TITLE: Colored balls puzzle
QUESTION [6 upvotes]: Imagine you have $n$ balls in a bag that are colored from $1$ to $n$.  At each turn you take two balls at random out that have different colors and color one the color of the other. You then put them both back in the bag.  What is the expected number of turns before all the balls have the same color?

REPLY [2 votes]: Suppose you have $a$ balls of colour 1, $b$ balls of colour 2 and so on.
Let $F(a,b,c)$ be the expected number of turns until they reach uniformity.
Then from small values, it looks like
$$F(a,b) = ab\\
F(a,b,c)=ab+ac+bc$$
from which I generalize to $\sum_{i<j}a_ia_j$.  This confirms $n$ choose $2$ as the answer to your question.
EDIT: Sorry, I didn't see @azimut's comment; the Gambler's Ruin argument gives a simple proof of this formula.