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We have a class containing 100 students. There are 5 projects through the semester in which students will work in randomly selected pairs. (Each project therefore has 50 pairs.)

If we randomly and independently create the pairs for each of the 5 projects, what is the expected number of repeat pairs throughout the course of the semester?

sffc
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1 Answers1

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Give all pairs of people a label. There are $N=\binom{100}{2}$ labels. Call them $1,2,3,\dots, N$. Let $X_i$ be the number of times pair $i$ repeat. The meaning of $X_i$ depends on how one counts multiple repeats by two particular individuals.

Then the total number $Y$ of repeats is given by $Y=X_1+X_2+\cdots+X_N$. By the linearity of expectation, we have $$E(Y)=E(X_1)+E(X_2)+ \cdots +E(X_N)=\binom{100}{2} E(X_1).$$

Calculating $E(X_1)$ is an easy problem, one we know the meaning of $X_1$. For let the two people in Pair $1$ be called $A$ and $B$. The probability $B$ is paired with $A$ on any particular project is $\frac{1}{99}$.

Added: OP has given the interpretation that being together $3$ times is $2$ pairs. Presumably that means being together $4$ times is $3$ pairs, and so on.

The probability that A and B are together exactly twice is $\binom{5}{2}p^2(1-p)^3$ where $p=1/99$. This makes a contribution of $\binom{5}{2}p^2(1-p)^3$ to the expectation of $X_1$.

The probability that A and B are together exactly three times is $\binom{5}{3}p^3(1-p)^2$. Under OP's interpretation, this makes a contribution of $2\binom{5}{3}p^3(1-p)^2$ to the expectation of $X_1$.

Deal similarly with the cases A and B are together four times, five times, and write down the contributions to the expectation of $X_1$.

Finally, add up to find $E(X_1)$, and multiply by $\binom{100}{2}$.

André Nicolas
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  • WolframAlpha gives $:$980149401/200000000$;$. $;;;$ –  Sep 09 '13 at 05:22
  • Nice to know. I had no clear intuition about what the approximate number would be, except that the expectation will not be terribly small. – André Nicolas Sep 09 '13 at 05:46
  • Thank you for the elaboration! The decimal form of 980149401/200000000 is 4.901. I wrote a little script to simulate the situation, and 10,000 simulations yields 4.947… within reasonable error of the exact solution! – sffc Sep 09 '13 at 07:21
  • Already the cases of exactly $2$ matches between A and B get us close ($4.899$), since multiple matches are rare. – André Nicolas Sep 09 '13 at 07:51