Revisiting an old Bridge puzzle posted in this blog about 9-11 months ago, now generalized to 8 variants.
Optimize: Get the total HCP held by a partnership as low as possible, such that there exists a pair of hands with a total of that much HCP and a) for any, b) for some configuration of hands that the other partnership is holding, and given a) best play, b) worst play from the opponents, you can always get a contract of a) 7S, b) 7NT.
My solution for aab (for any, best play, 7NT) was 19 HCP. I’ll try to get all solved.
