# A very complex angle on a very simple chance experiment

A bag contains 5 black stones and 5 white stones. Two stones are drawn randomly from the bag without replacement. What is the probability that two black stones are drawn?

This is the kind of chance experiment you would introduce in Year 9. You would expect students to draw and label tree diagrams and calculate probabilities for other outcomes.

In preparing a lesson plan involving this experiment, I noticed the general result that replacement increases the probably of drawing two black or two white stones but decreases the probability of drawing one of each. I decided to make that the main line of inquiry in the lesson.

In a two-draw chance game with any number of black and white stones, which rule gives the highest probability of drawing both a black stone and a white stone: replacement or no replacement?

Then I noticed (thanks to a hastily made spreadsheet) that when you have 15 black stones and 10 white stones, the probability of drawing both a black stone and a white stone is 0.5 exactly. This immediately made me curious. I set myself the “obviously” simple task of finding a rule $y=f(x)$ such that $P(X=1)=0.5$, where $x$ and $y$ represent the number of black and white stones respectively in a two-step, no-replacement draw and the random variable $X$ is the number of black stones drawn (which could be 0, 1 or 2).

After constructing a tree diagram with expressions in $x$ and $y$, I arrived at this relationship.

$\displaystyle\frac{2xy}{(x+y)(x+y-1)}=0.5$

That does not look like it will simplify well. But it does give a conic curve (rotated parabola) on which you can identify integer solutions.

Unlike the integer solutions to $y=x^2$, which are intergers and square numbers, the integer solutions of this equation follow a more interesting pattern.

Taking the lower path, the coordinates are $(1,0)$$(3,1)$$(6,4)$$(10,6)$, and $(15,10)$. They are pairs of adjacent triangular numbers.

So that’s that.

In this investigation, I was aided by Wolfram|Alpha and Google searches of the number sequence—because a lazy mathematician is an efficient mathematician.

Incidently, it seems that the solution curve is a transformation of $y=x^2$, involving a vertical enlargement of scale factor $\sqrt{2}$ and a rotation 45° clockwise about the origin $(0,0)$. Somewhere in there is the reason why $y=\sqrt{2}x^2$ has no integer solution pairs, “obviously”, but when you rotate the solution curve 45° you get integer solution pairs that are adjacent triangular numbers!