**Poker Digest Vol. 2, No. 17, August 13 - 26, 1999**

In the first three parts of this series, we have explored the use
of the principle of inclusion-exclusion. Now we get to the reason for
introducing the topic in the first place; namely, we want to determine
the probability no one is a dealt a pair in hold'em. This certainly is
the most sophisticated of the twenty-seven articles to appear thus far
in *Poker Digest*, but each step
can be followed if you keep in mind the basic context for
inclusion-exclusion and the reason each step makes sense. First, let's
establish the scenario in which inclusion-exclusion becomes the tool of
choice.

First, we are going to derive a general formula so let's assume there
are *n* players. Normally, *n* will be 9, 10 or 11, but if the game
is shorthanded, *n* may be smaller.

Let *P*_{i} denote the property player *i* is dealt a pocket pair. We
are interested in counting the number of ways no one has a pair, that
is, none of the properties
holds. This is exactly
the classical setting for applying inclusion-exclusion. We also
require counting the number of ways in which a prescribed
subset of properties can hold to be easy. We shall see this is the case.

For this problem the players are labelled 1,2,...,*n* because we are
defining our properties in terms of particular players. Recall that
semi-deals refer to dealing cards to players without regard to which
players get which hands. In the present case, we are interested in
which players get which cards and, thus, we are
interested in the number of deals rather than the number of semi-deals.
In most cases counting problems involving labelled objects is easier
than counting problems with
unlabelled objects and that is the case here too. In other words, we
want to know how many ways *n* labelled players may be dealt two cards each.
The first player can receive two cards in *C*(52,2) ways, the second player
in *C*(50,2) ways and so on until the *n*-th player can receive two
cards in
*C*(50-2*n*+2,2) ways. Multiplying these numbers yields

ways

Now we let *M*(*S*) denote the number of deals in which a
given subset *S* of players receive a pair, that is, for each the property *P*_{i} holds. Under this notation,
denotes
the total number of ways of dealing two cards to each of *n* people.

The principle of inclusion-exclusion then tells us that the number of
deals in which none of the properties are satisfied, or the
answer to the question, is

where the sum is over all subsets

because there are

The preceding equation expresses the number of deals to *n* players
in which no one receives a pair. This reduces the problem to that of
calculating *M*(*r*). In other words,
the problem is reduced to determining how many deals give a fixed set of
*r* players pairs (where what the remaining *n*-*r* players get is not
restricted). There is a complication arising here because some of the *r*players may get pairs of the same rank. How do we handle this?

Remember we are working with an arbitrary value of *n*. In fact, what
are the restrictions on *n*? First, *n* must be at least 1 as it does
not make much sense to deal cards to fewer than 1. Second, *n* is at
most 26 because there are only 52 cards in the deck. Thus,
.

Suppose there are *t* pairs of players who receive pairs of the same rank.
What kinds of restrictions are there on *t*? It is easy to see that
because the *t* pairs of players comprise 2*t* players and
there are only *r* players available. This implies
.
(The notation
means the integer part of
*r*/2. For example,
.) There
are only 13 ranks from which to
choose pairs so that if ,
there must be at least *r*-13 ranks
which occur as pairs in two different hands. Since *t* is not negative,
we must have
.
This gives us the
restriction that *t* satisfies
,

Hence, for each *t* satisfying
,
there may be *t* pairs of people who receive pairs of the same rank
using all four cards of the same rank. From *r* people
there are *C*(*r*,2*t*) ways to choose the 2*t* people who will get pairs
of the same rank as somebody else. From amongst these 2*t* people we must
choose the pairs of people who will receive pairs of of the same rank.
We can choose the first pair of people in *C*(2*t*,2) ways, the next
pair of people in *C*(2*t*-2,2) ways and so on until only two people
are left who form the final pair. Multiplying all these numbers and
dividing by *t*! because it doesn't matter in which order they are
chosen yields

ways of choosing the pairs of people to get pairs of the same rank.

(Stay tuned for Part V)

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