Timeline for Why is the probability of getting a pair in a five-card poker hand so complicated?
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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 |