RAINBOW FLUSH QUESTION

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New2vp
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Joined: Mon Sep 11, 2006 4:02 am

RAINBOW FLUSH QUESTION

Post by New2vp »

Mr. Ryan asked the following questions on another forum. Bill, I hope you don't mind that I answered here instead.

"Can anyone give the frequency of getting 1)a rainbow flush-a hand with all four suits and 2)the same for all four suits and an Ace.
Would the second happen approx 1/13th of the first hand?"

If this concerned dealt hands with a standard 52-card deck, here is my response, with the last question answered first.

"Would the second happen approx 1/13th of the first hand?" This one is the easiest. My guess before doing the work would be: No, it probably occurs a little more than 1/3 of the time. If we were talking about one-card hands, aces would indeed occur 1/13 of the time. But with each card added to the hand, up to 49, the odds of getting at least one ace in the hand increases. For 2-card hands, about 15% have at least one ace, for 3-card, it moves up to 22%, and so on. As many know, there are COMBIN(52,5) = 2,598,960 possible dealt 5-card hands. Subtracting the COMBIN(48,5) = 1,712,304 hands with no aces leaves you with 886,656 hands with at least one ace. So the chance of getting at least one ace in a dealt hand is 886,656/2,598,960 or about 0.3411. This does not exactly answer the question, but it likely is a reasonable approximation.

"give the frequency of getting 1)a rainbow flush-a hand with all four suits" I interpret this to mean a 5-card hand with two of one suit and one of each of the other three. There are COMBIN(13,2) = 78 different 2-card combinations in a suit. Each of the other 3 suits can be any card, so there are 13 x 13 x 13 of these 3-card combinations. Finally, there are 4 possible suits to be the 2-card suit. So, the total number of Rainbow Flushes are 4 x COMBIN(13,2) x 13^3 = 685,464 hands. This is 26.37% of the total.

"the same for all four suits and an Ace." It is easier to calculate the number of hands with no aces and a 2-1-1-1 suit distribution, so I will do that and subtract to get the answer. If we do it this way, we can use the same formula as above except substituting 12 cards per suit rather than 13 (12 cards because we are eliminating the aces from the deck). So, we get 4 x COMBIN(12,2) x 12^3 = 456,192 such hands with no ace. Subtracting this number from what we got in the previous paragraph leaves us with 229,272 rainbow flushes with at least one ace or 8.82% of all dealt hands. Now, finishing up the last question, division of 229,272 by 685,464 will get us 0.3345, which is fairly close to what we estimated.

Note, that the previous paragraph says "at least one ace." Included in that total will be several hands with a pair of aces, trip aces, and all 48 hands with quad aces. I will leave it as an exercise if you want to exclude those hands and count only hands with a single ace. It seems more involved to do that.

Also, if you are interested in counting drawn hands, it would depend very much on the pay schedule and would require more specialized programming than the pencil-and-paper statistics that I used here.

I would be interested in your reason for the request. But, there you have it.

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