Sousem Hands

Brian Alspach

21 October 1998

Abstract:

We count the number of 5-card poker hands with sousem rules, namely, 4-card straights, 4-card flushes and 4-card straight flushes are ranked hands. We assume a standard deck of 52 cards.

A poker variant sometimes used is to allow 4-card straights and 4-card flushes. This implies 4-card straight flushes also should be allowed. Properly ranking the resulting hands is more complicated because some hands will have two characteristics such as containing a pair and a 4-card straight. In order to see clearly what is going on, we count all possible combinations. Ranking is then a simple matter.

The total number of possible hands is

\begin{displaymath}{{52}\choose{5}} = \frac{52!}{5!47!}= 2,598,960.\end{displaymath}

We first break the possible hands into all possible types. The possibilities are

5-card straight flush.
These are easy to count because once the smallest card is known the entire hand is known. There are 40 cards which can be the smallest card so there are 40 5-card straight flushes.
4-of-a-kind.
There are 13 possible ranks, only one way to choose the 4 cards, and any of 48 cards to fill out the hand. This yields $13\cdot 48 = 624$ 4-of-a-kind hands.
full house.
There are 13 choices for the rank of the trips, 12 choices for the rank of the pair, 4 choices for the trips of the given rank, and 6 choices for the pair of the given rank. This yields $13\cdot 12\cdot 4\cdot 6 = 3,744$ full houses.
5-card flush containing a 4-card straight flush.
There are 4 choices for suit. Within a given suit, a 4-card straight flush can begin with any of 11 cards (ace through jack). The 4-card straight flush may be completed to a 5-card flush which is not a 5-card straight flush with any of 8 cards if the smallest card is either an ace or a jack, or with any of 7 cards in all other cases. This gives $2\cdot 8+9\cdot 7=79$ flushes in the suit containing a 4-card straight flush. Multiplying by 4 yields 316 5-card flushes containing 4-card straight flushes.
5-card flush not containing a 4-card straight flush.
In a given suit there are ${{13}\choose{5}} = 1,287$ flushes . Of these, 10 are 5-card straight flushes and 79 contain 4-card straight flushes but are not 5-card straight flushes. Subtracting 89 from 1,287 gives 1,198 flushes not containing a 4-card straight flush. Multiplying by 4 yields 4,792 5-card flushes not containing a 4-card straight flush.
5-card straight containing a 4-card straight flush.
There are precisely 8 4-card straight flushes beginning with either an ace or a jack. Each of these can be completed to a 5-card straight by adding any of 3 cards since we cannot choose the appropriate card in the same suit. Any of the remaining 36 4-card straight flushes can have any of 6 cards added to yield a 5-card straight. This produces $3\cdot 8+6\cdot 36 = 240$ 5-card straights containing 4-card straight flushes.
5-card straight not containing a 4-card straight flush.
A straight has the form x,x+1,x+2,x+3,x+4. The rank of x can be any of 10 values from ace, deuce through 10. There are then 4 choices for each of the cards giving $10\cdot 4^5=10,240$ of this form. However, we must remove the 40 straight flushes plus the 240 containing 4-card straight flushes. This leaves 9,960 5-card straights not containing a 4-card straight flush.
3-of-a-kind.
There are 13 choices for the rank of the 3-of-a-kind. The remaining 2 ranks are chosen in ${{12}\choose{2}}=66$ ways. There are 4 ways to choose the 3-of-a-kind of the given rank and 4 choices for each of the cards of the other 2 ranks. Altogether we obtain $13\cdot 66\cdot 4^3 = 54,912$ 3-of-a-kinds.
2 pairs.
There are ${{13}\choose{2}}=78$ choices for the ranks of the 2-pairs, there are 6 choices for each of the pairs of the given ranks, and there are 44 choices for the remaining card. This produces $78\cdot 36\cdot 44=123,552$ hands with 2-pairs.
4-card straight flush with a pair.
There are 44 4-card straight flushes as we saw earlier. For each card in the straight flush, there are 3 cards in other suits which can pair it. Thus, there are $44\cdot 12= 528$ 4-card straight flushes with a pair.
4-card straight flushes without a pair.
Again there are 44 4-card straight flushes. We now want to choose the last card so that it forms neither a pair, nor a 5-card flush, nor a 5-card straight as these have been counted already. If the 4-card straight flush begins with an ace or a jack, then there are 8 ranks which do not make a 5-card straight. We can choose any of 3 cards of these ranks so as not to form a 5-card flush. This gives us 24 cards to choose together with 8 choices for the ace or jack. If the 4-card straight flush begins with any of the other ranks, there are only 7 ranks one can choose to avoid forming a 5-card straight. Thus, there are 21 cards to choose in this case along with the 36 choices for the first card in the 4-card straight flush. This gives us $8\cdot 24 + 36\cdot 21 = 948$ hands of this kind.
4-card straight and 4-card flush.
We want to count hands which are both 4-card straights and 4-card flushes but NOT 4-card straight flushes. Such a hand must have the form x,x+1,x+2,x+3,y, where $y\neq x+4$, $y\neq x-1$, and y is of the same suit as 3 of the cards from x,x+1,x+2,x+3. Break this into two parts according to the rank of x. If x is ace or J, this gives 2 choices. There are 4 choices for the suit of the flush, 3 choices for the other suit, 4 choices for which subset of 3 cards will be in the flush suit, and 8 choices for the rank of y in the flush suit. This gives $2\cdot 4\cdot 3\cdot 4\cdot 8= 768$. If x is not an ace or a J, there are 9 choices for its rank. The rest of the numbers are the same except y can now be only any 7 ranks. This yields $9\cdot 4\cdot 3\cdot 4\cdot 7= 3,024$. Adding them together gives 3,792 hands of this type.
4-card flush with a pair.
There are ${{13}\choose{4}}=715$ 4-card flushes in a given suit of which 11 are 4-card straight flushes. This leaves 704 4-card flushes which have not been counted earlier. For each of the 4-card flushes, there are 12 cards which can pair one of the 4 cards. Since there are 4 suits to choose from, this gives us $4\cdot 704\cdot 12=33,792$ 4-card flushes containing a pair.
4-card flush without a pair.
As in the previous case, there are 704 flushes in a given suit which have not been counted before. However, there is an additional complication here. A 4-card flush which has only a single gap, such as x,x+2,x+3,x+4, cannot have any x+1 added to it as that results in a 5-card straight. So we have to break the 704 4-card flushes into those having a single gap and those not having a single gap. The forms of the 4-card flushes having a single gap are x,x+2,x+3,x+4, or x,x+1,x+3,x+4, or x,x+1,x+2,x+4. Since the rank x can be anything from ace through 10, there are 30 single gap 4-card flushes in a given suit. This leaves 674 4-card flushes with more than a single gap. A 4-card flush with a single gap may have any of 24 cards added to the hand and not get a hand counted earlier. A 4-card flush with more than a single gap may have any of 27 cards added to the hand. This gives $30\cdot 24+ 674\cdot 27=18,918$ 4-card flushes being counted in a given suit. Multiplying by 4 yields 75,672 4-card flushes without a pair. However, some of these also have a 4-card straight so we must remove them. Subtracting 3,792 leaves 71,880.
4-card straight with a pair.
A 4-card straight has the form x,x+1,x+2,x+3. There are 11 choices for the rank x, there are 4 choices of the rank to be paired, and there are 6 choices for the pair of of that rank. There are 4 choices for each of the remaining 3 cards except not all 3 can be chosen in the same suit as either of the paired cards as this would give a 4-card straight flush. Thus, there are 43- 2 = 62 choices for the remaining 3 cards. This gives us $11\cdot 4\cdot 6\cdot 62=16,368$ 4-card straights which also have a pair.
4-card straight without a pair.
A 4-card straight has the form x,x+1, x+2,x+3,y, where $y\not\in\{x-1,x+4\}$. If x is an A or J, there are 8 choices for the rank of y giving 16 sets of this type. If x is from $2,3,\cdots,10$, then there are 7 choices for y. This gives 63 sets of the latter type. To each card we can give any of 4 values yielding $79 \cdot 4^5=80,896$ choices. However, if all are chosen from the same suit, we have a 5-card flush. This removes 4 choices. If 4 are chosen from the same suit, we obtain a 4-card flush or 4-card straight flush so we must remove them too. There are 4 choices for the flush suit, 3 choices for the other suit, and 5 choices for which 4 cards are in the same suit. This removes another 60 choices. So altogether we have 79(45-64)= 75,840.
single pair.
A hand containing a single pair has 4 ranks represented in the hand. Altogether there are ${{13}\choose{4}}=715$ sets of 4 ranks chosen from 13 ranks. However, we do not wish to count 4-card straights so we eliminate the 11 rank sets of the form x,x+1,x+2,x+3 leaving 704 sets of 4 ranks not containing a 4-card straight. Given a fixed set x,y,z,w of 4 ranks, there are 4 choices for the rank of the pair and 6 choices for the pair of the chosen rank. Each of the remaining 3 cards can be any of 4 cards except we cannot choose all 3 of the same suit as either card in the pair as this would give us a 4-card flush. Thus, the 3 cards may be chosen in 43- 2 = 62 ways. We then have $704 \cdot 4\cdot 6\cdot 62=1,047,552$ hands with 1 pair.
no pair.
There are ${{13}\choose{5}} = 1,287$ ways of choosing 5 ranks from 13. There are 10 sets of ranks corresponding to 5-card straights and these must be removed. In addition, we must remove those sets of ranks containing 4 consecutive values as these would represent 4-card straights. A set containing x,x+1,x+2,x+3 allows any of 8 other values if x is either an ace or a jack. If x is any of the other ranks, 7 other values are allowed. This gives us $2\cdot 8+9\cdot 7=79$ more sets which must be removed. Altogether we have 1,287 - 10 - 79 = 1,198 sets of 5 ranks which do not contain any kind of straight.

Consider a given set containing ranks x,y,z,u,w. We can choose any of 4 cards for each of the ranks, but we do want to avoid both 5-card flushes and 4-card flushes. There are 4 5-card flushes of these given ranks. To determine the number of 4-card flushes, first observe there are 5 choices for the subset of 4 ranks in the same suit, there are 4 choices for the suit, and there are 3 choices for the suit of the remaining card. This gives 60 4-card flushes with the given ranks. Thus, there are 45- 64 = 960 hands of the given ranks which are not 4-card flushes. Hence, the number of no pair hands is $1,198\cdot 960=1,150,080$.

This completes the counting of the hands of each type. The sum of the numbers is 2,598,960 as it should be. Now we rank the hands under 3 different scenarios. First we suppose that the types of hands allowed are the usual poker hands plus a 4-card straight and a 4-card flush. The ranking of the hands is shown in the next table.

Type of Hand Number of Hands
5-card straight flush 40
4-of-a-kind 624
full house 3,744
5-card flush 5,108
5-card straight 10,200
3-of-a-kind 54,912
4-card straight 97,476
4-card flush 105,672
2 pairs 123,552
1 pair 1,047,552
high card 1,150,080

Note that 4-card straight flushes have been counted as 4-card straights, and hands which are both 4-card flushes and 4-card straights but not 4-card straight flushes also are counted as 4-card straights. In spite of this, a 4-card straight still is a better hand than a 4-card flush. Another possible surprise for some people is that both are better than 2 pairs.

The next scenario is allowing a 4-card straight flush to be a type of hand. It is surprising how strong the latter hand is.

Type of Hand Number of Hands
5-card straight flush 40
4-of-a-kind 624
4-card straight flush 2,032
full house 3,744
5-card flush 4,792
5-card straight 9,960
3-of-a-kind 54,912
4-card straight 96,000
4-card flush 105,672
2 pairs 123,552
1 pair 1,047,552
high card 1,150,080

For the last scenario we also shall allow the hand which contains both a 4-card straight and a 4-card flush, but not a 4-card straight flush. In the table below we simply call it a 4-card straight & 4-card flush.

Type of Hand Number of Hands
5-card straight flush 40
4-of-a-kind 624
4-card straight flush 2,032
full house 3,744
4-card straight & 4-card flush 3,792
5-card flush 4,792
5-card straight 9,960
3-of-a-kind 54,912
4-card straight 92,208
4-card flush 105,672
2 pairs 123,552
1 pair 1,047,552
high card 1,150,080

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