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A bag has a number of green balls and red balls. Two balls will be picked successively from it. When does the probability of picking two different color balls become equal to picking two same color balls? Is it when the bag has the same number of each color?
Note: I had this question when I was eating M & M with two colors picking two at a time.

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Assume that the number of green balls is $x_1$, the number of red balls is $x_2$ and the total number is $t$. When picking the first ball, the probability of picking a green ball is $\frac{x_1}t$ and the probability of picking a red ball is $\frac{x_2}t$. For the second ball, The probability of green ball if the first was green is $\frac{x_1-1}{t-1}$, and if the first was red is $\frac{x_1}{t-1}$. The same for the red ball, if the first was red is $\frac{x_2-1}{t-1}$, and if the first was green is $\frac{x_2}{t-1}$.
Hence, the probability of picking two different color balls is $\frac{2x_1x_2}{t(t-1)}$, the probability of picking two green balls is $\frac{x_1(x_1-1)}{t(t-1)}$, and the probability of picking two redballs is $\frac{x_2(x_2-1)}{t(t-1)}$, so the probability of picking two same color balls is $\frac{x_1^2-x_1+x_2^2-x_2}{t(t-1)}$.
By equating the two probabilities; of same color and different colors:
$\frac{2x_1x_2}{t(t-1)}=\frac{x_1^2-x_1+x_2^2-x_2}{t(t-1)}$
$2x_1x_2=x_1^2-x_1+x_2^2-x_2$
$x_1+x_2=x_1^2-2x_1x_2+x_2^2=(x_1-x_2)^2$
as $x_1+x_2=t$
$t=(t-2x_2)^2=t^2-4tx_2+4x_2^2$
$4x_2^2-4tx_2+t^2-t=0$
$x_2=\frac{4t+\sqrt{16t^2-16t^2+16t}}8$ or $x_2=\frac{4t-\sqrt{16t^2-16t^2+16t}}8$
so the values of $x_1$ and $x_2$ are $\frac{t+\sqrt t}2$ and $\frac{t-\sqrt t}2$ interchangeably.
Examples:
$t=1$, $x_1=1$, $x_2=0$, the probability of picking two same color or different colors are zero.
$t=4$, $x_1=3$, $x_2=1$
$t=9$, $x_1=6$, $x_2=3$
$t=100$, $x_1=55$, $x_2=45$
Note: I thought at the beginning that if the two colors are of the same number $x$, but found that this make the probability of picking two different colors are higher than the same color.
$\frac{2x^2}{t(t-1)}>\frac{2x^2-x}{t(t-1)}$

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    $\begingroup$ nice and clear answer $\endgroup$ – G Cab Dec 29 '19 at 22:05

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