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A school has $r$ groups. Each group has $n$ girls and $n$ boys. Any boy and girl know each other with probability $p$ if they belong to the same group, and with probability $q$ if they belong to different groups. In random order, a boy marries a girl to whom he is connected to by an edge (if they know each other). If is the turn of a guy who knows no girl or all the girls he knows are taken, he remains alone.

What is the probability that a guy will actually stay single?

The answer is similar to the one of a random bipartite graph admitting a perfect matching, and to the answer in this thread. The best answer will be awarded 100 bounty points, which I cannot add before 2 days.

fox
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    In the regime where $p,q$ are constant, we can argue that the answer is $\Theta(\frac1{rn})$ in exactly the same way as before, replacing edge probabilities with $\max{p,q}$ or $\min{p,q}$ as appropriate. Are you interested in a different asymptotic behavior of $p,q$, greater precision, or both? – Misha Lavrov May 21 '18 at 15:32
  • I am looking to understand the probability that the greedy algorithm selects a perfect matching, both for $p,q$ constant or functions of $n$. I am also interested in the diameter of the corresponding graph, which has been suggested must be $\log n/ \log degree$ (proposition 4 http://www.people.fas.harvard.edu/~bgolub/papers/homophily.pdf) – fox Jun 01 '18 at 17:47
  • I think one can use the coupon colelctor problem, each man proposes to a girl at one step, the girl rejects him if she has a bette rpartner already, there are log(rn) proposals made in total, subject to a minor modification https://people.eecs.berkeley.edu/~jfc/cs174/lecs/lec7/lec7.pdf – fox Jun 01 '18 at 18:03

1 Answers1

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Assuming a particular boy is being discussed. Following solution involves basic principles of probability and properties of the Bernoulli distribution.

If the boy knows certain h girls from his group and certain (g - h) from other groups,

    • Probability that the boy knows h girls from his group (using Bernoulli distribution) =

    P1 : $\left(\begin{array}{c}n\\ h\end{array}\right)p^h(1-p)^{n-h}$

    • Similarly for (g - h) girls of other group =

    P2 : $\left(\begin{array}{c}n(r-1)\\ g-h\end{array}\right)q^{g-h}(1-q)^{n(r-1)-(g-h)}$

  2. Probability that boy knows total g girls i.e. h + (g - h) = $\sum_{h=0}^g$ P1 × P2 =

    P3 : $\sum_{h=0}^g\left(\left(\begin{array}{c}n\\ h\end{array}\right)p^h(1-p)^{n-h}\right)\times\left(\left(\begin{array}{c}n(r-1)\\ g-h\end{array}\right)q^{g-h}(1-q)^{n(r-1)-(g-h)}\right)$

Let probability that boy (or girl since it's symmetric) is single = $X$

  1. Probability that the certain g girls are not single (complimentary event) =

    P4 : $(1-X)^g$

  2. Probability that the boy knows g girls and none is single = P3 X P4 =

    P5 : $(1-X)^g\times\sum_{h=0}^g\left(\left(\begin{array}{c}n\\ h\end{array}\right)p^h(1-p)^{n-h}\right)\times\left(\left(\begin{array}{c}n(r-1)\\ g-h\end{array}\right)q^{g-h}(1-q)^{n(r-1)-(g-h)}\right)$

  3. Probability that the boy knows (maybe $0$ ) girls and none is single = $\sum_{g=0}^{n\times r}$ P5 =

    P6 : $\sum_{g=0}^{n\times r}(1-X)^g\times\sum_{h=0}^g\left(\left(\begin{array}{c}n\\ h\end{array}\right)p^h(1-p)^{n-h}\right)\times\left(\left(\begin{array}{c}n(r-1)\\ g-h\end{array}\right)q^{g-h}(1-q)^{n(r-1)-(g-h)}\right)$

Probability that the boy is single = P6 =

$X$ = $\sum_{g=0}^{n\times r}(1-X)^g\times\sum_{h=0}^g\left(\left(\begin{array}{c}n\\ h\end{array}\right)p^h(1-p)^{n-h}\right)\times\left(\left(\begin{array}{c}n(r-1)\\ g-h\end{array}\right)q^{g-h}(1-q)^{n(r-1)-(g-h)}\right)$

Note that value of $X$ evaluated from above expression will be independent of g and h.

tejasvi
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