# Monthly Archives: February 2012

## Probability musings

From Futility Closet:

A bag contains 16 billiard balls, some white and some black. You draw two balls at the same time. It is equally likely that the two will be the same color as different colors. What is the proportion of colors within the bag?

A solution appears on the page, but it doesn’t suggest how it was determined or if the solution is unique.  I’ll provide that below plus an extension.

Assume there were w white balls originally, then the number of ways to draw two white is $\displaystyle \frac{w*(w-1)}{2!}$, two black is $\displaystyle \frac{(16-w)*(15-w)}{2!}$, and one of each is $\displaystyle \frac{w*(16-w)}{1!1!}$.  As the likelihood of drawing one of each color is to be the same as drawing only one color,

$\displaystyle \frac{w*(w-1)}{2}+ \frac{(16-w)*(15-w)}{2}= w*(16-w)$

This is a quadratic equation with solutions $w=\{6,10\}$, verifying Futility Closet’s solution, and establishing those solutions as the only ones possible.

Extension:  For what total number(s) of white and black billiard balls is the likelihood  for two drawn balls to be one of color the same as them being one of each?

Let n be the total number of balls.  Repeating the procedure above, the number of ways to draw two white is $\displaystyle \frac{w*(w-1)}{2!}$, two black is $\displaystyle \frac{(n-w)*((n-1)-w)}{2!}$, and one of each is $\displaystyle \frac{w*(n-w)}{1!1!}$.  Because the likelihoods are still equal,

$\displaystyle \frac{w*(w-1)}{2}+ \frac{(n-w)*(n-1-w)}{2}= w*(n-w)$

I solved this with my CAS (below) to get $\displaystyle w=\frac{n}{2}\pm\frac{\sqrt{n}}{2}$.

I noticed two things from this solution.

1. Because w must be a whole number, the denominators require n to be an even number and to guarantee that, the square root means n must be 4 times any perfect square.
2. For any value of n, there are only two solutions for any of these situations, a point that could have been obvious before seeing the solution if you noticed the equation was quadratic, but something I suspect many (most?) students would initially miss.

The original problem gave $n=4*4=16$ with solutions $\displaystyle w=\frac{16}{2}\pm\frac{\sqrt{16}}{2}=\{10,6\}$ confirming the patterns predicted by the general solution.  For another possible solution, let $n=4*25=100$ with solutions $\displaystyle w=\frac{100}{2}\pm\frac{\sqrt{100}}{2}=\{55,45\}$.

A further extension of the general solution would be to show that as $n\rightarrow\infty$, the likelihoods become essentially equivalent for drawing either two different colors or both the same color.

Pedagogical side note:  Both equations in this problem are quadratic and could be solved “by hand”.  But, why?  The point of these questions focuses on probability and not symbolic manipulation, so I argue that use of a CAS is entirely appropriate and perhaps should be the ideal approach in this problem. Keep the focus appropriately on the problem.  If a teacher needs to assess a student’s ability to solve quadratic equations, that should be done in a different question, not here.