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My brother brings a certain number of his marbles to play with in my room. Each marble is distinct. He has 8 total marbles that are either red or blue. One day, I spotted two red marbles in my room. The probability that any two of his marbles (of those that he plays in my room), randomly chosen, both being red is 1/2. How many marbles does he bring into my room?

I tried doing this:

let x = number of red marbles

So $(x/8)$ = probability of picking red marble

and then $(x-1)/(8 - 1)$ = probability of picking second red marble.

$(x/8)(x-1)/(7) = 1/2$, but I got x to be a decimal which is not possible.

EDIT: I kept guessing and checking $\frac{x}{b}\cdot\frac{x-1}{b-1}=\frac{1}{2}$, where $x =$ number of red balls, and $b=$ number of balls he brings into my room to get that $b=4$ and $x=3$, but unsure how to get this solution formally.

Harold Smith
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2 Answers2

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Assume your brother brings $x+2$ red and $y$ blue marbles, and leaves $2$ marbles, where $0\leq x+y\leq 6$.

Then the probability that those two marbles are both red is: ${\binom{x+2}{2}/\binom{x+2+y}{2}}$, which is claimed to be $1/2$
$$\dfrac{(x+2)(x+1)}{(x+y+2)(x+y+1)}=\dfrac{1}{2}$$

Find the integer solution.

Hunt and seek is a viable method.

Note: Increasing the number of red marbles ($x$) always raise the probability closer to one, if not already there, while increasing the number of blue marbles ($y$) will always decrease the probability.   Start at $x=0,y=0$ and searching with this as a guide.

So, indeed your brother brought three red and one blue marble into the room. $$\dfrac{(1+2)(1+1)}{(1+1+2)(1+1+1)}=\dfrac{6}{12}$$

Graham Kemp
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  • The OP already got the solution by trial and error... – Richard Ambler Oct 17 '19 at 04:09
  • thank you for writing this solution! You found specifically which marbles (red and blue) are brought into the room. But could your answer be simplified to just find "How many marbles does he bring into my room?" without any need to find the specific colored marbles he brought? – Harold Smith Oct 17 '19 at 04:19
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Since you have already found your answer, here are just some follow-up comments.

You are right the key is to solve ${x \over b} {x -1 \over b-1} = \frac12$ or equivalently, $b(b-1) = 2x(x-1)$ for positive integers $b, x$.

This is one equation with two unknowns, so it can have multiple solutions, and indeed it does: the first few pairs are $(b,x) = (4,3), (21, 15), (120, 85), (697, 493), (4060, 2871), (23661, 16731)$. Of course, all of them except the first pair is out of range for this problem (since the brother has $8$ marbles max).

I am not an expert on this kind of equation, but I noticed that, for large values, we have $b^2 \approx 2 x^2$ or $b \approx \sqrt{2} x$, so I just wrote some code to loop through $b$ and search for a few $x$ values near $b / \sqrt{2}$. The above $6$ pairs are all I found for $b\le 100000$.

And lo and behold, the sequence $4, 21, 120, 697, \dots$ is in OEIS where this exact probability puzzle is mentioned and there is a wealth of other information. Enjoy!

antkam
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