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A drawer contains red socks and black socks. When two socks are drawn at random, the probability that both are red is 1/2.

a) How small can the number of socks in the drawer be?

b) How small if the number of black socks is even?

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Solution:

Just to set the pattern, let us do a numerical example first. Suppose there were 5 red and 2 black socks; then the probability of the first sock’s being red would be 5/(5+2). If the first were red, the probability of the second’s being red would be 4/(4+2), because one red sock has already been removed. The product of these two numbers is the probability that both socks are red:

This result is close to 1/2, but we need exactly 1/2. Now let us go at the problem algebraically.

Let there be $r$ red and $b$ black socks. The probability of the first sock’s being red is $\frac{r}{r+b}$; and if the first sock is red, the probability of the second’s being red now that a red has been removed is $\frac{r-1}{r+b-1}$. Then we require the probability that both are red to be $\frac{1}{2}$, or

One could just start with $b=1$ and try successive values of $r$, then go to $b=2$ and try again, and so on. That would get the answer quickly. Or we could play along with a little more mathematics. Notice that

Therefore, we can create the inequalities

Taking the square roots, we have, for $r>1$.

From the first inequality we get

or

From the second we get

or all told

For $b=1$, $r$ must be greater than 2.414 and less than 3.414, and so the candidate is $r=3$. For $r=3, \ b=1$, we get

And so, the smallest number of socks is 4.

Beyond this, we investigate even values of $b$.

$b$	$r$ is between	eligible $r$	$P\left(2 \ \text{red socks}\right)$
2	5.8, 4.8	5	$\frac{5\left( 4 \right)}{7\left( 6 \right)} \neq \frac{1}{2}$
4	10.7, 9.7	10	$\frac{10\left( 9 \right)}{14\left( 13 \right)} \neq \frac{1}{2}$
6	15.5, 14.5	15	$\frac{15\left( 14 \right)}{21\left( 20 \right)} = \frac{1}{2}$

 

And so, 21 is the smallest number of socks when b is even. If we were to go on and ask for further values of r and b so that the probability of two red socks is 1/2, we would be wise to appreciate that this is a problem in the theory of numbers. It happens to lead to a famous result in Diophantine Analysis obtained from Pell’s equation. Try r = 85, b = 35.

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