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I was trying to solve this question combinatorically:

There are $n$ green balls and $n$ yellow balls, each numbered $1$ - $n$. We divide the $2n$ balls randomly into pairs. Compute the expected value of the number of pairs that have the same number.

I tried to compute the probabilities and possible values of the random variable $X$, which is the total of the same number pairs, but it did not work. I specifically would like to request an explanation for how to solve the problem this way.

Also, the answer was computed by indicators $X_i$ such that $X_i$ has the value of 1 if the pair $X_i$ has the same number. Thus, every $X_i$, according to the answer, has the value $1/(2n-1)$. I understand where the expression $1/(2n-1)$ comes from, but I couldn't understand why all values are the same; I would be glad for an explanation why this answer is right. I mean, we have less and less balls to pair each draw of pairs and that changes the probability, right?

The final answer is the sum of all of these indicators: $n/(2n-1)$. Thank you in advance! :)

bittahProfessional
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discreteAsker
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    Don't forget: expectation is linear, even if the variables are dependent. That's the big point of linearity. – lulu Dec 17 '21 at 15:16
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    $E[X+Y]=E[X]+E[Y]$ regardless whether or not $X$ and $Y$ are dependent or independent etc... Similarly, $E[\alpha X] = \alpha E[X]$. That is what we mean when we say that expectation is linear. If you want proof, see here. – JMoravitz Dec 17 '21 at 16:30
  • The question remains what the distribution looks like for the random variable $X$. Straight away I can tell that its minimum is $0$ and its maximum is $n$. I can also see that once $n-1$ pairs have been formed, there must be $n$ pairs total, because the last two balls that remain must be a pair. So the distribution has those properties, but I'm curious what the shape of the entire distribution looks like. EDIT: according to the answer each $X_i$ has an expected value of $1/(2n-1)$, which doesn't make sense to me for the reason I just pointed out, that $X_{n-1}$ should have to be $0$. Stumped. – bittahProfessional Dec 17 '21 at 19:01
  • After you say that the probability of $X_1$ being $1$ is $\frac{1}{2n-1}$, there is no big "reveal" of the value of $X_1$. So when we go on to the value of $X_2$, we have no additional applicable information from $X_1$. In fact we might as well have just asked about $X_2$ first. The probability is again $\frac{1}{2n-1}$. – paw88789 Dec 17 '21 at 22:08

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