Flushing Meadows: Part I

Brian Alspach

Poker Digest Vol. 2, No. 4, February 12 - 25, 1999

I walked into the club and headed for a seat at one of the service tables. The tea I ordered arrived about the same time as Bib Ladder who took a break from a 20-40 Omaha game to say a few words.

After exchanging greetings he started, ``I was down in the state of Washington playing at one of the casinos near Seattle. The only game going was $4-$8 hold'em so I joined in. Man, it was my first experience with one of those no fold'em hold'em games I've read about. Everyone, and I mean everyone, was calling and raising with any two cards. I tried playing only quality hands and got eaten alive. It seemed as if a straight or flush made it every hand. I don't think I've ever seen so many flushes.

So it got me to thinking about the chances of a board occurring which allows a flush. To me I see all that green felt as a meadow on which we romp and play this game of poker. So I think of boards which allow flushes as flushing meadows. Here are my calculations. Can you take a look at them?''

``Very poetic terminology, Bib. Sure, I'll look at what you did.''

``There are two live ones in that Omaha game who seem prepared to keep me in groceries for the next month or so. I think I'll return to the game. See ya later.''

Below are Bib's calculations.

We are going to determine the probability of a flushing board given no information on what any of the players is holding. This means we must determine the total number of flushing boards possible from 52 cards. The total number of possible boards is

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

Let's express the type of suit distribution of the board in the form (a,b,c,d). The possible types are (0,0,0,5), (0,0,1,4), (0,0,2,3), (0,1,1,3), (0,1,2,2), and (1,1,1,2). For example, the type (0,0,0,5) indicates the board is made up of five cards in the same suit. It is easy to see the first four types describe flushing boards and the last two types describe boards which are not flushing.

The type (0,0,0,5).: There are four choices for the suit in the board and ${{13}\choose{5}}=1,287$ choices for the five cards of that suit. This yields $4\cdot 1,287 = 5,148$ boards of type (0,0,0,5).
The type (0,0,1,4).: There are four choices for the suit with four cards in the board, three choices for the remaining suit, ${{13}\choose{4}}=715$ choices for the four cards of the one suit, and 13 choices for the card in the other suit. This yields $4\cdot 3\cdot 715\cdot 13 = 111,540$ boards whose type is (0,0,1,4).
The type (0,0,2,3).: There are four choices for the suit with three cards, three choices for the suit with two cards, ${{13}\choose{3}}=286$ choices for the three cards of the one suit, and ${{13}\choose{2}}=78$ choices for the two cards of the other suit. This produces $4\cdot 3\cdot 286\cdot 78 = 267,696$ boards of type (0,0,2,3).
The type (0,1,1,3).: There are four choices for the suit with three cards, three choices for the two remaining suits, ${{13}\choose{3}}=286$ choices for the three cards of the one suit, and 13 choices for each of the cards of the other two suits. This gives us $4\cdot 3\cdot 286\cdot 13^2 = 580,008$ boards of type (0,1,1,3).
The type (0,1,2,2).: There are four choices for the suit with one card, ${{3} \choose{2}}=3$ choices for the suits with two cards apiece, 13 choices for the card from the first suit, and ${{13}\choose{2}}=78$ choices for the two cards of each suit. This produces $4\cdot 3\cdot 13\cdot 78^2= 949,104$ boards of type (0,1,2,2).
The type (1,1,1,2).: There are four choices for the suit with two cards of that suit, ${{13}\choose{2}}=78$ choices for the card of that suit, and 13 choices for each of the other cards. This yields $4\cdot 87\cdot 13^3 = 685,464$ boards of type (1,1,1,2).

Adding the numbers of boards of the six types yields 2,598,960 boards as it should.

Boards of types (0,0,0,5), (0,0,1,4), (0,0,2,3), and (0,1,1,3) are flushing boards. The number of boards of these types is the sum

5,148 + 111,540 + 267,696 + 580,008 = 964,392.

This means the probability of a flushing board occurring given no information about players' hands is

$\begin{displaymath}\frac{964,392}{2,598,960} = 0.371.\end{displaymath}$

So a little more than one-third of the time a flush will be possible.

After I had checked Bib's calculations, I had a chance to talk with him again.

``Bib, your calculations are correct. Did you think about seeing what happens to these numbers from the standpoint of an individual player? There are two possibilities for the player: either she has two suited cards or two unsuited cards.''

``No, I didn't think about that. It would be interesting to see if there are some differences.''

``Let's do it!''

In the next article we explore flushing board probabilities from an individual player's viewpoint.

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