FOLLOWING ORIGINAL RULE

The casino challenge requires combining two hole cards with five community
cards, making a hand of four 9s for one player, and 9-8-7-6-5 diamond
straight flush for another, among six players.   The community cards must
therefore include 9 of diamonds ("9D"), one of [9S, 9H, 9C], two diamonds
chosen from [8D, 7D, 6D, 5D].   I refer to the Other 44 cards in the deck
as Os.  The community cards contain one O.

The probability of these exact cards: 9D, 8D, 7D, 6D, 5D, 9S, 9H, 9C,
followed by nine Os, being dealt, in that order, from a well-shuffled deck,
is 1*1*1*1*1*1*1*1*44*43*42*41*40*39*38*37*36 divided by
52*51*50*49*48*47*46*45*44*43*42*41*40*39*38*37*36 = 3.29...x 10^-14.  The
numerators are merely permuted when the order of dealing the prescribed
cards is permuted.  The probability of winning the challenge is therefore
3.29 x 10^-14, times the number of winning arrangements of [first hole
cards, times second hole cards, times  community cards], times the 4! = 24
permutations of [8D, 7D, 6D, 5D], times the 3! = 6 permutations of [9S, 9H,
9C].  I treat the 9D as nonequivalent to the other 9s, because it must
appear in the community cards.

An example winning arrangement starts with 9S, 7D, and four Os in the first
round of hole cards.  There are 6 x 5 = 30 ways to place the 9S and the 7D.

In the second round of hole cards, say 9C, 6D, and four Os are dealt.  The
same two players must receive the second 9 and the second diamond, to meet
the casino challenge, so the number of ways to place the 9C and the 6D is 1.

The winning community cards are 9D, 9H, 8D, 5D, and one O.  There are
5*4*3*2/2 = 60 ways to deal the four significant cards (considering 8D and
5D positionally equivalent here, to avoid subsequent double-counting).  So,
the probability of the example sequence is 3.29 x 10^-14 x 30 x 1 x 60 =
5.9 x 10^-11.  Finally multiplying by 6 X 24 permutations of 9s and
diamonds, the probability that a random deal to six players wins the casino
challenge is  8.5 x 10^-9.


WITH ALTERNATIVE RULE - NOT REQUIRING USE OF BOTH HOLE CARDS

The solutions from the strict rules are still valid.  Here are additional
allowed arrangements of the eight specified cards.

THIRD 9 REPLACING "O" IN THE COMMUNITY CARDS: An example winning hole card
sequence is 5D and 7D aligned to one player, and 9H dealt to another
player.  There are 6 placements of 5D in the first round of hole cards.
After aligning the 7D to 5D, there are 5 placements of 9H in the second
round of hole cards.  There are 5!/2!/2! = 30 orderings of the community
cards (considering 6D equivalent to 8D, 9C equivalent to 9S).  Probability
of this example is 6 x 5 x 30 x 3.29 x 10^-14.  Appending a factor of 2 for
the 9H appearing in either round of hole cards, 4! for permutations of
diamonds, and 3! for permutations of 9s, the combined added probability is
8.5 x 10^-9.

THIRD SIDE DIAMOND REPLACING "O" IN THE COMMUNITY CARDS: An example winning
hole card sequence is 9C and 9S aligned to one player, 6D dealt to another
player.  There are 6 placements of 9C in the first round of hole cards.
After aligning 9S to 9C, there are 5 placements of 5D in the second round
of hole cards.  There are 5!/3! = 20 orderings of the community cards
(considering 5D, 7D, 8D equivalent).  Probability of this example is 6 x 5
x 20 x 3.29 x 10^-14.  With a factor of 2 for the diamond in either round
of hole cards, x 4! x 3!, the combined added probability is 5.7 x 10^-9.
With the relaxed rules, the probability of a random deal satisfying the
casino challenge has very modestly increased to 22.7 x 10^-9.