I have a bag with $ x $ red balls and y blue balls. I randomly draw two balls at the same time without replacement, then keep repeating this process until the bag is empty. Call a pair of balls "colorful" if it is made up of one red ball and one blue ball. The expected number of colorful pairs is $ 2020.$ What is the number of possible pairs of $(x,y)$?
I was thinking to set up an equation, like $ 2xy/(x+y)^2-x-y = 2020, $ but couldn't get any farther than that. Could someone help me answer this?
The probability of getting RB or BR on the first draw is $$p=\frac{2xy}{(x+y)(x+y-1)}$$ This is also the probability of getting RB or BR on any draw.
There are $n=\frac12 (x+y)$ pairs in each run. The expected number of colourful pairs in any run is $$np=\frac{xy}{x+y-1}=2020$$ $$y=\frac{2020(x-1)}{x-2020}=2020+\frac{2.2.5.101.3.673}{z}$$
$z=x-2020$ must be a factor of the numerator. We must have $x, y \gt 0$ so $z \gt 0$. Also $x+y$ must be even. Assuming $x \le y$ possible values are :
\begin{array}{|c|c|c|c|} \hline z & x & y \\ \hline 2 & 2022 & 2041210 \\ \hline 6 & 2026 & 681750 \\ \hline 10 & 2030 & 409858 \\ \hline 30 & 2050 & 137966 \\ \hline 202 & 2222 & 22210 \\ \hline 606 & 2626 & 8750 \\ \hline 1010 & 3030 & 6058 \\ \hline 1346 & 3366 & 5050 \\ \hline \end{array}
A further 8 pairs are generated by switching $x, y$.
Therefore there are $16$ possible pairs of values $x,y$.