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Nov 16, 2021 at 1:10 vote accept EH86055
Nov 16, 2021 at 1:10 vote accept EH86055
Nov 16, 2021 at 1:10
Nov 12, 2021 at 15:03 review Close votes
Nov 14, 2021 at 17:40
Nov 12, 2021 at 14:36 audit Low quality posts
Nov 12, 2021 at 14:44
Oct 26, 2021 at 13:50 vote accept EH86055
Nov 16, 2021 at 1:10
Oct 25, 2021 at 22:19 history removed from network questions Asaf Karagila
Oct 25, 2021 at 20:39 history edited Rodrigo de Azevedo
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Oct 25, 2021 at 18:45 history became hot network question
Oct 25, 2021 at 11:10 comment added Peter @JaapScherphuis OK, I got it. Thank you.
Oct 25, 2021 at 11:09 comment added Jaap Scherphuis @Peter The first term in the brackets is all possible three card sets that do not have any card matching the pair you already have, so extending the existing pair is already excluded. It is only the addition of an extra pair or triple that remains to be subtracted. That said, Tommik's answer is a much easier way to count all the valid possibilities of three additional cards.
Oct 25, 2021 at 11:05 comment added Peter @JaapScherphuis If the given pair is extended to three of a kind with the other three cards , the result need not be a full-houese. Is this considered in the above formula ?
Oct 25, 2021 at 11:00 comment added Jaap Scherphuis It's complicated because you don't want the other three cards to give you another pair or a three-of-a-kind (making the hand a full house). Those are exactly the two cases being subtracted here.
Oct 25, 2021 at 10:59 answer added tommik timeline score: 14
Oct 25, 2021 at 10:50 history edited MJD CC BY-SA 4.0
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S Oct 25, 2021 at 10:39 review First questions
Oct 25, 2021 at 10:44
S Oct 25, 2021 at 10:39 history asked EH86055 CC BY-SA 4.0