Probability Question on Speed Dating Suppose I have 2 groups of 20 people. 20 Male and 20 Female.
After doing some speed dating, each Man writes down 2 women he fancies. And Each women writes down 2 men she fancies. Everybody has a uniform probability to like anyone else.
I consider it a 'Match' when a man and women put down each other on their respective lists. 
What is the probability that exactly 5 matches occur? 
Again assuming there is a uniform distribution of who people 'like' each other (each person has the same probability to like each other person).
I've coded a simulation and ultimately the probability converges to roughly 17.3%
But I am having a hard time rectifying the solution in a theoretical sense though. Would love any help!
Another note: given the inherent 'pick and replace' nature of the liking process. I believe it is roughly 39/380 for a given man M to like female F (and vice versa)
EDIT: ONLY MEN LIKE WOMEN AND VICE VERSA. It is a separate problem for same gender. An interesting one I agree. But my question is 'hetero-centric'.
 A: The probability that a given man (m) matches with exactly 1 woman (w) is given by $P_1 = \sum_{k=1}^{19} P(m\:chooses\:w\:\cap k\: w\:chooses\:m)$.
The probability that k women choose a given man is: $\binom{20}{k}\big(\frac{1}{10}\big)^{k}\big(\frac{9}{10}\big)^{20-k}$, while the probability that the man chooses exactly one woman in that scenario is $\frac{k}{20}+\frac{20-k}{20}\frac{k}{19}$. So the probability that a man matches with exactly 1 woman can be evaluated as $$P_1 = \sum_{k=1}^{19}\Big(\frac{k}{20}+\frac{20-k}{20}\frac{k}{19}\Big)\Big[\binom{20}{k}\Big(\frac{1}{10}\Big)^{k}\Big(\frac{9}{10}\Big)^{20-k}\Big]$$ = 0.19
The probability that a given man matches with exactly 2 women can be calculated in a similar fashion: 
$$P_2 = \sum_{k=2}^{20}\Big(\frac{k}{20}\frac{k-1}{19}\Big)\Big[\binom{20}{k}\Big(\frac{1}{10}\Big)^{k}\Big(\frac{9}{10}\Big)^{20-k}\Big]$$ = 0.01
The probability that a man does not get any matches is then given by $1-P_1-P_2$. 
To have exactly five matches, there are three cases to consider. 
Case 1: Five men each match with exactly one woman. The probability for this case is: $$\binom{20}{5}{P_1}^5(1-P_1-P_2)^{15}$$
Case 2: Three men each match with exactly one woman, and one man matches with two women. The probability for this case is: $$\binom{20}{3}{P_1}^3 17 P_2 (1-P_1-P_2)^{16}$$
Case 3: One man matches with one woman, and two men match with two women. The probability for this case is: $$20 P_1 \binom{19}{2} {P_2}^2 (1-P_1-P_2)^{17}$$
Summing the probabilities from each case, we find that the total probability of five matches is 0.17395. 
