Suit and Rank Distributions for Boards

Brian Alspach

25 August 2000

Abstract:

We examine properties of boards with respect to ranks and suits.

It is well known that a set contains distinct elements, but there are times when it is convenient to have an element repeated. In this case we use the term multiset. The multiplicity of an element is the number of times it appears in the multiset.

The concept of multiset comes into play when discussing boards in either hold'em or Omaha. The rank multiset of a board is the multiset of all ranks in the board, whereas, the rank set of a board is the set of all ranks in the board. An example makes this clear. Suppose the board consists of the cards $2\clubsuit,2\diamondsuit, 6\heartsuit,6\clubsuit,J\clubsuit$ . Then the rank multiset of the board is $\{2,2,6,6,J\}$ and the rank set is $\{2,6,J\}$ .

We first examine boards with respect to ranks. There are six types of rank multisets for boards.

Type A.: Type A is a rank multiset of the form $\{x,y,z,u,w\}$ , that is, the ranks are distinct. The number of Type A rank multisets is given by ${{13}\choose{5}}=1,287$ . To obtain the number of Type A boards, multiply 1,287 by 4⁵ because there are 4 choices for each of the 5 ranks. Doing so produces 1,317,888 Type A boards.
Type B.: Type B is a rank multiset of the form $\{x,x,y,z,u\}$ , that is, the board has a single pair. The number of Type B rank multisets is given by $13{{12}\choose{3}}=2,860$ because there are 13 choices for the rank of the pair and the remaining 3 ranks are being chosen from 12 ranks. To obtain the number of Type B boards, multiply the preceding number by $6\cdot 64$ because there are 6 choices for the pair of the given rank, and 4 choices for each card of the remaining 3 ranks. This leads to 1,098,240 Type B boards.
Type C.: Type C is a rank multiset of the form $\{x,x,y,y,z\}$ , that is, the board has two-pair. The number of such rank multisets is given by $13{{12}\choose{2}}=858$ because there are 13 choices for the rank of the singleton and we are choosing the 2 ranks of the pairs from 12 ranks. The number of Type C boards is obtained by multiplying the preceding number by $36\cdot 4$ because there are 6 choices for each of the pairs of the given ranks, and 4 choices for the card of the singleton rank. This yields 123,552 boards.
Type D.: Type D is a rank multiset of the form $\{x,x,x,y,z\}$ , that is, the board is a 3-of-a-kind hand. The number of Type D rank multisets is $13{{12}\choose{2}}=858$ because of 13 choices for the rank of the 3-of-a-kind, and 2 choices from 12 for the ranks of the 2 singletons. To obtain the number of Type D boards, we multiply the preceding number by $16\cdot 4$ because there are 4 choices for the 3-of-a-kind of the given rank, and 4 choices for each of the cards of the other 2 ranks. This produces 54,912 Type D boards.
Type E.: Type E rank multisets have the form $\{x,x,x,x,y\}$ , that is, the board contains quads. There are 13 choices for the rank of the quads and 12 choices for the rank of the singleton. This yields 156 Type E rank multisets. There are 4 choices for the card of the singleton rank leading to $4\cdot 156=624$ Type E boards.
Type F.: Type F rank multisets have the form $\{x,x,x,y,y\}$ , that is, there is a full house on board. There are 156 Type F rank multisets with 13 choices for x and 12 choices for y. There are 4 choices for the trips of rank x and 6 choices for the pair of rank y. Hence, there are 3,744 boards of Type F.

The preceding information is contained in the following table. The column head ``prob'' is the probability of a board occurring of that particular type.

RANK DISTRIBUTION

Type Rank Multisets Boards Prob

A 1,287 1,317,888 .507

B 2,860 1,098,240 .423

C 858 123,552 .048

D 858 54,912 .021

E 156 624 .0002

F 156 3,744 .0014

Now we move to considering suit distributions. We shall label these boards using a vector notation.

(2,1,1,1).: A Type (2,1,1,1) suit multiset is a multiset in which 2 cards are in the same suit and the remaining 3 cards are distributed with 1 from each of the remaining 3 suits. There are 4 choices for the suit which appears twice so that there are 4 multisets of this type. There are ${{13}\choose{2}}=78$ choices for the cards of the suit with 2 cards and 13 choices for each of the remaining cards. Thus, there are $78\cdot 13^3= 171,366$ such choices. Multiplying by 4 yields 685,464 boards of Type (2,1,1,1).
(2,2,1,0).: A Type (2,2,1,0) suit multiset has 2 suits represented with 2 cards apiece and 1 suit with a singleton. There are 6 choices for the 2 doubleton suits and 2 choices for the singleton suit. Thus, there are 12 multisets of Type (2,2,1,0). There are 78 choices for the 2 cards of a doubleton suit, and 13 choices for the card of the singleton suit. This gives 79,092 choices. Multiplying by 12 yields 949,104 boards of Type (2,2,1,0).
(3,1,1,0).: A Type (3,1,1,0) suit multiset has 3 cards of 1 suit and a card each of 2 other suits. There are 4 choices for the suit with 3 cards and 3 choices for the remaining 2 suits chosen from 3. This gives us 12 multisets of Type (3,1,1,0). There are ${{13}\choose{3}}=286$ choices for the 3 cards from the given suit and 13 choices for each of the remaining 2 cards. That produces $286\cdot 13^2=48,334$ choices. Multiplying by 12 gives us 580,008 Type (3,1,1,0) boards.
(3,2,0,0).: A Type (3,2,0,0) suit multiset has 3 cards from 1 suit and 2 cards from another suit. There are 12 such suit multisets. There are 286 choices for the 3 cards of the one suit, and 78 choices for the 2 cards of the other suit. This gives us $78\cdot 286\cdot 12=267,696$ Type (3,2,0,0) boards.
(4,1,0,0).: A Type (4,1,0,0) suit multiset has 4 cards from 1 suit and another card from a different suit. There 12 multisets of this form. There are ${{13}\choose{4}}=715$ choices for the 4 cards from the same suit and 13 choices for the remaining card. Altogether this produces $12\cdot 13\cdot 715=111,540$ boards of Type (4,1,0,0).
(5,0,0,0).: Finally, a Type (5,0,0,0) suit multiset has all 5 cards from the same suit. There are 4 such multisets and ${{13}\choose{5}}=1,287$ choices for the 5 cards from the suit. This yields 5,148 Type (5,0,0,0) boards.

As before we place the preceding information in a table. The column headed `prob' is the probability of a board of that type occurring.

SUIT DISTRIBUTION

Type Suit Multisets Boards Prob

(2,1,1,1) 4 685,464 .264

(2,2,1,0) 12 949,104 .365

(3,1,1,0) 12 580,008 .223

(3,2,0,0) 12 267,696 .103

(4,1,0,0) 12 111,540 .043

(5,0,0,0) 4 5,148 .002

website by the Centre for Systems Science
last updated 25 August 2000

Type	Rank Multisets	Boards	Prob
A	1,287	1,317,888	.507
B	2,860	1,098,240	.423
C	858	123,552	.048
D	858	54,912	.021
E	156	624	.0002
F	156	3,744	.0014

Type	Suit Multisets	Boards	Prob
(2,1,1,1)	4	685,464	.264
(2,2,1,0)	12	949,104	.365
(3,1,1,0)	12	580,008	.223
(3,2,0,0)	12	267,696	.103
(4,1,0,0)	12	111,540	.043
(5,0,0,0)	4	5,148	.002