Calculating Poker Odds

calculating poker odds with wild cards?

I need to know how to calculate the odds of poker hands with wild cards. I actually want to know how to calculate, not what the odds are.
At the moment I only know how to calculate odds with 1+ standard decks (equal number of ranks in each suit), but I don't know how to calculate odds when there are cards which don't fall into the standard ranks and suits (i.e wild cards).

Not sure whether you meant ordinary five-card poker or holdem - assuming you meant five-card.
I asked the same question a while ago and got no proper answer (http://answers.yahoo.com/question/index?qid=20100512172402AAzfJFX)

You have to consider each hand type, and enumerate (piecewise) the permutations it could be formed both without joker ('woj') and with the joker ('wj').
(We also make the commonsense assumption that if there is a joker, it will always be used to make the strongest possible hand, which is not always immediately apparent in the enumeration - see below.)
For the mid-ranking hands in particular, you also have to enumerate and exclude unwanted higher-ranking hands (e.g. exclude FullHouse from TwoPair or ThreeOfAKind, exclude StraightFlush from Flush or Straight).
All this excluding special cases gets to be a gigantic pain in the ass, e.g. we also have to exclude hands where the joker could be used to bridge a gap of one to form a straight from a near-straight, so it seems people like WizardOfOdds.com compute things of this complexity level by program instead.

Let's compute a few:

Total # of hand permutations: C(53,5) = 2,869,685

HighCard: cannot have joker, pick 5 non-consecutive ranks (also exclude wheel straight), and pick any 5 (non-flush) suits (i.e. exclude the case where all 5 are of same suit).
[C(13,5) - 9-1] * [5^4 -1] = [1277]*[624] = 796,848
=> P(HighCard (when deck contains joker)) = 0.278

Pair ('woj'): pick a rank for the pair, any two suits for the pair cards, any three other ranks and any three suits for those.
C(13,1)*C(4,2) * C(12,3)* C(4,1)³
Pair ('wj'): similar to joker+HighCard hand: pick joker + 4 non-consecutive ranks (exclude both straights and gap-one near-straights), and 4 (non-flush) suits.
1* [C(13,4) - 9-1 -near-straights] * [4^4 -1]

As you can see it gets real complicated real fast. For holdem with N players it's dimensions worse - it also differs depending on whether the joker is in one player's hand ('JP') or the community cards ('JC'), or neither ('NJ'). 'JC' is obviously the most complex case.

and of course remember our extra possible hand type:
FiveOfAKind is formed by Quads + joker
P(FiveOfAKind) = 13 / 2,869,685 = 0.0000045

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