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Think about Loose Coupling

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For a hand with a 5 to avoid scoring 2 points, it must have no pair, and no subset that sums to 15. Since that means we can only have one of each of the pairs (1, 9), (2, 8), (3, 7), (4, 6) that sum to 10, the hands to consider consist of the 16 ways we can choose one from each of those pairs.

If we have 9 we cannot have 6, so we must have 4, so we cannot have 2, so we must have 8, so we cannot have 7, so we must have 3. But 3+4+8 = 15, so we cannot have 9.

So we must have 1; if we have 8 we cannot have 7, or 6, so we must have 3 and 4. So we have 3+4+8 = 15 again, which means we cannot have 8.

So we must have 1 and 2, which means we cannot have 7, so we must have 3. But 1+2+3+5+4 = 15, and 1+3+5+6 = 15. So there is no hand including a five which doesn't include at least 2 points.


In reply to Re: Generating all 5-card hands by hv
in thread Generating all 5-card hands by thor

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