting the numbers of each type. You are looking at just two balls that were at some point drawn from the box and comparing it to all possible ways of drawing two balls from the box. $\endgroup$ – SlipEternal Jun 13 '14 at 22:03
  • $\begingroup$ @SlipEternal I know it doesn't matter for one particular drawing, but if we draw the first pair, then the second outcome will be effected by it, and we are counting the expected value of all, why can we apply first to all? $\endgroup$ – CYC Jun 13 '14 at 22:11
  • $\begingroup$ Because the outcomes of these "future" pairs are affected uniformly. So, in one particular pairing, the first pair affected the outcome of future pairs. But over all possible pairings, picking a particular pair is the same as picking a random pair from all possible pairs. $\endgroup$ – SlipEternal Jun 13 '14 at 22:28
  • $\begingroup$ @CYC - YES what happened with the first pair does affect the probability of what will happen with the second pair, BUT you're asking about "expected number", and expected values can add even if the variables are DEPENDENT. e.g. let A = 1 or 0 depending on if the first drawn pair is m-w, and B = 1 or 0 depending on if the second drawn pair is m-w, then E[A+B] = E[A] + E[B] = Prob(A=1) + Prob(B=1), even though A & B are dependent (as you correctly pointed out). this is a well known (and much loved :) ) result that can be a bit anti-intuitive. see en.wikipedia.org/wiki/Expected_value#Linearity $\endgroup$ – antkam Apr 2 '18 at 15:11
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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.

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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 $\dfrac{\binom{y}{2}}{\binom{x+y}{2}} = \dfrac{y(y-1)}{(x+y)(x+y-1)}$

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    $\begingroup$ Well, we were givem that $x,y$ are large, so I used $x-1\approx x$ (and so did Eric) $\endgroup$ – Hagen von Eitzen Jun 13 '14 at 21:29
  • $\begingroup$ Good point. Should I delete this answer then? $\endgroup$ – SlipEternal Jun 13 '14 at 21:52
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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 $2\cdot\frac x{x+y}\cdot \frac y{x+y}$

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  • $\begingroup$ Same confusion as my comment below Eric. If $x_{1}$ is paired with $x_2$, then $x_2$ must be paired with $x_{1}$, wouldn't it effect the answer? $\endgroup$ – CYC Jun 13 '14 at 21:12
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$$(x+y)^2 = x^2 + 2 x y + y^2$$

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.

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  • $\begingroup$ But if $x_{1}$ is paired with $x_{2}$, then $x_{2}$ must be paired with $x_{1}$, wouldn't it effect the answer?(I am a little confusing) $\endgroup$ – CYC Jun 13 '14 at 21:09

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