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?

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
$$x'_n+y'_n\sqrt6=(1+\sqrt6)(5-2\sqrt6)^n\ .$$
