**Poker Digest Vol. 2, No. 18, August 27 - September 29, 1999**

In the last issue,
recall that we are determining the probability no one is a dealt a pair
in hold'em when there are *n* hands dealt. We are using the *Principle
of Inclusion-Exclusion* to count the number of ways no one is dealt a
pair from which we shall derive the probability. We let *P*_{i} denote
the property player *i* is dealt a pocket pair.
We saw that

is the number of deals for which no one receives a pair, where

As a first step towards determining *M*(*r*), we saw that
for each *t* from
through the integer part of
*r*/2,
there are
ways of choosing *t* pairs of
people (subsequently referred to as couples) to receive pairs of the same
rank. That is where we stopped in the last issue.

Once the couples have been formed, there are *C*(13,*t*) ways of picking
which ranks of cards will be assigned to the *t* couples. These *t*ranks can be assigned in *t*! ways to the *t* couples. Once the ranks
have been assigned to the couples, there are six ways to split the four cards
of the same rank amongst a given couple. Hence, there are 6^{t} ways
for the pairs to be distributed amongst the *t* couples. Therefore, there are

deals in which precisely 2

The remaining *r*-2*t* people must get pairs of distinct ranks. There are
*C*(13-*t*,*r*-2*t*) choices for the ranks. Each choice of ranks can be assigned
in (*r*-2*t*)! ways to the players and each assignment can have any of six
pairs of a given rank yielding 6^{r-2t} deals for each assignment of
ranks. Finally, the remaining *n*-*r* people
can be dealt any two cards from the remaining 52-2*r* cards. This
can be done in
ways. Altogether, this yields

as the number of deals in which some subset of precisely 2

We substitute this for *M*(*r*) in the above answer for the question and
simplify. This yields

as the answer to the basic question of the number of ways in which two cards can be dealt to each of two people so that no one receives a pair. In order to compute the probability of such a deal occurring, simply divide the preceding number by the number of 2-card hands that can be dealt to

In fact, with a little extra effort one can use the numbers obtained
in the main formula to do much more. Let's illustrate with *n*=4.
Using the formula one obtains
*M*(0) = 1,896,396,138,000 which
is the total number of ways of dealing two cards to each of four players.
Continuing one obtains
*M*(1) = 446,210,856,000,
*M*(2) = 39,885,786,720,
*M*(3) = 1,604,266,560, and
*M*(4) = 24,480,144. The number *M*(4) is
the number of deals in which all four players receive a pair.

Suppose we would like to know the probability of precisely three players
being dealt a pair. The number *M*(3) is the number of deals in which
three players receive a pair. However, for each of the
*C*(4,3) = 4subsets of three properties, we also are including the number of deals
in which all four players receive a pair. Therefore, we must subtract
4*M*(4) from *M*(3) to obtain the number of deals in which precisely
three players receives a pair. This yields 1,506,345,984 deals
(we essentially are doing inclusion-exclusion with the 3-subsets as our
base set) for which precisely three players receive pairs.

Let's now do the same for a subset of two players. Once this example
is understood the process should be clear. First, there are
*C*(4,2) =
6 subsets with two players each. Now *M*(2) gives us the total number
of deals for which the subsets of two players are dealt pairs. As
before, this includes deals with pairs going to three and four players.
There are two 3-subsets containing a given 2-subset and there is the
single 4-set containing it. This means we have a contribution from 12
3-subsets (obtained from )
which must be subtracted. Since
*M*(3) includes deals for four 3-subsets, we subtract 12/4 = 3 times
*M*(3). The 4-subset is counted six times in *M*(2) and four times in
*M*(3). Thus, it is counted negative six times in
*M*(2)-3*M*(3). So
we must add 6*M*(4) in order to have deals with all four players
receiving a pair contributing zero to the count. This gives us
*M*(2)-3*M*(3)+6*M*(4) = 35,219,867,904 deals in which precisely two
players are dealt a pair. (Again it is inclusion-exclusion with the
2-subsets as the base set.)

The preceding technique allows us easily to obtain the following table of probabilities. The columns correspond to the exact number of players receiving a pair and the rows correspond to the number of players being dealt. For example, the probability exactly one player is dealt a pair when dealing to four people is .196. A blank entry means an entry is inappropriate and a neg entry means the probability is so small it is negligible.

0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | |

1 | .941 | .059 | ||||||||||

2 | .886 | .111 | .0035 | |||||||||

3 | .834 | .156 | .0099 | .0002 | ||||||||

4 | .785 | .196 | .019 | .0008 | neg | |||||||

5 | .739 | .230 | .029 | .0019 | .0001 | neg | ||||||

6 | .700 | .259 | .042 | .0032 | .0003 | neg | neg | |||||

7 | .655 | .285 | .054 | .0058 | .0004 | neg | neg | neg | ||||

8 | .617 | .306 | .068 | .0086 | .0007 | neg | neg | neg | neg | |||

9 | .581 | .324 | .082 | .012 | .0012 | .0001 | neg | neg | neg | neg | ||

10 | .547 | .339 | .096 | .016 | .0019 | .0001 | neg | neg | neg | neg | neg | |

11 | .515 | .351 | .110 | .021 | .0027 | .0003 | neg | neg | neg | neg | neg | neg |

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