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?

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:

\frac{5}{5+2}\times \frac{4}{4+2}=\frac{5\left( 4 \right)}{7\left( 6 \right)}=\frac{10}{21}

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

\frac{r}{r+b}\times \frac{\:r-1}{r+b-1}=\frac{1}{2}

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

\frac{r}{r+b}>\frac{\:r-1}{r+b-1}

Therefore, we can create the inequalities

\left(\frac{r}{r+b}\right)^2>\frac{1}{2}>\left(\frac{\:r-1}{r+b-1}\right)^2

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

\frac{r}{r+b}>\frac{1}{\sqrt{2}}>\frac{\:r-1}{r+b-1}

From the first inequality we get

r>\frac{1}{\sqrt{2}}\left( r+b \right)

or

\begin{align*}
r & >\frac{1}{\sqrt{2}-1}b \\ \\
r & > \left( \sqrt{2}+1 \right)b
\end{align*}

From the second we get

\left( \sqrt{2}+1 \right)b>r-1

or all told

\left(\sqrt{2}+1\right)b+1>r>\left(\sqrt{2}+1\right)b

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

P\left(2\:\text{red socks}\right)=\frac{3}{4}\cdot \frac{2}{3}=\frac{1}{2}

And so, the smallest number of socks is 4.

Beyond this, we investigate even values of b.

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.