# A fair game with triangular numbers?

Definition 1

A ball game is a state where you have $n$ white balls and $m$ black balls. The rule is that you remove first one ball from the cup. And without returning the first ball, you pick another.

• $P(A)$ is defined as the probability of drawing two balls with opposite colors

• $P(B)$ is defined as the probability of drawing two balls with the same color

A game is defined as fair if $P(A)=P(B)$.

Conjecture 1

A game is fair if and only if $n$ and $m$ are consecutive triangular numbers.

$(n,m) = (1,3) \ , \ (3,6) \ , \ (6,10) \ \ldots$

Image a cup with 2 white balls and 2 black balls. If you draw two balls with the same color, you win. If on the other hand you draw two balls with opposite colors, I win. Note that once a ball is drawn the ball is gone.

Is this a fair game? No, ofcourse not. After you pull a ball from the cup there are two of my colors, and only one of yours. Giving me a $P = 2/3$ chance of winning. This can be illustrated in the following diagram

Following a doted line means I win. Following a whole line, you win. There are more doted lines, than whole; hence I win. One can make this game fair by changing the balls

just count the lines, or do the simple math. Now a fun generalization is to find all configurations that allow a fair game. Suprisingly this is always two consecutive triangular numbers. I want to explain this to my class in an intuitive way, perhaps let them explore it.

• Is there a way to use the diagrams or else to obtain a intuitive explenation why the solutions are always two consecutive triangular numbers?
• One can use the pictures to set up the Diophantine equation. Solving is mechanical, but I think not particularly intuitive. – André Nicolas Nov 19 '14 at 18:26
• I love this question ! – mick Nov 19 '14 at 22:51

I think the diagram may not generalise in an obviously triangular way. For example with $$3$$ and $$6$$ the diagram has $$36$$ lines ($$72$$ if you count them in both directions) and the best I could do was something like
I think the most you can say is that this suggests for a fair game you have $$mn = \tfrac12m(m-1) +\tfrac12n(n-1)$$ i.e. $$2mn = m(m-1)+n(n-1)$$
I do not see how to see directly from the diagram that the pair $$\frac12k(k-1),\frac12k(k+1)$$ provides a solution for positive $$k$$, or that there are no other essentially different solutions
• Use the quadratic formula to show $$m=n +\frac{1\pm\sqrt{8n+1}}{2}$$
• $$8n+1$$ is an odd integer so its square root is either irrational or another odd integer
• $$\sqrt{8n+1}=2k+1$$ has the solution $$n=\frac12k(k+1)$$, i.e. a triangular number
• $$n=\frac12k(k+1) \implies m=\frac12k(k-1)$$ or $$\frac12(k+1)(k+2)$$ and you are done