Re: Generating all 5-card hands

[hv]

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.

Hugo

[Anonymous Monk]

I thought you would be interested to know that I added a link to this discussion at Cribbage Corner, in the section on hand distribution and probabilities. Thank you for taking the trouble to work this out!