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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?

Colonel Panic
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  • New question about tweaking the problem to have unique solution at https://math.stackexchange.com/questions/655950/designing-a-probability-question-with-two-unknowns-but-a-unique-solution – Colonel Panic Jan 29 '14 at 16:57

2 Answers2

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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.

heropup
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  • Cool thanks. How did you get those solutions? – Colonel Panic Jan 28 '14 at 23:24
  • Also, do you think it's possible to tinker with the problem (change the values) so there's only one solution, or will there always be lots because there's two unknowns and one equality? – Colonel Panic Jan 28 '14 at 23:26
  • I was about to respond with a more complete answer, but David (see above) beat me to it. – heropup Jan 28 '14 at 23:27
  • Re: unique solution, I'm not hugely well up in this but I think that a Pell-type equation always has infinitely many solutions, or none at all. There are certainly cubic Diophantine equations with unique solutions, perhaps you could "translate" one of them into a probability question. – David Jan 28 '14 at 23:31
  • Agreed: in general, Diophantine equations with unique solutions are neither easy to construct nor prove the uniqueness of the solution. That's well beyond the difficulty of solving a simple hypergeometric probability question. – heropup Jan 28 '14 at 23:34
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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 $$x'_n+y'_n\sqrt6=(1+\sqrt6)(5-2\sqrt6)^n\ .$$

David
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  • The line $x_n+y_n\sqrt6=(1+\sqrt6)(5+2\sqrt6)^n$ gives infinitely many $(x_n,y_n)$ because $n$ can be any positive integer. And the following lines confirm that they are all solutions of $x^2-6y^2=-5$. – David Feb 13 '14 at 23:50
  • Thanks sorry I was confused by the two $x^2 - 6 y^2$ equations with different right hand sides. – Colonel Panic Feb 13 '14 at 23:55
  • This all makes sense now. Very cool. – Colonel Panic Feb 13 '14 at 23:55