Solution to puzzle of white tailed cats

The following solution is reproduced from Prof Ian Stewart’s book “Cabinet of Mathematical Curiosities”.


Suppose that there are c cats, of which w have white tails. There are c(c-1) ordered pairs of distinct cats, and w(w-1) ordered pairs of white tailed cats. (You can choose the first cat of the pair in c ways, but the second in only c-1 ways since you have used up one cat. Ditto for white tailed cats. By ordered, what is meant is choosing first cat A and then cat B is considered to be different from B first and then A. If you don’t like that, then both formulae have to be halved — with the same result.)

This means that the probability of both cats having white tails is \frac{w(w-1)}{c(c-1)} and this must be 1/2. Therefore, c(c-1)=w(w-1) with c and w being whole numbers. The smallest solution is c=4, and w=3. The next smallest turns out to  be c=21 and w=15. Since Ms Smith has fewer than 20 cats, she must have 4 cats, of which three have white tails.

More fun later,

Nalin Pithwa

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