Ratios of binomial coefficients 
Trillian has $n$ mice, of which $w$ are white. She chooses four at random. The probability that two are white is equal to the probability none are white.

This gives an equation in binomial coefficients
$$ {w \choose 2} { n-w \choose 2 } = {n-w \choose 4} $$ 
I know of one solution: $n=8, w=2$. Is it unique?
 A: No, the answer is not unique for the given binomial identity.  It is not too difficult to see that if $n = w+1$, the identity holds trivially, as well as if $n = w$:  in both cases, the probability of getting exactly two white mice is zero, as is the probability of getting no white mice.
There are nontrivial solutions as well, corresponding to generalized solutions of a Pell-type Diophantine equation.  For instance,  $$(n,w) \in \{ (15,4), (53,15), (122,35), (498,144), (1181,342), \ldots \}$$ are also solutions.  Interestingly, characterizing the complete solution set is actually a topic in number theory, not combinatorics or probability.
A: Doing a bit of algebra, your equation becomes
$$x^2-6y^2=-5$$
where
$$x=2n-2w-5\quad\hbox{and}\quad y=2w-1\ .$$
This is related to Pell's equation.  You can see by trial and error that $x_0=1$, $y_0=1$ is a solution (though it is meaningless in terms of your problem).
Consider now the new equation
$$x^2-6y^2=1\ ,$$
which has a solution $x=5$, $y=2$.
Define integers $x_n,y_n$ by
$$x_n+y_n\sqrt6=(1+\sqrt6)(5+2\sqrt6)^n\ .$$
Then
$$\eqalign{x_n^2-6y_n^2
  &=(x_n+y_n\sqrt6)(x_n-y_n\sqrt6)\cr
  &=(1+\sqrt6)(5+2\sqrt6)^n(1-\sqrt6)(5-2\sqrt6)^n\cr
  &=(-5)(1)^n\cr
  &=-5\ ,\cr}$$
so this gives infinitely many solutions to the original equation
$$(x,y)=(17,7),\,(169,69),\ldots$$
and hence
$$(n,w)=(15,4),\,(122,35),\ldots\ .$$
Further solutions can be generated from the conjugate surd by setting
$$x'_n+y'_n\sqrt6=(1+\sqrt6)(5-2\sqrt6)^n\ .$$
