Rod Pemberton wrote:
>> assuming 52(m) cards to be distributed in junks of five(n)
>> for the first five cards:
>> m*(m-1)*(m-2)*(m-3)*(m-4)/fac(n) ;the other meaning of 'FAC' yet ;)
>> 52*51*50*49*48 / 2*3*4*5 = 2598960 variants
>> the next set of five:
>> 47*46*45*44*43 / 2*3*4*5 = 1533939 variants
>> and so on ...
> until...
> 5*4*3*2*1 / 2*3*4*5 = 1 variant
I'd have only two cards left at the end :)
> Houston, we have a problem...
> It appears that's combinations (un-ordered set).
> A deck order of 1,2,3,4,5 is distinctly different from a deck order of
> 2,5,4,1,3. The game needs to keep track of the deck order. The
remaining
> five cards could be 1,50,23,37,9. I'd think we'd want permutations
> (ordered sets). For 52 cards:
> 52!=52*51*... = (big number Wolf posted)
> For five cards remaining:
>
> 5*4*3*2*1 / 1! = 120
> So, I asked: if 1,2,3,4,5 is ranked as first of 120 permutations of
> 1,2,3,4,5, and all permutations are ranked from 1 to 120, what
permutation
> out of 120 is 2,5,4,1,3? (I used example value 33.) Is there a way to
> generate the permutation from this rank number? Or, vice versa?
I think the order of a set of five isn't relevant at all in a cards game,
so we may always see and sort it ascending (like most players do).
But if you want an order oriented numbering it will depend on the
chosen way if you can recalculate an order by their given number.
>> but I hope you don't mean to make an array for all variants :)
> Nope, I took it that Alex stated one could generate different deck
orders
> for an un-ordered set from an equation - without generating all possible
> ordered sets...
I'm not too familiar with gambling code, but I think it is possible
to create an algo which works both directions.
__
wolfgang
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