From a 52 card deck it is possible to deal 2,598,960 different five-card poker hands.

probability/card/poker.hands New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

From a 52 card deck it is possible to deal 2,598,960 different
five-card poker hands.  Of these 1,088,240 will contain a pair.  Other
possibilities are 4 royal flushes, 36 straight flushes, 624
four-of-a-kind hands, 3,744 full houses, 10,200 straights, 54,912
three-of-a-kind hands and 123,552 two-pair hands.  How do you come up
with these numbers?

-- Dave Turner <smilindave@webtv.net> (December 1997)

Show solution

Here (at UCIrvine), we call this Math 6A.

5 card hands: 52 C 5 (read 52 choose 5) - that is, there are 52
              possibilities for the first card, times 51 for the
              second, times 50, 49, and 48 for the last 3, but the
              order of the cards is not important so we divide by the
              number of orders (5! or 5*4*3*2*1).  So this value should
              be (52*51*50*49*48)/(5*4*3*2*1) = 2598960

4 Royal Flushes should be clear as we can list them (Spades, Hearts,
              Diamonds and Clubs).

For the Straight Flushes, there are 4 suits and 10 possible high
              cards: 5, 6, 7, 8, 9, 10, J, Q, K, A so there are
              4*10 = 40 SF's.  Four of these (the ones topped by the
              A's) are RF's and presumably we want to make a
              distinction, thus leaving 36.

4 Of a Kind:  We have 13 possible values for the 4 of a kind (4 queens,
              4 sevens, etc.) and 48 remaining cards to fill out our
              hand.  Thus there are 13*48 = 624

Full House:   A Full House has a group of 3 of the same value and a
              group of 2 of the same value.  Since there are 13
              possible values (2 - A) for the 3 of a kind and 12
              remaining for the pair, there are
              13*12*(4 C 3 = 4)*(4 C 2 = 6) = 3744

Flush:        A Flush is a set of 5 cards all of the same suit.  There
              are 4 suits to choose from and 13 C 5 ways to pick 5
              cards from a suit.  But 10 of these will be straights
              which would be a better hand, so we don't count those.
              Thus there are: 4*[(13 C 5) - 10] = 5108 (I know you
              didn't ask, but it seemed like you might want it).

Straights:    A straight is 5 numerically consecutive cards not all of
              the same suit.  There are 10 possible starting points for
              our straight.  Each card can be one of four suits so
              there are 4^5 suit patterns.  40 of these will be SF's or
              RF's so we don't want to count those.  Thus there are
              10*4^5-40 = 10200 straights.

3 of a Kind:  This is a bit more subtle.  As before there are
              13*(4 C 3) ways to get the set of 3.  The 2 remaining
              cards cannot be the same value as the 3 (giving us 4 of
              a kind) or each other (a Full House).  Thus there are 48
              possibilities for the 4th card and 44 for the 5th.  Since
              the order of these last two does not matter, we divide by
              2! = 2.  Thus there are: 13*(4 C 3 = 4)*48*44/2 = 54912
              3 of a kinds.

Two Pair:     Again we must be careful.  There are (13 C 2) ways to
              pick the values for the pairs and (4 C 2) ways to pick
              the suits for each.  The remaining card cannot be either
              of these 2 values thus leaving 44 cards to choose from.
              There are, therefore, (13 C 2)*(4 C 2)*(4 C 2)*44 =
              123,552 two pair hands.

One Pair:     Last, and least, one pair.  There are 13 values this pair
              can have.  For each value there are (4 C 2 = 6) ways to
              pick the suits.  The remaining 3 cards cannot match this
              pair or each other.  Thus there are 48, 44, and 40
              possible cards for the remaining 3 in our hand.  Since
              3H, 3S, 5C, 7D, KS is the same as 3H, 3S, KS, 5C, 7D, we
              must divide these last 3 by 3! = 6.  This gives
              13*(4 C 2 = 6)*48*44*40/6 = 1098240.  Note: your value
              has a typo.

Garbage:      The number of ways to have junk in your hand is just the
              total number of hands minus all of the hands with a pair
              or better.  This is left as an exercise :)

-- Matt <mroozee@math.uci.edu>