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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

UnfairDraw4balls

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

FairDraw4balls

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?
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  • $\begingroup$ One can use the pictures to set up the Diophantine equation. Solving is mechanical, but I think not particularly intuitive. $\endgroup$ – André Nicolas Nov 19 '14 at 18:26
  • $\begingroup$ I love this question ! $\endgroup$ – mick Nov 19 '14 at 22:51
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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

enter image description here

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

The mechanical approach would be

  • 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
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