We are asked to find the odds of a hand of four 9's and a hand of a 9-high diamond straight flush both occurring in a single deal of Texas Hold-'em, with six players. We want to know the odds both with the requirement that each hand contains both hole cards, and without such a restriction. To reduce clutter in the explanations, we use the following terminology: Cards 8D, 7D, 6D, and 5D are "sfd's" (straight-flush diamonds). The three 9's other than 9D are "nd9's" (non-diamond 9's). 9D, the nd9's, and the sfd's are the eight "target cards." Note that the answer to this problem does not depend on the order in which we deal. So for ease of explication, we deal out the five community cards first, and then, pairwise, the 12 hole cards going to the six players. We make the following observations: 1) The 9D must be in community because a single player cannot have both four 9's and a straight flush simultaneously, that requiring at least 8 cards. 2) Community must contain at least one 9 in addition to 9D, since a player has at most two 9's as hole cards. 3) Community must contain at least two sfd's, besides 9D, since a player has at most two sfd's as hole cards. 4) Besides the minimum four target cards required to fulfill 1 through 3 above, community will have a fifth card that is an nd9, an sfd, or a non-target card. We will calculate three probabilities, one for each of these situations. 5) To satisfy the condition that each hand uses both hole cards, there cannot be two nd9's or three sfd's in addition to 9D in community, because then a single 9 or a single sfd would be sufficient to complete one of the hands being sought. Thus, in this case, community must contain a non-target card. The rationale for our calculations is that we deal out the five community cards and take the probability of the required cards being contained therein. Then, given that the community cards contain those cards, we take the probability that the next twelve cards, dealt as six pairs of hole cards, contain the cards that complete the two hands being sought. We multiply those together to get the joint probability of the proper community cards being dealt and the proper hole cards being dealt. We do this calculation for each of three situations, depending on what the "final" community card is, as set forth in observation 4 above. We then manipulate these three results to arrive at the final answers. Situation A ( Community consists of 9D, one nd9, three sfd's ) Probability of this community being dealt in first five cards: 1/52 * 3/51 * 4/50 * 3/49 * 2/48 * 5! / 3! (There are 5! / 1!1!3! different "configurations" of dealing out a 9D, one nd9, and three sfd's, each configuration having equal probability. We calculate the probability of one such configuration and multiply that by the number of configurations.) Probability of next 12 cards ( dealt out as 6 pairs of hole cards) containing a pair of hole cards consisting of a single sfd and a non-target card, and another consisting of a pair of nd9's: ( (1/47 * 44/46) + (44/47 * 1/46) ) * 2/45 * 1/44 * 6! / 4! (The first pair of hole cards' being the one containing the one remaining sfd can occur in two equally probable and mutually exclusive ways: the sfd comes first, with a non-target card coming second, and vice-versa. There are four pairs of hole-cards that consist of all non-target cards. This results in 6! / 4! configurations of these six pairs of hole cards.) Situation B: ( Community consists of 9D, two nd9's, two sfd's ) Probability of this community being dealt in first five cards: 1/52 * 3/51 * 2/50 * 4/49 * 3/48 * 5! / 2!2! Probability of next 12 cards ( dealt out as 6 pairs of hole cards) containing a pair of hole cards consisting of a single nd9 and a non-target card, and another consisting of a pair of sfd's: ( (1/47 * 44/46) + (44/47 * 1/46) ) * 2/45 * 1/44 * 6! / 4! Situation C: ( Community consists of 9D, one nd9, two sfd's, one non-target card ) Probability of this community being dealt in first five cards: 1/52 * 3/51 * 4/50 * 3/49 * 44/48 * 5! / 2! Probability of next 12 cards ( dealt out as 6 pairs of hole cards) containing a pair of hole cards consisting of a pair of nd9's and another consisting of a pair of sfd's: 2/47 * 1/46 * 2/45 * 1/44 * 6! / 4! Cranking out the calculations, we arrive at the following probabilities for obtaining a 9-high diamond straight flush and four 9's: Situation A: 5.6950127 * 10 ^ -9 Situation B: 8.542519 * 10 ^ -9 Situation c: 8.542519 * 10 ^ -9 With the restriction that both hole cards must be used, the probability is that of Situation C by itself, 8.542519 * 10 ^ -9 Without that restriction, the probability is that of all three situations added together, or 2.278 * 10 ^ -8