Sousem Still Wows 'Em: Part II

This is a continuation of a session with Bib Ladder about determining the appropriate ranking for poker hands under sousem rules. In my previous article, the types of hands we are going to count are listed.

``Alright, Bib, let's get down to counting the hands I listed. First, the total number of five-card hands is C(52,5) = 2,598,960. There are 40 five-card straight flushes because once the smallest card is known the entire hand is known and there are 40 cards which can be the smallest card.

For four-of-a-kind there are 13 possible ranks, only one way to choose the four cards, and any of 48 cards to fill out the hand. This yields four-of-a-kind hands.

Let's count full houses next. There are 13 choices for the rank of the trips, 12 choices for the rank of the pair, four choices for the trips of the given rank, and six choices for the pair of the given rank. This yields full houses.

Next we treat five-card flushes containing a four-card straight flush. Within a given suit, a four-card straight flush can begin with any of 11 cards (ace through jack). The four-card straight flush may be completed to a five-card flush which is not a five-card straight flush with any of eight cards if the smallest card is either an ace or a jack, or with any of seven cards in all other cases. This gives flushes in the suit containing a four-card straight flush. Multiplying by four yields 316 five-card flushes containing four-card straight flushes.

We now count five-card flushes not containing a four-card straight flush. There are C(13,5) = 1,287 flushes in a given suit. Of these, 10 are five-card straight flushes and 79 contain four-card straight flushes but are not five-card straight flushes. Subtracting 89 from 1,287 gives 1,198 flushes not containing a four-card straight flush. Multiplying by four yields 4,792 five-card flushes not containing a four-card straight flush. How are you doing so far, Bib?''

``No problem, professor. I've come a long way with this counting stuff.''

``Great! Let's look at five-card straights containing a four-card straight flush. There are eight four-card straight flushes beginning with either an ace or a jack. Each of these can be completed to a five-card straight by adding any of three cards since we cannot choose the appropriate card in the same suit as that would give us a five-card straigh flush. Any of the remaining 36 four-card straight flushes can have any of six cards added to yield a five-card straight. This produces five-card straights containing four-card straight flushes.

Next comes five-card straights not containing a four-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 four choices for each of the cards giving of this form. However, we must remove the 40 straight flushes plus the 240 containing four-card straight flushes. This leaves 9,960 five-card straights not containing a four-card straight flush.

Now we look at three-of-a-kind hands. There are 13 choices for the rank of the three-of-a-kind. The remaining two ranks are chosen in C(12,2) = 66 ways. There are four ways to choose the three-of-a-kind of the given rank and four choices for each of the cards of the other two ranks. Altogether we obtain three-of-a-kind hands.

Next in line are two pair. There are C(13,2) = 78 choices for the ranks of the two pair, there are six choices for each of the pairs of the given ranks, and there are 44 choices for the remaining card. This produces hands with two pair.''

``Professor, are you avoiding sousem type hands?''

``Not really, Bib. I am saving the best to last. Let's look at four-card straight flushes with a pair. There are 44 four-card straight flushes as we saw earlier. For each card in the straight flush, there are three cards in other suits which can pair it. Thus, there are four-card straight flushes with a pair.

For four-card straight flushes without a pair, there are 44 four-card straight flushes. We now want to choose the last card so that it forms neither a pair, nor a five-card flush, nor a five-card straight as these have been counted already. If the four-card straight flush begins with an ace or a jack, then there are eight ranks which do not make a five-card straight. We can choose any of three cards of these ranks so as not to form a five-card flush. This gives us 24 cards to choose. If the four-card straight flush begins with any of the other ranks, there are only seven ranks one can choose to avoid forming a five-card straight. Thus, there are 21 cards to choose in this case. This gives us four-card straight flushes without a pair.

Perhaps the most interesting hand of all is the hand which contains both a four-card straight and a four-card flush, but no four-card straight flush. Let's call this type of hand a double sousem. The form of a double sousem must be x,x+1,x+2,x+3,y, where y is neither an x+1 nor an x-1 and is in the same suit as three of the other cards. If x is an A or J, then the rank of y allows eight choices; in all other cases there are seven choices for the rank of y. There are four choices for the suit of y, four choices for which of x,x+1,x+2,x+3 is not in this suit and three choices for that suit. This yields double sousem hands.

Next on our list is four-card flushes with a pair. There are C(13,4) = 715 four-card flushes in a given suit of which 11 are four-card straight flushes. This leaves 704 four-card flushes which have not been counted earlier. For each of the four-card flushes, there are 12 cards which can pair one of the four cards. Since there are four suits to choose from, this gives us four-card flushes containing a pair.

For four-card flushes 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 four-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 five-card straight. So we have to break the 704 four-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; x,x+1,x+3,x+4 and x,x+1,x+2,x+4. Since the rank x can be anything from ace through 10, there are 30 single gap four-card flushes in a given suit. This leaves 674 four-card flushes with more than a single gap. A four-card flush with a single gap may have any of 24 cards added to the hand and not get a hand counted earlier. A four-card flush with more than a single gap may have any of 27 cards added to the hand. This gives four-card flushes being counted in a given suit. Multiplying by four yields 75,672 four-card flushes without a pair. However, some of these also have a four-card straight so we must remove them. Subtracting 3,792 leaves 71,880.

A four-card straight with a pair has the form x,x+1,x+2,x+3. There are 11 choices for the rank x, there are four choices of the rank to be paired, and there are six choices for the pair of of that rank. There are four choices for each of the remaining three cards except not all three can be chosen in the same suit as either of the paired cards as this would give a four-card straight flush. Thus, there are 4³ - 2 = 62 choices for the remaining 3 cards. This gives us four-card straights which also have a pair.

A four-card straight without a pair has the form x,x+1, x+2,x+3,y, where y is neither x-1 nor x+4. If x is an A or J, there are eight choices for the rank of y giving 16 sets of this type. If x is from , then there are seven choices for y. This gives 63 sets of the latter type. To each card we can give any of four values yielding choices. However, if all are chosen from the same suit, we have a five-card flush. This removes four choices. If four are chosen from the same suit, we obtain a four-card flush or four-card straight flush so we must remove them too. There are four choices for the flush suit, three choices for the other suit, and five choices for which four cards are in the same suit. This removes another 60 choices. So altogether we have 79(4⁵ - 64) = 75,840.

A hand containing a single pair has four ranks represented in the hand. Altogether there are C(13,4) = 715 sets of four ranks chosen from 13 ranks. However, we do not wish to count four-card straights so we eliminate the 11 rank sets of the form x,x+1,x+2,x+3 leaving 704 sets of four ranks not containing a four-card straight. Given a fixed set x,y,z,w of four ranks, there are four choices for the rank of the pair and six choices for the pair of the chosen rank. Each of the remaining three cards can be any of four cards except we cannot choose all three of the same suit as either card in the pair as this would give us a four-card flush. Thus, the three cards may be chosen in 4³ - 2 = 62 ways. We then have hands with one pair.

Finally, we consider hands with no pairs or better. There are C(13,5) = 1,287 ways of choosing five ranks from 13. There are 10 sets of ranks corresponding to five-card straights and these must be removed. In addition, we must remove those sets of ranks containing four consecutive values as these would represent four-card straights. A set containing x,x+1,x+2,x+3 allows any of eight other values if x is either an ace or a jack. If x is any of the other ranks, seven other values are allowed. This gives us more sets which must be removed. Altogether we have 1,287 - 10 - 79 = 1,198 sets of five 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 four cards for each of the ranks, but we do want to avoid both five-card flushes and four-card flushes. There are four five-card flushes of these given ranks. To determine the number of four-card flushes, first observe there are five choices for the subset of four ranks in the same suit, there are four choices for the suit, and there are three choices for the suit of the remaining card. This gives 60 four-card flushes with the given ranks. Thus, there are 4⁵ - 64 = 960 hands of the given ranks which are not four-card flushes. Hence, the number of no pair hands is ''.

Bib said, ``Let me add all those numbers together to see that we get the total number of five-card hands.''

That he did and found that the sum is correct. In the next and last article about sousem, we present the correct hand rankings for three sousem scenarios and discuss aspects of these rankings.