Spoiler for New Scientist Enigma 1695:

A set of snooker balls contains six coloured and 15 red balls. From such a set Harry took some coloured balls and a larger number of reds and put them in his bag. From the remainder Tom took some coloured balls and some reds (more of each than Harry took) and put them into his bag.

Each of them calculated that if he drew two balls at random out of his own bag simultaneously there was a 1 in X chance that they would both be coloured, X representing the same number for each person.

How many coloured balls and how many red balls from the set were not taken by either of them?

*Edited 12 May to correct an error right at the end.*

Pretty easy this week. I was a little taken aback by the word “simultaneously” but presumably that just means you don’t put one ball back before drawing the second. If there are *n*_{c} colored (excuse my spelling) balls and *n*_{r} red balls totalling *N* = *n*_{c} + *n*_{r} then the probability that both balls drawn are colored is (*n*_{c}/*N*)((*n*_{c}–1)/(*N*–1)) = (*n*_{c}^{2}–*n*_{c})/(*N*^{2}–*N*). We just have to find two pairs of values for *n*_{c} and *N* consistent with the problem statement such that the probabilities are equal.

First of all, notice that Harry must have taken 2 colored balls. No fewer, partly because it says “balls”, plural, and partly because if he took fewer than 2, his probability of drawing 2 colored balls would be zero, which can’t be expressed as “a 1 in X chance”. But if he took 3 then Tom could not take more colored balls than Harry did. So Harry took 2, leaving 4, of which Tom took either 3 or all 4. (Harry also took some red balls — no more than 7, because Tom took more — making a total of *N*_{H}, and Tom took some of the rest, making a total of *N*_{T}.)

That means Harry’s probability of drawing 2 colored balls is 2/(*N*_{H}^{2}–*N*_{H}) while Tom’s is 6/(*N*_{T}^{2}–*N*_{T}) or 12/(*N*_{T}^{2}–*N*_{T}). Either way it must be that *N*_{T}^{2}–*N*_{T} is divisible by 3, since it’s either 3 or 6 times *N*_{H}^{2}–*N*_{H}. This happens if *N*_{T} equals 0 or 1 modulo 3. *N*_{T} is at least 4 and no more than 18. So *N*_{T}^{2}–*N*_{T} is 12, 30, 42, 72, 90, 132, 156, 210, 240, or 306. Of these the only one that is 3 times a number of the form *N*_{H}^{2}–*N*_{H} is 90, and the only one that is 6 times a number of that form is 72. This means we have either *N*_{T} = 10 and *N*_{H} = 6, or *N*_{T} = 9 and *N*_{H} = 4. In the former case *n*_{cH} = 2, *n*_{rH} = 4, *n*_{cT} = 3, *n*_{rT} = 7, and in the latter case *n*_{cH} = 2, *n*_{rH} = 2, *n*_{cT} = 4, *n*_{rT} = 5. But the second case violates the requirement that Harry took “a larger number of reds” (larger than the number of colored balls he took). So they took a total of 5 colored balls and 11 reds, leaving 1 colored and 4 red.

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