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 2


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


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:


Combine the two inequalities so obtained to conclude

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


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