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#1
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Sorry if this seems like such a basic question, but how would you find the possible number of combinations a holding can be dealt? For example, if you believe a player has JJ, there are 6 ways that can be dealt? So in short, whats the equation for finding this?
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#2
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This is basic probability, pick up a high school probability maths book will explain everything.
But in essence, for JJ: #combinations = 4 choose 2 (which is labelled 4C2 in math-speak - with the 4 a superscript and the 2 a subscript). In fact, p choose r = #no of ways to choose R objects from a set of P elements = pCr = p! / (p-r)!r! where p! = p.(p-1).(p-2)....3.2.1 e.g. 4! = 4.3.2.1 = 12 So for the JJ scenario, p = 4, r = 2 4C2 = 4! / (4-2)!.2! = 12/2x2 = 12/4 = 6 |
#3
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So basically, out of four jacks, you can choose any of the four for the first card. For the second whole card, you are left with three, so that is 4x3, i got that, and you just divide by 2 always? and how would you find AK?
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#4
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You divide by two because you would be double counting otherwise.
J [img]/images/graemlins/club.gif[/img] J [img]/images/graemlins/heart.gif[/img] is the same as J [img]/images/graemlins/heart.gif[/img] J [img]/images/graemlins/club.gif[/img]. But 4x3 counts them both individually. When order matters, it's called a permutation. When it doesn't, it's a combination. With AK, it's (4x4)/2 = 8. Permutations and combinations explained. |
#5
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Wrong.
AK is (8 X 4) / 2 = 16 combinations. Thanks for playing. |
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