# Picking socks probability proof

I'm trying to follow the proof from question 1b out of the book "50 Challenging Problems in Probability" by Mosteller. The problem states:

A drawer contains red and black socks. When two socks are drawn at random, the probability that both are red is $1/2$. How small can the number of socks in the drawer be if the number of black socks is even?

The answer given starts as follows:

Let there be $r$ red and $b$ black socks. The probability of the first socks' 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 $$\frac{r}{r+b} \times \frac{r-1}{r+b-1} = \frac{1}{2}$$ Notice that $$\frac{r}{r+b} > \frac{r-1}{r+b-1} \quad \text{for b} >0$$ Therefore we can create the inequalities $$\left(\frac{r}{r+b}\right)^2 > \frac{1}{2} > \left(\frac{r-1}{r+b-1}\right)^2$$

This is where I'm confused. Why is it that $\left(\frac{r}{r+b}\right)^2 > \frac{1}{2}$? If $r=1, b=100$ then obviously $\left(\frac{1}{101}\right)^2 < \frac{1}{2}$. Am I missing some obvious assumption?

## 2 Answers

You're not missing an assumption, you seem to have written and then immediately forgotten an explicit statement:

$$\frac{r}{r+b} \cdot \frac{r-1}{r+b-1} = \frac{1}{2}$$

Given that is true, and that one of the fractions is larger than the other, it's clear that their squares must lie either side of $\frac{1}{2}$

(consider that in the initial equation you have a product of different terms; consequently one must be greater than the square root of the product, the other must be less than it; in this case we know which one is greater).

Note that in your example with $r=1, b=100$, it is NOT the case that $\frac{r}{r+b} \cdot \frac{r-1}{r+b-1} = \frac{1}{2}$ - if you ignore the condition, the result wouldn't be expected to hold.

An explicit way to see the inequality is as follows. Start with the inequality between the multiplied fractions

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

Multiply both sides by $\frac{r}{r+b}$:

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

The right hand side of the inequality is just $\frac{1}{2}$ from the initial probability equation. So

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

Similarly, if you multiplied the original inequality by $\frac{r-1}{r+b-1}$, you would get:

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

Combine the two inequalities so obtained to conclude

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