How many are there where we have no more than 2 of any suit?
What is the math behind the answer?
10-29-2014 08:50 PM
#1
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starting pineapple handsHow many are there where we have no more than 2 of any suit? | |
10-30-2014 01:28 PM
#2
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I worked this out for the 14 card draws a while ago, so this is much simpler, but the math is the same. | |
10-30-2014 01:37 PM
#3
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Summary of flushes on a 14 card FL draw in pineapple: | |
10-30-2014 01:38 PM
#4
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Hmm, that's a lot of hands. It's even more when we look at all the 3, 4 and 5 suited hands. I was asking because we have over 1 million pineapple hands in our db and we thought it might be cool to see which 5 card sets led to the most royalties. Maybe we don't have enough data for this type of thing yet? | |
10-30-2014 02:03 PM
#5
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You just included every possible hand in C(52,5) ~= 2.6 million | |
10-30-2014 04:28 PM
#6
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This counts a lot of similar hands, and you can probably ignore the 4 and the 12 in those equations, which is | |
10-30-2014 05:27 PM
#7
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Does this count the hands below as two different hands? | |
10-30-2014 08:03 PM
#8
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The 250,458 result counts those as the same. | |
10-30-2014 08:10 PM
#9
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OK, I see what you were saying. Here's the confusion. When I said this: | |
10-30-2014 10:20 PM
#10
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Ok, so when all aakkq hands with no more than 2 of a suit are counted as 1 then our total drops down to 250kish, right? In other words, we also count A K Q A K as the same hand, right? | |
10-30-2014 11:43 PM
#11
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10-31-2014 09:45 AM
#12
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Wow, I was tired last night. I think I got hung up on the card-values, but I wasn't seeing your point. | |