<s>[INST] Extracting two balls with the same color vs distinct color I am trying to find which has higher probability when extracting from a box without replacement: two balls of the same color, or two balls with different colors.
Let's assume that in the box are $a$ white balls and $b$ black balls.
$$P(\text{distinct color})=P(\text{first white})P(\text{second black/first white})+P(\text{first black})P(\text{second white/first black})$$
$$=\frac{a}{a+b}\frac{b}{a+b-1}+\frac{b}{a+b}\frac{a}{a+b-1}=\frac{2ab}{(a+b)(a+b-1)}$$
$$P(\text{same color})=P(\text{first white})P(\text{second white/first white})+P(\text{first black})P(\text{second black/first black})$$
$$=\frac{a}{a+b}\frac{a-1}{a+b-1}+\frac{b}{a+b}\frac{b-1}{a+b-1}=\frac{a^2+b^2-(a+b)}{(a+b)(a+b-1)}$$
But there is still needed to find which  is more probably  to happen, the case which I don't know how to proceed. Do I have to solve (I don't know how) for which domain of $a$ and $b$ does the following inequality hold true? 
$$a^2+b^2-(a+b)< 2ab \equiv (a-b)^2<a+b $$
This is like in the below answer, so I did it correctly, but still that doesn't help me to conclude an answer.. 
 [/INST] Your inequality
\begin{align*}
(a-b)^2< a + b
\end{align*}
can be expressed as
\begin{align*}
a - \sqrt{2a + \frac{1}{4}}+\frac{1}{2} < b < a + \sqrt{2a + \frac{1}{4}}+\frac{1}{2}
\end{align*}
or
\begin{align*}
\left|b - a - \frac{1}{2}\right| < \sqrt{2a + \frac{1}{4}}
\end{align*}
This domain isn't pretty in any way, but we can make statements about its behavior as $a,b \rightarrow \infty$. For example, dividing both sides by $a$,
\begin{align*}
\left|\frac{b}{a} - 1 - \frac{1}{2a}\right| < \sqrt{\frac{2}{a} + \frac{1}{4a^2}}
\end{align*}
Therefore, in order for the probability of distinct colors to be greater than probability of the same, we need $b = a$ for sufficiently large values of $a$.
</s>