Then 6 ways to choose a pair from each rank, (Good).įirst there are C(13,6) = 1716 ways to choose six ranks. Regarding the probability of dragons, without looking at how you figured it, here's how I'd do it: At this point I plan to add that URL to the FAQ. So, assuming I didn't completely botch that I come up with 2,856,863,152 6-pair hands which makes the odds only 221 to 1.ĭouble-check on the first (largest) number:Ħ ranks chosen from 13 = 1716 distinct ranksĮach rank can have two cards chosen from 4 = 6 choices of suit per rank, or 6^6 = 46,656 suit assignmentsĥ2 cards minus the 24 cards of the ranks chosen = 28 unmatched cardsĭragon is one card of every rank. But some of the cards you can pick for the 13th card match a pair, so it would be easy to double-count.Ħ pairs, no trips: (13C6)* 6^6 * 28 cards left over = 2,241,727,488 possibilitiesġ quad + 4 pairs, no trips: 13 * (12C4) * 6^4 * 32 cards = 266,872,320Ģ quads + 2 pairs, no trips: (13C2) * (11C2) * 6^2 * 36 cards = 5,559,840 The brute force approach: 6 pairs can be made up of either distinct pairs or quads. How did you calculate 6 pairs? I'm not able to come up with an easy way. Some of these will be dragons in addition to being 3-flushes, but obviously only a small fraction.

So, we have 5,705,516,392 hands that can be played as 3-flushes, giving odds of about 110:1. Percentage % of a A-K (including A-K suited): 0.01057%Ĭounting or not counting the 4 A-K suited hands doesn't make any significant difference.ģ-flushes are a bit harder because some of them could be two-flushes, although the effect of those cases is pretty small.

# of 13-card hands consisting of A-K with any suits = number of ways to pick one suit for each of 13 cards = 4^13 = 67,108,864

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