Conditional probabilities, urns I found this question interesting and apparently it has to do with conditional probabilities:
An urn contains six black balls and some white ones. Two balls are drawn simutaneously. They have the same color with probability 0.5. How many with balls are in
the urn?
As far as I am concerned I would say it is two white balls...

probability of 2 black balls: $\frac{6}{8}*\frac{5}{7}=\frac{30}{56}$
of 2 white balls: $\frac{2}{8}*\frac{1}{7}=\frac{2}{56}$
$\frac{30}{56}+\frac{2}{56}=\frac{32}{56}\neq\frac{1}{2}$
Assume the number of white balls is $n$:
2 black balls = $\frac{6}{6+n}*\frac{5}{5+n}=\frac{30}{n^2+11n+30}$
2 white balls = $\frac{n}{6+n}*\frac{n-1}{5+n}=\frac{n^2-n}{n^2+11n+30}$
total probability $=\frac{1}{2}$ (as per problem setup) and also $=\frac{30}{n^2+11n+30}+\frac{n^2-n}{n^2+11n+30}=\frac{n^2-n+30}{n^2+11n+30}$
Moving sides, $2(n^2-n+30)=n^2+11n+30$
$2n^2-2n+60=n^2+11n+30$
$n^2-13n+30=0$
$(n-10)(n-3)=0$
$n=\{10,3\}$
Thus, this problem is solved when there are either 3 or 10 white balls.
