Update: Just spotted one hole that I thought had been covered in this but seems to be there after all. For example, there are 48 ways to get four aces but only 20 7-high flushes. Therefore a holding of say full house has to take into account the number different ways to land on each rank above or below it or at least take into account the cumulative sum of all possible ways up to current rank and divide that by the (constant and known) number of all possible combinations.
But I could for example calculate the number of possible hands at each rank by a cut-down combination operation and feed that in instead of doing your ++ iterative approach - this changes things a bit so I need to do further work on this algorithm even so.
More update - actually that wasn't quite the right example. I should have been talking about exact rank, i.e. for identical hands. Here is a better one: There are only four combinations (and 480 permutations) of 7 5 4 3 2 suited, but there are 1020 combinations (and 122400 permutations) of the same hand unsuited. These might be unique by the reduced iteration but do not therefore have equal likelihood.
-M
Free your mind
In reply to Re^2: Optimising combinatorial iterations by filtration
by Moron
in thread Optimising combinatorial iterations by filtration
by Moron
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