Exclamation Points

Brian Alspach

Poker Digest, Vol. 5, No. 5, February 22 - March 7, 2002

Over the past three years, I have received sporadic enquiries regarding the meaning of the expression ``$n!!$'' So once and for all, I wish to make clear what this notation means, and why it occurs naturally in hold'em counting problems.

Let me ask a seemingly unrelated question and go from there. Suppose we have a league consisting of $2m$ teams, and $m$ venues for competition to take place. How many possible pairings are there for $m$ simultaneous competitions to take place?

If there are two teams, it is clear there is only one possible pairing. If there are four teams, a moments reflection will lead you to conclude there are three possible ways the four teams may be paired.

Once we move to six teams, we begin to realize some kind of systematic approach is required. There are several ways to arrive at a solution. Here is one such way.

Label the six teams 1, 2, 3, 4, 5 and 6. Team 1 must be paired with some team, and there are five choices for the pair involving team 1.

Once the first pair is removed from consideration, there are four teams left to pair up. We already have seen there are three ways to form pairs from four teams. Thus, there are $5\cdot 3 = 15$ ways to form pairs from six teams.

With eight teams, there are seven ways to pair team 1, and 15 ways to pair the six teams left over. Thus, the answer for eight teams is $7\cdot 5\cdot

A clear pattern has emerged from these small examples. For two teams the answer is 1, for four teams the answer is 3, for six teams the answer is $5\cdot 3$, and for eight teams the answer is $7\cdot 5\cdot

Thus, it seems a solid conjecture is that for $2m$ teams, we take the product of successively decreasing odd numbers starting with $2m-1$ and concluding with 3. (More accurately, we conclude with 1 in order to accommodate the answer for two teams, but the 1 serves no other purpose.)

The motivational discussion above contains the germ of a proof. We know the correct answer for small values, and we assume the conjecture is true for $2m-2$ teams. When considering $2m$ teams, team number 1 may be paired with any of $2m-1$ other teams. This leaves $2m-2$ teams unpaired, but we have assumed the answer for $2m-2$ to be a falling product of successive odd numbers starting with $2m-3$. Multiplying by $2m-1$ gives the answer we conjectured for $2m$ teams and we conclude our conjecture is, in fact, a theorem.

The preceding proof technique is called induction. Some mathematicians (not me) have an aversion to induction proofs. If a direct proof is available, they will use it even when it is more cumbersome than the induction proof. In this case, there is a proof which looks nothing like induction. I like this proof too.

Recall that $C(2m,2)$ denotes the number of ways of choosing two objects from $2m$ objects. Also, recall that $C(2m,2) = 2m(2m-1)/2$.

Start with the $2m$ teams. Arbitrarily choose two of the teams to be paired. This can be done in $2m(2m-1)/2$ ways and leaves $2m-2$ teams.

From the remaining $2m-2$ teams, again arbitrarily choose two teams to be paired. This can be done in $(2m-2)(2m-3)/2$ ways.

Continue in the same way. After $m$ steps, all teams will be paired. In fact, the last step is trivial because, at that point, only two teams remain unpaired leaving us no choice. We take the product of all the numbers and notice the product has a very nice form. The numerator is $(2m)!$ and the denominator is 2 raised to the $m$-th power.

However, the number just given is not the correct answer because we are forming the pairs in a particular order with $m$ steps. This means we could get the same pairing of teams in any of $m!$ orders. So we also have to divide the preceding number by $m!$.

Upon dividing by $m!$, some wonderful cancellation takes place. We have 2 to the $m$-th power in the demoninator and we have $m$ successive even numbers -- $2m, 2m-2, \ldots, 4, 2$ -- in the numerator. Divide one 2 into each of these even numbers. This gets rid of the 2s in the denominator, and changes the product of the successive even numbers into $m!$. The two $m!$ terms now cancel and all that is left is the falling product of the successive odd numbers starting with $2m-1$. Isn't that a nice little proof?

This number $(2m-1)(2m-3)\cdots 5\cdot 3\cdot 1$ is denoted by $(2m-1)!!$ How does it arise naturally in hold'em counting problems? As an aside, let me point out that when $n$ is even, then $n!!$ denotes the product of successively decreasing even numbers starting with $n$ and finishing with 2.

Each player in hold'em is dealt two cards. There are a variety of questions people ask about the probabilities of certain outcomes with, say, 10 players. For some questions, it is important which players have certain hands; that is, the order in which the hands are dealt is important. I call these ``deals''.

There are other questions for which the order of the hands is unimportant. What is important is whether or not a hand is somewhere. I call these ``semideals.'' For example, we might ask how many possible player semideals are there in 10-handed hold'em.

We now have the notation to express this number very compactly even though the number itself is huge. First of all, there are $C(52,20)$ ways of choosing the 20 cards to be dealt to the players.

Once the 20 cards are chosen, the 10 hands are simply a pairing of the 20 cards; just the same as pairing 20 teams to play at 10 venues. We have seen that the number of possible pairings is 19!!

So the total number of player semideals is $C(52,20)19!!$ As I said, that is a nice compact way of writing it. Just for fun, let's see what the actual number is. First, $C(52,20) = 125,994,627,894,135$. Second, $19!! = 654,729,075$. Their product is 82,492,346,176,096,206,475,125 and this is the total number of player semideals in a 10-handed hold'em game.

If you want to know the total number of semideals, you must multiply by $C(32,5)$ because that is the number of boards possible from the remaining 32 cards. Of course, in actual play, order does matter which greatly increases the number of possibilities again.

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