(21/45)(6/44)(5/43)(4/42)(3/41)(2/40)(1/39) ?
The number of different dealt hands for one player on Stud is:
52 × 51 × 50 × 49 × 48 × 47 × 46 / 2 / 3 / 4 / 5 / 6 / 7 =
133,784,560
For 2 players i come to this number of different hands:
133,784,560 × ( 45 × 44 × 43 × 42 × 41 × 40 × 39 / 2 / 3 / 4 / 5 / 6 / 7 ) / 2 = 3,035,546,247,333,600
Looks good. Very big number.
For the number of Flushs in the 13 cards of one suite i have now:
13 x 12 x 11 x 10 x 9 / 2 / 3 / 4 / 5 = 1,287
Straight - and Royal Flush included.
And then are 38 cards in deck for card 6 + 7 of the 2 players, but not the cards what make the players Flush better.
All what i can say in moment. I have now not enough time to go on. Not easy.
Forget what I wrote before; it’s rubbish! I got mixed up and stared counting a 7 card flush.
OK, agree with your figures:
133,784,560 possible hands for P1
1,287 possible flushes for P1
So odds (P1: flush) = 1,287 / 133,784,560
Given P1 has a particular 5-card flush, we now build the same flush for P2 (in a different suit):
For P2’s first card we can choose any of P1’s 5 ranks, in any of 3 remaining suits, from 45 remaining cards: (15/45) = (1/3)
The remaining 4 cards are now fixed, but can be chosen in any order. Thus: (4/44)(3/43)(2/42)(1/41)
So out of all possible scenarios for 2 players in 7-card stud, the odds of both being dealt the exact same flush (in different suits) are:
(1,287 / 133,784,560) * (1/3) * (4/44) * (3/43) * (2/42) * (1/41)
I think?? XD
It’s really complicated. If the first player only has a 5 or 6 card flush, then they can sometime have blockers to other suits. You can figure it out, but the equation is really complicated.
Someone ran a heads up simulation, and the odds are about 1 : 500,000
so actually not that crazy, but there can’t be many people who have played enough hand of 7-card stud to have seen that happen.
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This specifically is wrong. There (13 choose 5) = 1287 possible flushes in one suit with 5 cards, but when you dealing with 4 suits and seven cards, that’s 4 * (13 choose 5) * (47 choose 2), but that double counts some 6 and 7 card flushes. To do the math properly you have to compute 5,6 and 7 card flushes separately for P1
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I looked to it again, had a app/simulation running. The result:
I checked it and i’m sure it is ok. For a chance of 1:500,000 i would need more Flushs, i can get them only when i’m not looking to the cards 6 and 7 what changes the Flush to a better Flush. But this would not be ok.
This is my guess for 5 card stud
Remove 5 cards
47 cards left
1,533,937
only 3 meet requirements of flush tie
.00000000387
Approximately 1 in 258 million
Give or take millions… lol
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1287 x (42 x 41 / 2) x (40 x 39 / 2) = 864,323,460
I don’t think we need to divide by 2 here, because that’s taken care of in the next step (using 6 suit combos instead of 12 if ordering mattered). You’re not quite exactly a factor of two out though either, so maybe I’ve just misunderstood what you’re actually calculating? The methodology seems fine though.
The number 435,629,698 is counted, not calculated.
My App:
It deals all the flushs of Suite 1 to player 1. It starts at flush 2-6, goes later the flushs up, the last is then the Royal Flush. 1.287 Flushs on Suite 1.
It deals player 2 the same flush, but in suite 2.
It saves the result/cards of the flushs.
It deals the first 2 of 42 cards to player 1, card 6 and 7 on hand. Later all other 2 of 42 combinations.
It looks has the flush of player 1 changed.
If not changed, it deals all 2 of 40 card combinations to player 2.
It looks has the deald cards 6 and 7 changed the flush of player 2. If not, then it counts it.
Do … loop
On end i have then the number of all 2x the same flush in 2 suites. 435,629,698