Suppose I have a set $\mathcal{S}$ of $N$ distinct items. Now consider the set $\mathcal{P}$ of all possible pairs that I can draw from $S$. Naturally, $|\mathcal{P}| = \binom{N}{2}$. Now when I draw $k$ items (pairs) from $\mathcal{P}$ with a uniform distribution, what is the expected number of distinct items from $S$ in those $k$ pairs?

P.S.: I also asked this question over at stats, but got no answers so far, so I am trying here. Thanks for your time!

Edit I pick the pairs without replacement.


For choosing without replacement, here is an exact answer. Assuming $n \geq 2$, so that there is at least one pair, and $1 \leq k \leq \binom{n}{2}$, so that you're choosing at least one pair but not more than the total number of pairs, the expected value is

$$n - \left(\frac{n^2 - 3n - 2k + 4}{n-1}\right) \frac{\binom{\binom{n}{2} - n + 1}{k-1}}{\binom{\binom{n}{2} - 1}{k-1}}.$$

We can assume that we are choosing pairs in order. Let $X_k$ be the number of distinct items from $S$ through $k$ pairs. Let $Y_i$ be the number of items in the $i$th pair that did not appear in any of the previous pairs. So $X_k = \sum_{i=1}^k Y_i$.

Now, $Y_i$ is either 0, 1, or 2. Since there are $\binom{n}{2} - n + 1$ pairs that do not contain a given item and $\binom{n}{2} - 2n + 3$ pairs that do not contain either of two given items, we have $$P(Y_i = 1) = \frac{\binom{\binom{n}{2} - n + 1}{i-1} + \binom{\binom{n}{2} - n + 1}{i-1} - 2 \binom{\binom{n}{2} - 2n + 3}{i-1}}{\binom{\binom{n}{2} - 1}{i-1}}$$ and $$P(Y_i = 2) = \frac{\binom{\binom{n}{2} - 2n + 3}{i-1}}{\binom{\binom{n}{2} - 1}{i-1}}.$$ Thus $$E[Y_i] = 2\frac{\binom{\binom{n}{2} - n + 1}{i-1}}{\binom{\binom{n}{2} - 1}{i-1}}.$$

It can be proved by induction that $$\sum_{i=0}^k \frac{\binom{M}{i}}{\binom{N}{i}} = \frac{(N+1)\binom{N}{k} - (M-k)\binom{M}{k}}{(N+1-M)\binom{N}{k}}.$$

Thus $$E[X_k] = \sum_{i=1}^k E[Y_i] = 2\sum_{i=1}^k \frac{\binom{\binom{n}{2} - n + 1}{i-1}}{\binom{\binom{n}{2} - 1}{i-1}} $$ $$= 2\frac{(\frac{n(n-1)}{2}-1+1)\binom{\binom{n}{2}}{k-1} - (\frac{n(n-1)}{2} - n + 1 - k + 1)\binom{\binom{n}{2} - n + 1}{k-1}}{(\binom{n}{2} - 1+1-\binom{n}{2} + n - 1)\binom{\binom{n}{2} - 1}{k-1}}$$ $$= \frac{n(n-1)\binom{\binom{n}{2}}{k-1} - (n^2 - 3n - 2k + 4)\binom{\binom{n}{2} - n + 1}{k-1}}{(n - 1)\binom{\binom{n}{2} - 1}{k-1}}$$ $$= n -\frac{(n^2 - 3n - 2k + 4)\binom{\binom{n}{2} - n + 1}{k-1}}{(n - 1)\binom{\binom{n}{2} - 1}{k-1}}.$$

  • $\begingroup$ @ Mike: Cool. I think your answer is correct! $\endgroup$ – user17762 Nov 14 '10 at 5:16
  • $\begingroup$ +1 for the Latex labor alone. I questioned this answer at first because when I plugged in small values of $n$ and $k$, the formula wanted to divide by zero. For $k = 0$, and then for $(n,k) = (1,1)$ and even $(n,k) = (2,2)$. But it settled down and started producing answers that match a monte carlo simulation. Nicely done. $\endgroup$ – I. J. Kennedy Nov 14 '10 at 6:06
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    $\begingroup$ @I. J. Kennedy: Thanks for verifying the formula! I was a little concerned that I had missed something somewhere. As far as the small values, I assumed $n \geq 2$ (as otherwise you can't construct pairs), $k \geq 1$ (so that you're drawing at least one of the pairs), and $k \leq \binom{n}{2}$ (since you can't draw more pairs than there are). (The last case rules out the $n = 2$, $k = 2$ scenario.) Without those assumptions, you're right; the formula can yield a division by zero. $\endgroup$ – Mike Spivey Nov 14 '10 at 6:51

You are picking edges from a complete graph and looking for the expected number of vertices that get picked.

Consider the expected number of vertices that don't get picked.

The probability that a vertex gets picked in 1 try is $\dfrac{2}{n}$.

Thus it does not get picked in all tries is $(1-\dfrac{2}{n})^k$.

Thus the expected number of vertices that don't get picked is $n(1-\dfrac{2}{n})^k$

and thus the expected number of vertices that get picked is

$n(1 - (1-\dfrac{2}{n})^k)$

As Ross points out, this assumes you pick the pairs with replacement.

  • $\begingroup$ I think so. I made a mistake in the comment. It must be that $X = \sum_{i=1}^n X_i$ where $X_i = 1$ if vertex $i$ was picked at some try, and 0 otherwise. $\endgroup$ – Naga Nov 13 '10 at 15:34
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    $\begingroup$ This is fine if the edges are picked with replacement. Otherwise, vertices that have not been picked are more likely to be picked on subsequent draws. $\endgroup$ – Ross Millikan Nov 13 '10 at 15:34
  • $\begingroup$ @Ross: You are right. I was assuming with replacement. $\endgroup$ – Aryabhata Nov 13 '10 at 15:37
  • $\begingroup$ Thanks, this is a very helpful answer. Just the idea of picking edges from a complete graph helps me understand the problem better! On the other point, I do indeed pick without replacement. $\endgroup$ – Björn Pollex Nov 13 '10 at 22:47

You can calculate the exact probability of drawing $t$ unique vertices using inclusion-exclusion. We can estimate this probability as

$\binom{n}{t} \binom{\binom{t}{2}}{k}$

but you're overcounting instances when less than $ t $ vertices actually appear; you can fix that using inclusion exclusion.

Given that, you can try to calculate the expectation directly.

You can also try to work your way from Moron's formula. His formula gives the expectation when k edges are drawn with replacement. We know that distribution of the number of distinct edges drawn, so we can express the expectation when k edges are drawn with replacement using the expectations when $k_0$ unique edges are drawn for $k_0 \leq k$. We get linear equations and maybe we can solve them to find the probability when k unique edges are drawn.


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