The last table contains the information on the number of semi-deals in which at least 2 sets of quads at least as big as some threshold rank or 2 straight flushes occur. In order to determine the probability that 4-of-a-kind or better loses, we still must count the number of semi-deals in which exactly 1 player makes 4-of-a-kind and exactly 1 player makes a straight flush. These are the only semi-deals not included in the table. There are many cases to wade through.

First, let's assume there are precisely 3 hearts in the board. The rank sets which work are , , , and doubling up on the middle 2 by letting the non-consecutive rank be smaller. There are 64 such rank sets. Since there are 4 choices of suit, there are 256 sets of 3 suited cards allowing a straight flush.

The completion of the 3 suited cards to the board is broken into subcases.
For the first subcase consider those boards having a pair for which neither
card is in the flush suit. There are 10 choices for ranks and 3 choices for
the cards comprising the pair. Altogether we obtain 7,680 boards of this
type. A player gets 4-of-a-kind only if the remaining 2 cards of the pair
end up with some player so there is no choice here. For some of the sets
of 3 suited cards, there is more than 1 way to get a straight flush. For
example, if the board has the 4,5,7 of hearts, then either the 3,6 of
hearts or the 6,8 of hearts produce a straight flush. We call this a
*board-player pair*. Going over the rank sets above leads to 400
board-player pairs so that the 7,680 boards actually produce 12,000
situations (from 30 times 400) in which there will be 2 cards producing
a straight flush and 2 cards producing 4-of-a-kind. Then we are
choosing 2 cards from 43 to complete one of the hands, 2 cards from 41 to
complete the other hand, 32 cards from 39 to distribute to the other
players, and the number of semi-deals in (5) for the 32 cards. Multiplying
all these numbers gives us

8,102,647,245,440,680,097,195,556,776,250,000,000 | (17) |

semi-deals with 3 suited cards for which exactly 1 player has 4-of-a-kind and at least 1 player has a straight flush. The ``at least'' in the preceding sentence indicates one small problem. The semi-deals counted in (17) include some for which 2 players have straight flushes. We must eliminate them. The only way this can happen with 3 suited cards is if the rank set has the form , where . There are 32 such rank sets and there are 30 choices for the pair giving us 960 boards. The 6 cards which give us the 3 special hands can be allotted in different ways because we are dealing with Omaha. One possibility is that 1 player has the 4 straight flush cards and a second player has the cards producing quads. So we are choosing 2 cards from 41 to complete the 1 hand, 32 from 39 to distribute to the other 8 players, and the number of semi-deals in (5). We take the product of the 4 numbers to get

717,842,502,364,622,821,456,970,700,000,000 | (18) |

semi-deals. This kind of semi-deal is counted twice in (17) but is counted nowhere else so we should subtract (18) from (17).

The other possibility for 2 players is that 1 player has the cards for a
straight flush and quads, while a second player has straight flush cards.
There are 2 choices for which straight flush cards go with the cards for
quads, 2 choices from 41 to complete the 1 hand, 32 choices from 39 to
be distributed to 8 players in (5) ways. The product we get in this
case is

1,435,685,004,729,245,642,913,941,400,000,000 | (19) |

semi-deals. These semi-deals are counted once in (17), but have been counted earlier so we must subtract (19) from (17).

Finally, there may be 3 players getting the special cards. We complete
1 hand in
ways, the second hand in
ways, and the third hand in
ways. For the rest of the
cards we are choosing 28 cards from 35 to be distributed in (6) ways.
The product is

34,456,440,113,501,895,429,934,593,600,000,000 | (20) |

semi-deals. These are counted twice in (17) and have been counted before so that we must subtract (20) twice from (17).

Making the appropriate corrections on (17) yields

8,031,580,837,706,582,437,871,316,676,950,000,000 | (21) |

semi-deals with exactly 1 player having a straight flush and exactly 1 other player having 4-of-a-kind when there are 3 suited cards on board, and the pair on board does not use any of the suited cards. Of course, we want these numbers to reflect a threshold for the rank of the 4-of-a-kind. Instead of doing all the computations over again for each rank, we observe that the semi-deals counted in (21) place no restriction on the rank. We can get the restriction in place by simply working with (21) and using proportions. If we examine the various rank sets, we find that the ranks occur this often as valid ranks for quads: A (80), 2 (78), 3 (68), 4 (58), 5 (48), 6 (48), 7 (48), 8 (48), 9 (48), 10 (50), J (60), Q (70), and K (80). The sum is 784 and this means that deuces appear as quads in 78/784 of the semi-deals counted in (21). Thus, we see how to verify for the various thresholds. The following table gives this information.

We now consider the subcase of the board containing a single pair with
one of the cards in the pair being in the flush suit. There are 9
choices for the card which pairs one of the 3 suited cards and there are
30 choices for the last card in the board. This then gives us 69,120
boards. However, again we must consider the board-player pairs which
leads to 108,000 situation where there are 2 cards giving a straight
flush and 2 cards giving 4-of-a-kind. We are choosing 2 cards from 43
to complete the 4-of-a-kind hand and 2 cards from 41 to complete the
straight flush. The remaining 32 cards are chosen from 39 and they are
distributed in (5) possible semi-deals. Multiplying the various numbers
gives rise to

72,923,825,208,966,120,874,760,010,986,250,000,000 | (22) |

semi-deals. Again we have to eliminate those which have 2 players ending up with straight flushes. There are 32 rank sets which allow 2 straight flushes and there are 270 choices for the other 2 cards. This gives us 8,640 boards allowing 2 straight flushes. If 1 player has 2 straight flushes and another player has 4-of-a-kind, then we are choosing 2 cards from 41 to complete the 1 hand, choosing 32 from 39 for the other players to be dsitributed in (5) ways. The product is

6,460,582,521,281,605,393,112,736,300,000,000 | (23) |

and these semi-deals with 1 player having 2 straight flushes and 1 player having 4-of-a-kind is counted twice in (22) but nowhere else. Thus, we need to subtract (23) from (22).

A second possibility is for 1 player to have quads and a straight flush,
and another player to have a straight flush. The numbers are the same as
the preceding subcase except that we have 2 choices for which straight
flush to put with the quads. Thus, we get exactly twice as many semi-deals
as (23) which is

12,921,165,042,563,210,786,225,472,600,000,000. | (24) |

The semi-deals counted in (24) have been counted before so that we need to subtract (24) from (22) as well.

Finally, the 3 good hands may be spread over 3 players in which case we
are choosing 2 from 41, 2 from 39, 2 from 37, 28 from 35 and distributing
the latter in the number of ways from (6). This product gives

310,107,961,021,517,058,869,411,342,400,000,000 | (25) |

semi-deals. They are counted twice in (22) and have been counted before. Thus, we need to subtract twice (25) from (22). Doing all these adjustments to (22) yields

72,284,227,539,359,241,940,841,850,092,550,000,000 | (26) |

semi-deals in which exactly 1 player gets a straight flush and exactly 1 player gets 4-of-a-kind when there are 3 suited cards on board and a single pair involving 1 of the suited cards.

The semi-deals counted in (26) have no restriction on the rank of the 4-of-a-kind. We can do so again using the proportionality idea. In this case, we have the following appearances for the ranks in pairs as follows: A (12), 2 (12), 3 (18), 4 (24), 5 (30), 6 (30), 7 (30), 8 (30), 9 (30), 10 (30), J (24), Q (18), and K (12). The following table gives us the number of semi-deals with exactly 1 player having quads at least as big as the qualifying rank and exactly 1 other player having a straight flush when the board has 3 suited cards and a single pair involving 1 of the suited cards.

The last subcase we consider here is 3 suited cards on board as well as
2 pairs. This implies both pairs must involve cards from the flush suit.
There are 256 sets of 3 suited cards and 27 ways of choosing 2 pairs from
the 3 ranks represented by the 3 suited cards. We again take 27 times
400 to take into account the board-player pairs giving us 10,800 boards
to play with. For each of them we have 2 choices for the pair to become
4-of-a-kind, we are choosing 2 from 43 to complete the straight flush
hand, we are choosing 2 from 41 to complete the 4-of-a-kind hand, and we
are choosing 32 from 39 to be distributed in (5) ways to the remaining 8
players. This then gives us

14,584,765,041,793,224,174,952,002,197,250,000,000 | (27) |

semi-deals. Now we have to consider duplicate counting which is messy in this case. The number of semi-deals with 2 straight flushes and 2 sets of quads is obtained by restricting ourselves to the 32 rank sets which allow 2 straight flushes. There are 27 ways of choosing the pairs giving us 864 boards. Now 8 of the cards are determined, but they can be allotted to different numbers of players. First, suppose only 2 players end up with the good hands. If 1 player has completions of both pairs and a second player has completions of both straight flushes, then we are choosing 32 from 39 and distributing them in (5) ways to the other 8 players. This gives us

787,875,917,229,464,072,330,821,500,000 | (28) |

semi-deals with 2 straight flushes and 2 sets of quads. They have been counted 4 times in (27) but nowhere else, so we need to remove 3 times (28) from (27).

Continuing with the preceding subcase, if each of 2 players has a completion to quads and a completion to a straight flush, then we get the same number as (28) again but there are 2 separate pairings possible and each is counted twice in (27). They have been counted before so that we must remove another 4 times (28) from (27) for this situation.

If 3 players end up with the good hands, 1 player must have 2 of them in
a single hand. This sets up 3 little subcases: 1 player gets both sets
of quads, 1 player gets quads and a straight flush, or 1 player gets both
staight flushes. All 3 of them involve the same numbers to multiply,
namely, 864,
,
,
,
and (6). Multiplying these numbers gives

37,818,044,027,014,275,471,879,432,000,000 | (29) |

semi-deals. Altogether they are counted 11 times in (27) and all have appeared in earlier counts. Thus, we need to remove 11 times (29) from (27).

If the good hands are spread over 4 players, then we are choosing 2 cards
successively from 39, 37, 35 and 33 for hand completion. The remaining
24 cards are chosen from 31 to distribute in (7) ways to 6 players. So
we get

1,588,357,849,134,599,569,818,936,144,000,000 | (30) |

semi-deals. Each of them is counted 4 times in (27) and have been counted before. Thus, we must remove 4 times (30) from (27).

The number of semi-deals with 2 straight flushes and at least 1 4-of-a-kind
is obtained by choosing the pair to put into the set of quads giving us
situations with 6 cards determined. We then get
subcases depending on how the good cards are allotted. If 2 players get
the good cards, then the number of semi-deals is the product of 1,728,
,
,
and (5). This product is

1,292,116,504,256,321,078,622,547,260,000,000. | (31) |

Let's look at the 2 scenarios for 2 players getting the good cards. If
1 player gets both straight flushes and another player gets the quads,
then the semi-deals counted in both (28) and (29) are each counted
twice in (31). So removing them gives us

1,214,904,664,367,833,599,534,126,753,000,000 | (32) |

semi-deals in which 1 player gets 2 straight flushes and exactly 1 other player gets quads. These are counted twice in (27) but counted nowhere else. This means we must subtract (32) from (27).

On the other hand, if 1 player gets quads and a straight flush, then
because there are 2 choices for which straight flush goes with the
quads we get 2 times (32) for the number of such semi-deals. There is
1 pattern of the type counted in (29) counted in this. So we must
remove them which then gives

2,546,414,964,485,627,881,773,215,088,000,000 | (33) |

semi-deals for which 1 player has both quads and a straight flush, and exactly 1 other player has a straight flush. These have been counted before so must be removed from (27).

We now continue with semi-deals having 2 straight flushes and at least 1
4-of-a-kind, where the good cards are spread over 3 players. The basic
number of semi-deals for this situation is given by multiplying 1,728,
,
,
,
,
and (6). Doing so yields

632,938,299,931,096,356,051,413,919,360,000,000 | (34) |

semi-deals. One of the types counted in (29) is counted here so we must remove them to give us

632,900,481,887,069,341,775,942,039,928,000,000 | (35) |

semi-deals with exactly 3 players having 2 straight flushes and quads amongst them. These are counted twice in (27) and have been counted earlier so that we must remove 2 times (35) from (27)

Let's now determine the number of semi-deals with 2 sets of quads and
exactly 1 straight flush. There are 10,800 boards taking into account
the board-player pairs (in other word, we have chosen the 2 flushing
cards to make a straight flush). Since we are assuming 2 sets of quads,
6 of the cards are determined and we must consider the different ways
they may be allotted to the players. As above, we first consider when
2 players share the goodies. The basic number of semi-deals is obtained
by taking the product of 10,800,
,
and the number in (5). Doing so gives us

8,075,728,151,602,006,741,390,920,375,000,000 | (36) |

semi-deals with 6 good cards held by 2 players. There are 2 subcases. If 1 player has both sets of quads, then one of the types counted in (29) is counted twice in (36). So removing 2 times (29) from (36) gives us

8,000,092,063,547,978,190,447,161,511,000,000 | (37) |

semi-deals in which 1 player makes 2 sets of quads and exactly 1 other player gets a straight flush. Such semi-deals are counted twice in (27) but have been counted nowhere else before. So we must subtract (37) from (27).

Now if 1 player holds quads and a straight flush while the other player
has quads, then there are 2 choices for pairing the quads with the
straight flush. One type of semi-deal counted in (29) is counted once
in (36) so we must subtract (29) from 2 times (36) to give us

16,113,638,259,176,999,207,309,961,318,000,000 | (38) |

semi-deals for which 1 player gets quads and a straight flush, and exactly 1 other player gets a straight flush. The semi-deals in (38) have been counted twice in (27) and have been counted earlier. Thus, we must subtract 2 times (38) from (27).

If the hands are spread over 3 players, there are

387,634,951,276,896,323,586,764,178,000,000,000 | (39) |

semi-deals. Types counted in (29) are counted 3 times in (39). So doing the appropriate subtraction gives us

387,521,497,144,815,280,760,348,539,704,000,000 | (40) |

semi-deals with exactly 3 players having 2 sets of quads and a straight flush amongst them. Such semi-deals are counted twice in (27) and have been counted earlier so that we need to subtract 2 times (40) from (27).

We are now ready to make the adjustments on (27). Performing the
appropriate subtractions gives us

12,493,160,602,010,112,248,091,363,201,205,500,000 | (41) |

semi-deals with exactly 1 player getting 4-of-a-kind and exactly 1 other player getting a straight flush, when the board has 3 suited cards and 2 pairs. No restriction has been placed on the rank of the quads so we again use proportions to bring in the restrictions. The appearance of the suited cards provides pair opportunities in the following proportions: A (12), 2 (12), 3 (18), 4 (24), 5 (30), 6 (30), 7 (30), 8 (30), 9 (30), 10 (30), J (24), Q (18) and K (12). The following table then contains the number of semi-deals with exactly 1 player holding quads and exactly 1 other player holding a straight flush, where the quads have some minimum rank.

We are now ready for the last case, that is, 4 suited cards in the board. In this case there can be at most 1 player with 4-of-a-kind and this is possible only if one of the 4 suited cards is paired. So the first step is to determine the number of rank sets of 4 suited cards which allow 1 or more straight flushes.

If there are 4 consecutive ranks, the smallest rank is ace through J
giving us 18 board-player pairs. If the rank set has the form
and it is not 4 consecutive ranks, then there are many subcases
which altogether produce
216 board-player pairs. Another possibility is a rank set of the form
,
where *y* is strictly bigger than *x*+2. Then *y* must
be either *x*+3 or *x*+4, there are 27 contributions to the
board-player pairs. There are many subcases with 1 set of 2 consecutive
ranks and they lead to 356 board-player pairs. Finally, there are 59
board-player pairs arising from rank sets with no successive ranks.
Summing gives us 676 board-pairs. There are 4 choices for suit, 4 ranks
for the pair, and 3 cards which may form the pair. The product is 32,448
which is the number of situations. Four of the cards are determined so
that we are choosing 2 cards from 43 to complete 1 hand, 2 cards from 41
to complete another hand, and 32 cards from 39 to be distributed in (5)
ways for the remaining 8 players. This product gives

21,909,558,151,671,598,982,816,785,522,980,000,000 | (42) |

semi-deals with exactly 1 player getting 4-of-a-kind and at least 1 player getting a straight flush with 4 suited cards and 1 pair on board. We must remove those which have 2 players getting a straight flush. The number of rank sets allowing 2 players to get straight flushes is 77. We multiply by 4 for the number of suits, 4 for the rank being paired, and 3 for the choice of the pairing card. This gives us 3,696 situations which may arise that we wish to eliminate. Now we have 6 cards making big hands. So we are choosing 2 cards from 41 to complete 1 of the hands, 32 from 39 to be distributed to the other players in (5) ways. This yields

2,763,693,634,103,797,862,609,337,195,000,000 | (43) |

semi-deals in which 2 players get the 6 good cards. If 1 player gets both straight flushes and the other player gets the quads, then this is being counted twice in (42) but has not been counted anywhere else. Thus, we subtract (43) from (42). On the other hand, if 1 player gets a straight flush and quads and the other player gets a straight flush, then this is being counted once in (42) but has been counted before. This means we must subtract another (43) from (42) so that altogether we subtract 2 times (43) from (42).

If the 3 good hands are spread over 3 players, then we take the product
of 3,696,
,
,
,
,
and (6). We obtain

132,657,294,436,982,297,405,248,185,360,000,000 | (44) |

semi-deals with 2 distinct players getting straight flushes and a third player getting quads. These have been counted twice in (42) and have been counted before so that we should subtract 2 times (44) from (42). Performing the adjustments on (42) yields

21,641,479,869,163,530,590,143,679,815,065,000,000 | (45) |

semi-deals with exactly 1 player getting a straight flush and exactly 1 player getting 4-of-a-kind when there are 4 suited cards and a pair in the board.

The number of semi-deals in (45) ahs no restriction on the rank of the
4-of-a-kind. Again we want to introduce the effect of the restriction.
We do it as before to get the following table.