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Suppose you have 52 cards and know that you are in a state where the only cards you draw are 2,3,4,5,6,7,8,9,10. You draw 4 cards from the 52 cards (36 considering the state you know you are in). What is the chance that the sum of black cards drawn is twice that of red cards?

One way to approach the problem is to split by the number of black cards drawn. For pairs of red and black cards (R,B) we can either have (2,2) or (1,3). We can then count, for each of the 9 values, each case for which the black cards are twice the sum of the red ones. The case (3,1) has no solutions. Is there a less tedious way to approach this?

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At least twice, or exactly twice? – mjqxxxx Jan 30 '14 at 16:55
Sorry, exactly twice. – gman Jan 30 '14 at 17:09
What have you tried? Where are you stuck? (Please answer by editing the question to show your thoughts on it.) – Barry Cipra Jan 30 '14 at 17:26
I doubt there's a less tedious way to go about it. – mjqxxxx Jan 30 '14 at 18:58
up vote 2 down vote accepted

I agree with the comment made by mjqxxxx, there doesn't seem to be any non-tedious approach. But here's a small start.

There are only $7$ possible red/black sums, with black being twice red: $(4,8)$, $(5,10)$, $(6,12)$, $(7,14)$, $(8,16)$, $(9,18)$, and $(10,20)$. Anything larger (e.g., $(11,22)$) would require $2$ red cards and $3$ black cards, and anything smaller (e.g., $(3,6)$) could only be formed with $1$ red card and $2$ black cards.

Each of the seven possibilities is achievable with either one or two red cards, but from here on out it looks like a matter of making lists. I'd be interested in seeing a slick way to organize things.

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