Re: Probability of getting different card hands.
- From: mike4ty4@xxxxxxxxx
- Date: 13 Oct 2005 20:26:44 -0700
Proginoskes wrote:
> mike4ty4@xxxxxxxxx wrote:
> > Hi.
> >
> > Suppose you have a deck of 52 playing cards, and you are going to play
> > an 8-card (yep, 8 cards!) extended (because we have more than five
> > cards) poker game. In this type of game, there would be the following
> > hands:
> >
> > Royal flush: AKQJ10987 all of the same suit
> > Straight flush: In numeric order (either A2345678910JQK or
> > 2345678910JQKA), all of the same suit, but not a royal flush
> > Flush: All of the same suit, but not straight or royal
> > Straight: Numeric order but not all the same suit
> > Dual quads: Two quads, like AAAAKKKK
> > Quads/four of a kind: Four cards of the same number: 22223KJ9
> > Quads and three of a kind: Ex. 4444KKKA
> > Quads and a pair: Ex. 10101010AA32
> > Dual three of a kind and a pair: Ex. 555JJJ1010
> > Dual three of a kind: Ex. 999AAA3J
> > Three of a kind & a pair: Ex. 777225KA
> > Three of a kind: Three of the same number: 555AKJ107
> > Four pair: Four pairs, ie. 88AA2266
> > Three pair: Three pairs, ie. KKJJ10106Q
> > Two pair: Two pairs, ie. JJ9952KA
> > One pair: Two cards of the same number, ie. AA32910KJ
> > No pair: No pairs and no other patterns at all, ie. 6JQ8K10A2
> >
> > What is the probability of each of these, if the game is played w/a
> > single deck? I figured that for a royal flush you ahve a probability of
> > 1 in 188,134,537.5, for a straight flush it's 1 in 31,355,756.25, and
> > for dual quads, 1 in 4,823,962.5. Are these right?
>
> The number of royal flushes is 4 (one per suit), and the number of ways
> to get 8 cards is C(52,8). The probability is thus 4/C(52,8), which
> matches what you have.
>
> The number of straight flushes (other than royal flushes) is 4 * 6, so
> your answer would be 24/C(52,8), and this also matches what you got.
>
> For dual quads, you have to choose the suits (C(13,2) ways, since the
> order isn't important, and you can't choose the same suit twice), and
> the hand is decided at that point. Answer: C(13,2)/C(52,8), which is 1
> in 9,647,925, so I suspect you counted AAAAKKKK as being different from
> KKKKAAAA.
>
Yep -- didn't notice that. I went and counted them with order as
mattering, getting 13 * 12 = 156 hands, but it should be half that (the
reversed form, ie. 22223333 is identical to the regular one) -- (13 *
12)/2 = 13 * 6 = 78 hands, or 13C2 (ie. order does not count).
> > What abou thte other
> > hands? What are the probabilities of those? The probabilities are
> > needed to figure out which hands are worth more. Like, for example,
> > dues dual three of a kind & a pair beat four pair, or the other way
> > around? Does four of a kind beat three pair? Thanks for any help.
>
> The best way to do these (correctly) is by using counting formulas.
> None of the hands look difficult to calculate, but I'd need more time
> to get all of them.
>
There's a few more hands I just realized also exist: quads & two pair,
three of a kind & two pair, pair-straight (four pair with all pairs in
numeric order), half-straight (4 cards in numeric order), half-flush (4
cards of the same suit), and half-straight flush (4 cards in numeric
order & all the same suit).
> --- Christopher Heckman
.
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