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.