I was playing a heads-up sit-and-go HORSE tournament with Cardgrrl a little while ago when this hand came up.
The diamonds just kept pouring in! If I've ever seen a seven-card flush in stud before, it didn't make enough of an impression that I remember it. This one even included four parts of a royal flush to boot.
How rare is it? Well, let's work it out. The first card can be anything, because I'm interested in the question of getting all seven cards from the same suit, not just in diamonds specifically. The second card obviously has to match the suit of the first, or the process is foiled from the get-go. There are 51 unseen cards left in the deck, of which 12 are of the same suit, so the probability of getting dealt a second one is 12/51. Similarly, to get the third card to match, it must be one of the now 11 remaining of that suit out of the 50 unknown cards.
Continuing that logic, the cumulative probability of getting the last six cards all to match the first card in suit is 12/51 x 11/50 x 10/49 x 9/48 x 8/47 x 7/46. That works out to 0.000051 (0.0051%), or about one time in 19,491 hands.
As a double-check that I didn't screw that up, I went at it a different way. There are 1716 different seven-card combinations that you can pull from a quarter-deck of 13 cards (e.g., all the diamonds), because C(13, 7) = 1716. There are four suits, so a total of 1716 x 4 = 6864 different seven-card combinations you can draw from a standard deck that will be seven-card flushes. The total number of seven-card combinations you can draw from a 52-card deck is given by C(52, 7), which Excel tells me is 133,784,560. Therefore, the probability of drawing seven cards at random and having it be one of the all-one-suit combinations should be 6864/133,784,560, which is--TA DA!--0.000051, exactly the same as with the first approach. That give me great confidence that I have the number right.
It's meaningless, of course, since there is no bonus for having a seven-card flush instead of a five-card flush. But it sure is pretty--and quite rare.
Tuesday, September 29, 2009
So many diamonds!
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3 comments:
Don't think your logic isn't correct.
As it was Stud you would have already known some of the other cards that were out. So the randomness of cards is less than you say.
although it is late here and I may be being just stupid ??
Sure, one could make a different situation-specific calculation, as opposed to the general case that I presented here. If, e.g., it were a full ring game and you saw three diamonds in your hand and 6 more out among your opponents' up cards, the chances of getting all of the remaining 4 diamonds is obviously smaller than if you saw no diamonds out. But here there was only one card to see at the beginning, and only 4 visible total even by the end, so the calculation wouldn't change by very much.
What would be even more spectacular, or tragic I suppose, would be not getting the ace of diamonds and loosing to a higher diamond flush.
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