Red and blue balls

Probability Level 2

$m$ red balls and $n$ blue balls are in a bag.

Two are pulled out at random.

There is an equal chance that the two are the same color or that they are different colors.

Is $\sqrt{m+n}$ rational?

Image credit: www.slideteam.net

It depends Yes No

1 solution

David Vreken
Jan 30, 2019

The probability of picking two balls of different colors is $\frac{\binom{m}{1}\binom{n}{1}}{\binom{m+n}{2}} = \frac{2mn}{(m+n)(m+n-1)}$ ,

and the probability of picking two balls of the same color is $\frac{\binom{m}{2}}{\binom{m+n}{2}} + \frac{\binom{n}{2}}{\binom{m+n}{2}} = \frac{m^2 + n^2 - m - n}{(m+n)(m+n-1)}$ .

For the chances to be equal, $\frac{2mn}{(m+n)(m+n-1)} = \frac{m^2 + n^2 - m - n}{(m+n)(m+n-1)}$ which simplifies to $m + n = m^2 - 2mn + n^2 = (m - n)^2$ .

Therefore, $\sqrt{m + n} = |m - n|$ which must be rational for positive integers $m$ and $n$ .

Nice solution, David!

Geoff Pilling - 2 years, 4 months ago

Nice problem and solution. Interesting to note that the solution pairs $(m,n)$ (taking $m \gt n$ without loss of generality) are all pairs of successive triangular numbers $(T_{n+1}, T_{n})$ .

Brian Charlesworth - 2 years, 4 months ago

Definitely!

I've been wondering what the general solution might be for N colors of balls where the probability is equal for them all being the same or all being different when you pick out N balls randomly.

Geoff Pilling - 2 years, 4 months ago

@Geoff Pilling – I'm not getting anywhere on a general solution, or even on solutions for N = 3. I've tried finding solutions $m,n,p \ge 2$ to the equation

$m(m - 1)(m - 2) + n(n - 1)(n - 2) + p(p - 1)(p - 2) = 6mnp$

without success. Have you had more luck?

Brian Charlesworth - 2 years, 4 months ago

@Brian Charlesworth – So far only for the trivial cases... 🤔. Although I must admit recently I haven't had much time to think about it. At some point I'll get out a piece of paper and a pencil and plow into it.

Geoff Pilling - 2 years, 4 months ago

Hi @Brian Charlesworth,

I'm curious... Have you found a site that builds a community around creating and solving great math problems? (Kinda like Brilliant but more focused on the building the community aspect?)

Geoff Pilling - 2 years, 4 months ago

@Geoff Pilling – No luck on that front. Someone mentioned reddit but that site is a bit of a dumpster fire. I'm waiting to see how the "enhanced Discussion experience" promised by staff pans out; maybe it could be called the "Brilliant Café", a designated page (either staff- or volunteer-monitored) where all can go to join in on whatever topics are brought up by the community. If that happens then it might bring @Michael Mendrin back for another go, unless he has already found a site more like Brilliant used to be.

Brian Charlesworth - 2 years, 4 months ago

@Brian Charlesworth – I sure hope something works out. Do you suppose that Brilliant are aware that they risk losing their customer base?

Geoff Pilling - 2 years, 4 months ago

@Geoff Pilling – Judging by comments from Silas Hundt, they are pleased with the activity generated by the changes they've made. They seem willing to sacrifice some of their "old" base (who were here more for the community aspect) to generate a "new" base (who are here more for the individual learning opportunities). There is more revenue to be had from the latter, so to survive as a business they have gone in that direction and in the process eliminated features that were expensive to operate and were being used by fewer and fewer members. They have committed to keeping a Community page but it may have to evolve to justify its existence (and expense).

Brian Charlesworth - 2 years, 4 months ago

Nice observation, @Brian Charlesworth ! You may have already worked out a proof for this, but here is mine:

Using the quadratic equation on $m + n = (m - n)^2$ leads to $m = \frac{2n + 1 \pm \sqrt{8n + 1}}{2}$ . Since $m$ and $n$ are integers, the numerator must be an even integer, and since $2n + 1$ is an odd integer, $\sqrt{8n + 1}$ must be an odd integer as well, so $8n + 1 = (2p + 1)^2$ for some integer $p$ , which solves to $n = \frac{p(p + 1)}{2}$ , a triangular number. Plugging this value back into $m$ gives $m = \frac{p(p - 1)}{2}$ or $m = \frac{(p + 1)(p + 2)}{2}$ , the previous or next triangular number.

David Vreken - 2 years, 4 months ago

It is a bit easier to simply solve $\frac{2mn}{(m+n)(m+n-1)} = \tfrac12$ from which the identity $m+n = (m-n)^2$ is almost immediate.

Mark Hennings - 2 years, 4 months ago

@Geoff Pilling had a solution posted similar to this, but I'm not sure what happened to it.

David Vreken - 2 years, 4 months ago

Red and blue balls

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