3.4.3 Two Pair Poker Hands: Video


PROFESSOR: So let’s
do a basic example of counting that
illustrates the use of these new generalized
rules of the Division Rule and the Generalized
Product Rule. And let’s count some
particular kind of poker hands called a 2-pair. So poker is a game where each
player is dealt five cards from a deck of 52 cards. And the definition
of a 2-pair hand is that there are 2
cards of some rank. The ranks are Ace,
Deuce, up through King, so the ranks are
13 possible ranks. Ace is 1, 2, 3, up through
10, and then Jack, Queen, King is 11, 12, 13. So there are 13 possible ranks. We’re going to choose 2
cards of some rank– that’s called a pair. Then we’re going to choose 2
cards of a different rank– a second rank. And finally, we’re
going to choose a card of still a third rank. So I get a pair,
and another pair, and another card that does
not match the ranks of either of the first two. And that is the
definition of a hand that, in poker, is called 2-pair. Let’s take an example. Here’s a typical 2-pair hand. I’ve got 2 Kings. They both have rank 13. One is a King of Diamonds,
the other is a King of Hearts. There are four of these suits,
so-called– Diamonds, Hearts, Spades, Clubs. There are 2 Aces,
a pair of Aces. One is an Ace of Diamonds,
the other’s an Ace of Spades. And finally, there hanging
loose, this third rank that doesn’t match the Kings or
the Aces– namely a 3 of Clubs. Now, the way that
I’m going to propose to count the number
of 2-pair hands is to think about it this way. I’m going to begin by choosing
the rank for the first pair, and there are 13 possible ranks
that the first pair might have. Once I’ve fixed the
rank for the first pair, the second pair has to
have a different rank. So there are 12 ranks left. And once I’ve picked the
ranks for the 2 pairs, then I have the rank
of the last card, which is 11 possible choices. Then, in addition,
once I’ve chosen the rank of the first pair,
the rank of the second pair, and the rank of the loose
card, the fifth card, I select a pair of suits
to go for the first pair. So let’s say if the first
pair, I’ve figured out we’re going to be 2 Aces. Which two aces should they be? Well, pick two of
the four suits. And there are four
choose two ways to choose the suits for the pair of aces. Likewise, there are four choose
two ways to choose the two suits for the pair of kings. And finally, there are
four possible suits I can choose for the
rank of the last card. So that says that I might,
for example, specify a two-pair hand by
saying, OK, let’s choose a pair of kings to
come first and a pair of aces to be the second pair and a
three to be the loose card. Let’s choose the set of
two elements diamonds and hearts for the
kings, the two elements diamonds and spades for the
aces, and a club for the three. This sequence of
choices specifies exactly the two-pair
hand that we illustrated on the previous slide,
namely two kings, a diamond and a hearts; two aces,
a diamond and a space; and the three of clubs. So I can count the number
of two-pair her hands fairly straightforwardly. There were 13 choices for
the rank of the first pair, 12 for the second, 11 for
the rank of the third card, four choose way to choose
the suits of the first pair, four choose way to
choose two ways to choose the suits of the second
pair, and four ways to choose the suits
for the last pair. So this is the total– 13 times
12 times 11 times 4 choose 2 twice times 4. And that’s not right. There’s a bug. What’s the bug? Well, the problem
is that what I’ve described in this number on the
previous slide, that number, is exactly the
set of six tuples, consisting of the first card
ranks and the second card ranks and the last card rank
and the first card suits and the second card suits
and the last card suit. That is, if it’s counting
the number of possible ranks for a first choice,
the number of possible ranks for a second
choice, third, and so on, this
set of six things are being counted
correctly by the formula on the previous page. The difficulty is that
counting these six tuples is not the same as counting
the number of two-pair hands. We’ve counted the number of six
tuples of this kind correctly, but not the number of
two-pair, because this mapping from six tuples to two-pair
hands is not a bijection. Namely, if I look
at the six tuple, choose kings and
then aces and a three with these suits and
those suits and final suit for the three, which
determines this hand– the king of diamonds, king of hearts,
ace of diamonds, ace of spades, three of clubs– there’s
another six tuple that would also yield the same hand. Namely, what I can
do is I’ll keep the three of clubs specified. But instead of choosing
the kings and their suits and the aces and
their suits, I’ll choose the aces and their suits
and the kings and their suits. So I’m just switching these two
entries and those two entries. And if I do that, here’s
a different six tuple that’s specifying the
same two-pair hand. That is, this
tuple is specifying a pair of aces and a pair of
kings, where the aces have suits diamonds spades and
the kings have suits diamonds hearts and the three
has suits clubs. So the bug in our reasoning
was that when we were counting and we said there
are 13 possible ranks for the first pair
and there are 12 possible ranks for
the second pair and we were distinguishing
the first pair from the second pair,
that was a mistake. There isn’t any first
pair and second pair. There are simply two pairs. And there’s no way to
tell which is first and which is
second, which is why we got two different ways
from our sextuples of mapping to the same two-pair,
depending in the sextuple which of the two-pair I
wanted to list first. So in fact, since either
pair might be first what I get is this map, from
six tuples to two-pair hands, is actually a
two-to-one mapping. It’s not a bijection,
because there’s no difference between the
first pair and the second pair. There’s just a couple of pair. If I do that, then I
can fix this formula. Now that I realize that
the mapping from these six tuples, which I’ve
counted correctly to the things I want to count–
namely, the two-pair hands– is two to one, then, by the
generalized product rule, or by the division
rule, all I need to do is divide this number by a half. And that is really the answer
of the number of two-pair hands.

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