Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We have a box containing red and black balls. If we draw two at random the probability of getting both of them red is $1/2$. Which basically means:

\begin{equation} \frac{r}{r+b} \cdot \frac{r-1}{r+b-1} = \frac{1}{2} \end{equation} Then we have for a positive number of red and black balls, $r$ and $b$ respectively:

\begin{equation} \frac{r}{r+b} > \frac{r-1}{r+b-1} \end{equation}

and the following inequality follows:

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

How can I derive this inequality?

share|cite|improve this question
up vote 4 down vote accepted

If $x$ and $y$ are distinct positive numbers with $xy = z$, then one of $x$ and $y$ must be bigger than $\sqrt z$ and one must be smaller. If both $x$ and $y$ were bigger than $\sqrt z$, their product would be bigger than $z$, and if both were smaller, their product would be smaller.

So if you have $x>y$, then you must have $x > \sqrt z > y$, and so $x^2 > z > y^2$.

Here you have $x={r\over r+b}$, $y = {r-1 \over r+b-1}$ and $z=\frac12$.

share|cite|improve this answer
Thanks for the answer, I liked all of them, but I chose yours since I think it gave the best intuition into why it holds. – Euclean Jul 19 '12 at 14:36
You are welcome. I hoped you might appreciate an intuitive answer rather than a more symbolic one. – MJD Jul 19 '12 at 14:39

If you multiply your first inequality by $\frac {r-1}{r+b-1}$ on both sides you get the left half of your second. And if you multiply it by (what?) you get the right half of your third. If you need to derive the first, cross multiply it and it is obvious.

share|cite|improve this answer

So we have $x > y$ such that $x \cdot y = \frac{1}{2}$. Thus $$ x^2 = x \cdot \frac{1}{2y} \stackrel{x>y}{>} \frac{1}{2} = y \cdot \frac{1}{2 y} \stackrel{y<x \implies \frac{1}{y} > \frac{1}{x}}{>} y \cdot \frac{1}{2 x} = y^2 $$

share|cite|improve this answer

Take the equation

r(r-1)/{(r+b) (r+b-1)}=1/2 for comparison and then for the inequality r/(r+b) > (r-1)/(r+b-1) and multiply both sides by r/(r+b) and you get [r/(r+b)]$^2$ >r(r-1)/{(r+b) (r+b-1)}=1/2.

Now for the other side of the inequality multiple both sides by (r-1)/(r+b-1) and you have

r(r-1)/{(r+b) (r+b-1)}=1/2 > [(r-1)/(r+b-1)]$^2$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.