Timeline for What are the odds of a at least a certain amount of aces being dealt depending on the number of players?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Feb 1, 2016 at 10:05 | vote | accept | Kriem | ||
| Feb 1, 2016 at 10:05 | comment | added | Kriem | Thanks! This thread has kept me from getting a good night's sleep ;) | |
| Jan 30, 2016 at 21:27 | comment | added | Daniel | To calculate at least 2 aces you sum (exactly4 + exactly3 + exactly2) | |
| Jan 30, 2016 at 21:23 | comment | added | Daniel | eg. 5 players = 8 cards delt. You care about number of card combinations - en.wikipedia.org/wiki/Combination 0 aces - possible combinations with no ace - (46!/((46-8)!*8!)) divided by the total possible combinations - (50!/((50-8)!*8!)) exactly 1 ace - combinations of chosing 1 ace = 4 multiplied by the combinations of choosing the other 7: 46!/((46-7)!*7!) divided by the total number of combinations(same as for 0) exactly 2 aces - (combos of choosing 2 aces = 6) x (46!/((46-6)!*6!)) / total:(50!/((50-8)!*8!)) exactly 3 aces - 4 x (46!/((46-5)!*5!)) / total | |
| Jan 30, 2016 at 20:10 | comment | added | paparazzo | Can you show your calcs? | |
| Jan 29, 2016 at 22:08 | comment | added | Dr.DrfbagIII | Well done, sir. | |
| Jan 29, 2016 at 21:20 | comment | added | Daniel | Third time's the charm :) | |
| Jan 29, 2016 at 21:19 | history | edited | Daniel | CC BY-SA 3.0 |
fixed calculation errors for 3-rd and 4-th columns
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| Jan 29, 2016 at 19:07 | comment | added | Dr.DrfbagIII | I finally tried out doing this whole table for myself and mine matches your exactly in the 0-aces, at-least-1-ace, and 4-aces columns but does not for the other two columns, especially as the number of players goes up. The thing that looks fishy now is that for the 10-players row, it's implied that there's only a 9% chance of there being exactly one ace, but 58% chance of there being exactly two? FYI, I tried it in Excel with the HYPEGEOMDIST function. | |
| Jan 29, 2016 at 18:44 | history | edited | Daniel | CC BY-SA 3.0 |
fixed content and clarification
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| Jan 29, 2016 at 17:43 | comment | added | Dr.DrfbagIII | I don't think this is quite right...I'd be interested in seeing how you did it. For the 2players-0aces, this should simply be (46/50)(45/49)(44/48)(43/47) = .709 which doesn't match your .828. Also, on the 10 players line, I see that you have the odds of 4 aces being dealt as higher than the odds of zero aces being dealt, but just intuitively it seems like that inequality should be go the opposite way since less than half the deck is being dealt out. | |
| Jan 29, 2016 at 16:56 | history | answered | Daniel | CC BY-SA 3.0 |