A combinatoric question

34 Views Asked by At

Given three types of balls, red, green and blue. The number of each type of balls is denoted as R,G, and B. The total number of balls N is even. We (uniformly) randomly group balls into pairs (so we have N/2 pairs). What is the expected number of red-green pairs?

I tried to run simulations on this, but ideally I want to find a closed form formula. Any ideas?

1

There are 1 best solutions below

0
On BEST ANSWER

Line the pairs up and label them.  Let $X_k$ be the indicator random variable that the $k$-th pair is red-green (that is, $X_k$ equals one if it is or zero if it is not).  The expected value of this random variable is therefore the probability that the $k$-th pair is red-green.  All such random variables have identical distribution (although not independent), therefore their expectations are equal.$$\forall k\in\{1..N/2\}~.~\mathsf E(X_k)=\mathsf E(X_1)$$

Let $X$ be the count for red-green pairs.  This will equal $\sum_{k=1}^{N/2} X_k$ and expectation is linear, so the expectation for this count is:$$\begin{align}\mathsf E(X)&=\mathsf E(\sum_{k=1}^{N/2}X_k)\\[0.5ex]&\vdots\end{align}$$