Flushing Meadows: Part IV

Brian Alspach

Poker Digest Vol. 2, No. 7, March 26 - April 8, 1999

``Bib, I think it is time to get down to the origin of this discussion about probabilities of flushing boards. You brought it up while telling me about a no fold'em hold'em game you tangled with south of the border in Washington state. You felt as though people were making flushes almost every time a flushing board was present. So let's determine the probability of at least one player having a flush when a flushing board is present. There are a variety of possible scenarios so let's restrict ourselves to one of them and say that the others can be dealt with in a way similar to how we handle this.

The scenario we shall examine is when the board has exactly three cards of some suit and the player has no cards of that suit. We also assume the game has 10 players.''

``Yeah, professor, this is what really interests me. I am curious how unnatural that game might have been.'' ``The important goal is for you to understand the logic of determining the answer. Once you understand that the rest is easy as it involves only arithmetic.''

Bib nodded and was eager to get started so I began.

``We are interested in the probability of somebody actually having made a flush. We don't care which particular player. This means when we consider a particular set of 18 cards to be chopped into nine hands, we don't care how the hands are distributed to the players; we care only what the hands are. This is what I call a semi-deal. It is a particular instance of what mathematicians call unordered partitions of sets.

So the first step we take is to count the total number of choices for the 18 cards making up a semi-deal. There are 45 unseen cards and we are choosing 18 at a time giving us ${{45}\choose{18}} = 1,715,884,494,940$ sets of 18 cards from the 45 unseen cards. In order to determine the total number of semi-deals, we must calculate how many semi-deals a given set of 18 cards produces. There are ${{18}\choose{2}}$ ways to choose two cards for the first player, ${{16}\choose {2}}$ choices for the second player, and so on. This gives us the product

$\begin{displaymath}{{18}\choose{2}}{{16}\choose{2}}{{14}\choose{2}}\cdots{{4}\choose{2}}{{2}\choose {2}}.\end{displaymath}$

We must divide this product by 9! since the same set of hands may be dealt in 9! different orders. If we perform the calculation, we find that the number of semi-deals from a given set of 18 cards is

$\begin{displaymath}17\cdot 15\cdot 13\cdot 11\cdot 9\cdot 7\cdot 5\cdot 3 = 34,459,425.\end{displaymath}$

We use the notation 17!! to denote a product of positive integers decreasing by 2 for each term. Multiplying the preceding number by the number of sets of 18 cards above, we find there are 59,128,393,062,047,809,500 semi-deals from 45 unseen cards.

One interesting aspect of determining probabilities is that sometimes it is easier to determine the probability of something not happening. That is the case here. Let me explain why. As an example, suppose the 18 cards to be partitioned into nine hands of two cards each has four cards from the potential flush suit and 14 cards from the remaining suits. The four cards from the flushing suit may be distributed so that two hands get two each, or one hand gets two while two other hands get one each, or four hands get one each. The first two situations result in somebody getting a flush and the latter does not produce a flush. If we were to count how many ways we could have a flush, we would have two subcases to consider, whereas, if we were to count how many ways we do not have a flush, there is only one case. As more cards from the flushing suit are amongst the 18 cards, there are more subcases to consider when counting the number of ways to have a flush. Meanwhile, there is always only one way to avoid a flush, namely, the flushing suit cards must be spread amongst different hands so no one gets two. Thus, we are going to count the number of semi-deals not producing a flush. Does this make sense, Bib?''

``You've convinced me, professor.''

``Let me now talk in terms of a general case, Bib. I think you are ready for this. Let's consider a set of 18 cards from the 45 unseen cards in terms of how many flushing cards it contains. A set of 18 cards has k flushing cards and 18-k cards from the other suits, where k may be any value between 0 and 10, inclusive. The number of ways of choosing the 18-k cards from the 35 non-flushing cards is ${{35}\choose{18-k}}$ and the number of ways of choosing k flushing suit cards from 10 is ${{10}\choose{k}}$ . The product of these two numbers gives us the number of sets of 18 cards containing precisely k cards from the flushing suit.

Once we have a set of 18 cards with k cards from the flushing suit, the only way we can avoid a flush is for the k cards to be spread over k different hands. Counting the number of ways of doing this with 18 given cards is the most subtle part of this computation. Think of the k separate cards of the flushing suit as waiting for a second card to join them to form k hands of two cards apiece. Since the k flushing cards have ranks they are distinguishable from each other. This means there are 18 - k choices for the card to join a particular rank, then 18 - 1 - k choices for the card to join another particular rank and so on until k cards have been chosen to join the k cards of the flushing suit. There are now 18 - 2k cards left to partition into hands of two cards each. This can be done in (18 - 2k - 1)!! ways. Thus, there are $(18 - k)(18 - 1 - k)(18 - 2 - k) \cdots(18 - 2k + 1)(18 - 2k - 1)!!$ semi-deals not producing a flush from a given set of 18 cards containing precisely k flushing cards, where the 18 cards have been chosen from the 45 unseen cards.''

``Well, professor, I'm goin' have to think about what you just showed me.''

``The hardest part, Bib, is seeing why choosing cards to join the k flushing cards works the way it does. Try it with a smaller example and you should be able to see why it works as I am saying.

Observe that if the semi-deal has 10 flushing cards in it, someone must have a flush because the 10 flushing cards, along with the remaining eight, are dealt to only nine players so at least one person gets two cards from the flushing suit. This means we need only consider through nine flushing cards. Thus, for each k between 0 and 9, inclusive, the number of semi-deals not producing a flush is given by

$\begin{displaymath}{{35}\choose{18-k}}{{8}\choose{k}}(18-k)(18-1-k)\cdot(18-2k+1)(18-2k-1)!!.\end{displaymath}$

When k = 9, the preceding formula makes no sense because it ends with (-1)!!, but if we consider where that term arises, we see there is no problem. When k = 9 and nine cards have been paired with the nine flushing cards, all cards have been dealt so we need not multiply by anything -- meaning we can treat (-1)!! as 1 -- whereas, for k<9, there are undealt cards and the term (18-2k-1)!! comes into play as it is the number of ways of dealing 18-2k cards in hands of two cards each.

If we evaluate the preceding for k running from 0 through 9, we obtain 37,813,174,028,347,929,750 semi-deals not allowing a flush. Dividing this number by the total number of semi-deals above yields a probability of 0.6395 that no one has a flush. In other words, the probability is 0.3605 that someone has a flush.

``So, professor, you are telling me when I have no hearts in my hand and there are three hearts in the board, I should expect to see someone with a flush more than one-third of the time? Well, the game I was telling you about was way out of whack!''

``Don't be so sure, Bib. Our memories are very selective. Also, Bib, these numbers are based on all players staying in until the river card has come. The actual behavior of people in the game can distort the figures considerably. Next time we shall address the issue of what you see versus these probabilities we have been computing''

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