primrose
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- #1
It's easy to come up with cases where you're guaranteed to have the strongest hand on the River, but what about having the weakest hand? This is a question I've asked myself once for some reason, and I think it makes for a nice puzzle.
Precise specification: choose 5 community cards and 2 whole cards such that the number N of other possible combinations of whole cards that beat you (beat, not tie!) is as high as possible. Note that there are 45*44/2 = 990 other possible hands, so 0 <= N <= 990.
Solution:
Precise specification: choose 5 community cards and 2 whole cards such that the number N of other possible combinations of whole cards that beat you (beat, not tie!) is as high as possible. Note that there are 45*44/2 = 990 other possible hands, so 0 <= N <= 990.
Solution:
You can get a perfect N = 990 by having whole cards 
with a board of . You have 3 high and only one 3 is left, which means that any other hand has at least 4 high. Alternatively, you can exchange the roles of 3s and 2s, and of course the suits don't matter. Other than that, I believe the solution is unique.
with a board of . You have 3 high and only one 3 is left, which means that any other hand has at least 4 high. Alternatively, you can exchange the roles of 3s and 2s, and of course the suits don't matter. Other than that, I believe the solution is unique.
