1
00:00:00,970 --> 00:00:02,790
PROFESSOR: So let's
do a basic example

2
00:00:02,790 --> 00:00:06,230
of counting that
illustrates the use

3
00:00:06,230 --> 00:00:09,870
of these new generalized
rules of the Division

4
00:00:09,870 --> 00:00:12,140
Rule and the Generalized
Product Rule.

5
00:00:12,140 --> 00:00:16,040
And let's count some
particular kind of poker hands

6
00:00:16,040 --> 00:00:17,590
called a 2-pair.

7
00:00:17,590 --> 00:00:21,980
So poker is a game where each
player is dealt five cards

8
00:00:21,980 --> 00:00:25,260
from a deck of 52 cards.

9
00:00:25,260 --> 00:00:27,800
And the definition
of a 2-pair hand

10
00:00:27,800 --> 00:00:33,030
is that there are 2
cards of some rank.

11
00:00:33,030 --> 00:00:37,190
The ranks are Ace,
Deuce, up through King,

12
00:00:37,190 --> 00:00:41,090
so the ranks are
13 possible ranks.

13
00:00:41,090 --> 00:00:44,120
Ace is 1, 2, 3, up through
10, and then Jack, Queen, King

14
00:00:44,120 --> 00:00:45,910
is 11, 12, 13.

15
00:00:45,910 --> 00:00:48,050
So there are 13 possible ranks.

16
00:00:48,050 --> 00:00:50,450
We're going to choose 2
cards of some rank-- that's

17
00:00:50,450 --> 00:00:51,860
called a pair.

18
00:00:51,860 --> 00:00:55,110
Then we're going to choose 2
cards of a different rank--

19
00:00:55,110 --> 00:00:56,800
a second rank.

20
00:00:56,800 --> 00:00:59,190
And finally, we're
going to choose

21
00:00:59,190 --> 00:01:01,410
a card of still a third rank.

22
00:01:01,410 --> 00:01:05,400
So I get a pair,
and another pair,

23
00:01:05,400 --> 00:01:09,220
and another card that does
not match the ranks of either

24
00:01:09,220 --> 00:01:10,420
of the first two.

25
00:01:10,420 --> 00:01:12,630
And that is the
definition of a hand that,

26
00:01:12,630 --> 00:01:15,960
in poker, is called 2-pair.

27
00:01:15,960 --> 00:01:17,790
Let's take an example.

28
00:01:17,790 --> 00:01:19,620
Here's a typical 2-pair hand.

29
00:01:19,620 --> 00:01:21,130
I've got 2 Kings.

30
00:01:21,130 --> 00:01:22,820
They both have rank 13.

31
00:01:22,820 --> 00:01:26,330
One is a King of Diamonds,
the other is a King of Hearts.

32
00:01:26,330 --> 00:01:29,680
There are four of these suits,
so-called-- Diamonds, Hearts,

33
00:01:29,680 --> 00:01:32,090
Spades, Clubs.

34
00:01:32,090 --> 00:01:34,547
There are 2 Aces,
a pair of Aces.

35
00:01:34,547 --> 00:01:36,880
One is an Ace of Diamonds,
the other's an Ace of Spades.

36
00:01:36,880 --> 00:01:40,250
And finally, there hanging
loose, this third rank

37
00:01:40,250 --> 00:01:43,905
that doesn't match the Kings or
the Aces-- namely a 3 of Clubs.

38
00:01:46,510 --> 00:01:48,370
Now, the way that
I'm going to propose

39
00:01:48,370 --> 00:01:50,820
to count the number
of 2-pair hands

40
00:01:50,820 --> 00:01:52,110
is to think about it this way.

41
00:01:52,110 --> 00:01:56,790
I'm going to begin by choosing
the rank for the first pair,

42
00:01:56,790 --> 00:02:00,970
and there are 13 possible ranks
that the first pair might have.

43
00:02:00,970 --> 00:02:03,790
Once I've fixed the
rank for the first pair,

44
00:02:03,790 --> 00:02:06,370
the second pair has to
have a different rank.

45
00:02:06,370 --> 00:02:09,449
So there are 12 ranks left.

46
00:02:09,449 --> 00:02:15,110
And once I've picked the
ranks for the 2 pairs,

47
00:02:15,110 --> 00:02:18,370
then I have the rank
of the last card,

48
00:02:18,370 --> 00:02:22,200
which is 11 possible choices.

49
00:02:22,200 --> 00:02:24,067
Then, in addition,
once I've chosen

50
00:02:24,067 --> 00:02:26,400
the rank of the first pair,
the rank of the second pair,

51
00:02:26,400 --> 00:02:31,910
and the rank of the loose
card, the fifth card,

52
00:02:31,910 --> 00:02:37,241
I select a pair of suits
to go for the first pair.

53
00:02:37,241 --> 00:02:39,490
So let's say if the first
pair, I've figured out we're

54
00:02:39,490 --> 00:02:40,810
going to be 2 Aces.

55
00:02:40,810 --> 00:02:42,600
Which two aces should they be?

56
00:02:42,600 --> 00:02:45,000
Well, pick two of
the four suits.

57
00:02:45,000 --> 00:02:48,650
And there are four
choose two ways to choose

58
00:02:48,650 --> 00:02:52,460
the suits for the pair of aces.

59
00:02:52,460 --> 00:02:55,480
Likewise, there are four choose
two ways to choose the two

60
00:02:55,480 --> 00:02:58,630
suits for the pair of kings.

61
00:02:58,630 --> 00:03:01,160
And finally, there are
four possible suits

62
00:03:01,160 --> 00:03:06,440
I can choose for the
rank of the last card.

63
00:03:06,440 --> 00:03:12,770
So that says that I might,
for example, specify

64
00:03:12,770 --> 00:03:15,290
a two-pair hand by
saying, OK, let's

65
00:03:15,290 --> 00:03:18,800
choose a pair of kings to
come first and a pair of aces

66
00:03:18,800 --> 00:03:21,950
to be the second pair and a
three to be the loose card.

67
00:03:21,950 --> 00:03:24,170
Let's choose the set of
two elements diamonds

68
00:03:24,170 --> 00:03:27,140
and hearts for the
kings, the two elements

69
00:03:27,140 --> 00:03:31,380
diamonds and spades for the
aces, and a club for the three.

70
00:03:31,380 --> 00:03:33,970
This sequence of
choices specifies

71
00:03:33,970 --> 00:03:37,130
exactly the two-pair
hand that we illustrated

72
00:03:37,130 --> 00:03:39,503
on the previous slide,
namely two kings, a diamond

73
00:03:39,503 --> 00:03:42,700
and a hearts; two aces,
a diamond and a space;

74
00:03:42,700 --> 00:03:45,950
and the three of clubs.

75
00:03:45,950 --> 00:03:49,750
So I can count the number
of two-pair her hands fairly

76
00:03:49,750 --> 00:03:50,910
straightforwardly.

77
00:03:50,910 --> 00:03:53,280
There were 13 choices for
the rank of the first pair,

78
00:03:53,280 --> 00:03:56,410
12 for the second, 11 for
the rank of the third card,

79
00:03:56,410 --> 00:03:59,290
four choose way to choose
the suits of the first pair,

80
00:03:59,290 --> 00:04:01,490
four choose way to
choose two ways to choose

81
00:04:01,490 --> 00:04:04,280
the suits of the second
pair, and four ways

82
00:04:04,280 --> 00:04:06,680
to choose the suits
for the last pair.

83
00:04:06,680 --> 00:04:10,620
So this is the total-- 13 times
12 times 11 times 4 choose 2

84
00:04:10,620 --> 00:04:12,930
twice times 4.

85
00:04:12,930 --> 00:04:13,990
And that's not right.

86
00:04:13,990 --> 00:04:15,650
There's a bug.

87
00:04:15,650 --> 00:04:18,110
What's the bug?

88
00:04:18,110 --> 00:04:21,050
Well, the problem
is that what I've

89
00:04:21,050 --> 00:04:25,260
described in this number on the
previous slide, that number,

90
00:04:25,260 --> 00:04:30,160
is exactly the
set of six tuples,

91
00:04:30,160 --> 00:04:33,900
consisting of the first card
ranks and the second card ranks

92
00:04:33,900 --> 00:04:36,140
and the last card rank
and the first card suits

93
00:04:36,140 --> 00:04:38,370
and the second card suits
and the last card suit.

94
00:04:38,370 --> 00:04:44,500
That is, if it's counting
the number of possible ranks

95
00:04:44,500 --> 00:04:46,870
for a first choice,
the number of possible

96
00:04:46,870 --> 00:04:48,690
ranks for a second
choice, third,

97
00:04:48,690 --> 00:04:53,540
and so on, this
set of six things

98
00:04:53,540 --> 00:04:56,090
are being counted
correctly by the formula

99
00:04:56,090 --> 00:04:57,740
on the previous page.

100
00:04:57,740 --> 00:05:03,160
The difficulty is that
counting these six tuples

101
00:05:03,160 --> 00:05:07,890
is not the same as counting
the number of two-pair hands.

102
00:05:07,890 --> 00:05:10,790
We've counted the number of six
tuples of this kind correctly,

103
00:05:10,790 --> 00:05:14,440
but not the number of
two-pair, because this mapping

104
00:05:14,440 --> 00:05:19,060
from six tuples to two-pair
hands is not a bijection.

105
00:05:19,060 --> 00:05:22,440
Namely, if I look
at the six tuple,

106
00:05:22,440 --> 00:05:25,220
choose kings and
then aces and a three

107
00:05:25,220 --> 00:05:27,840
with these suits and
those suits and final suit

108
00:05:27,840 --> 00:05:31,321
for the three, which
determines this hand-- the king

109
00:05:31,321 --> 00:05:33,820
of diamonds, king of hearts,
ace of diamonds, ace of spades,

110
00:05:33,820 --> 00:05:37,470
three of clubs-- there's
another six tuple that

111
00:05:37,470 --> 00:05:40,360
would also yield the same hand.

112
00:05:40,360 --> 00:05:43,060
Namely, what I can
do is I'll keep

113
00:05:43,060 --> 00:05:46,350
the three of clubs specified.

114
00:05:46,350 --> 00:05:50,140
But instead of choosing
the kings and their suits

115
00:05:50,140 --> 00:05:51,790
and the aces and
their suits, I'll

116
00:05:51,790 --> 00:05:56,400
choose the aces and their suits
and the kings and their suits.

117
00:05:56,400 --> 00:05:59,800
So I'm just switching these two
entries and those two entries.

118
00:05:59,800 --> 00:06:03,170
And if I do that, here's
a different six tuple

119
00:06:03,170 --> 00:06:09,300
that's specifying the
same two-pair hand.

120
00:06:09,300 --> 00:06:12,150
That is, this
tuple is specifying

121
00:06:12,150 --> 00:06:15,080
a pair of aces and a pair of
kings, where the aces have

122
00:06:15,080 --> 00:06:19,330
suits diamonds spades and
the kings have suits diamonds

123
00:06:19,330 --> 00:06:22,830
hearts and the three
has suits clubs.

124
00:06:22,830 --> 00:06:26,690
So the bug in our reasoning
was that when we were counting

125
00:06:26,690 --> 00:06:29,660
and we said there
are 13 possible ranks

126
00:06:29,660 --> 00:06:31,760
for the first pair
and there are 12

127
00:06:31,760 --> 00:06:33,950
possible ranks for
the second pair

128
00:06:33,950 --> 00:06:35,860
and we were distinguishing
the first pair

129
00:06:35,860 --> 00:06:38,080
from the second pair,
that was a mistake.

130
00:06:38,080 --> 00:06:40,900
There isn't any first
pair and second pair.

131
00:06:40,900 --> 00:06:43,660
There are simply two pairs.

132
00:06:43,660 --> 00:06:45,560
And there's no way to
tell which is first

133
00:06:45,560 --> 00:06:47,380
and which is
second, which is why

134
00:06:47,380 --> 00:06:52,450
we got two different ways
from our sextuples of mapping

135
00:06:52,450 --> 00:06:59,430
to the same two-pair,
depending in the sextuple which

136
00:06:59,430 --> 00:07:01,880
of the two-pair I
wanted to list first.

137
00:07:01,880 --> 00:07:04,920
So in fact, since either
pair might be first

138
00:07:04,920 --> 00:07:09,790
what I get is this map, from
six tuples to two-pair hands,

139
00:07:09,790 --> 00:07:12,670
is actually a
two-to-one mapping.

140
00:07:12,670 --> 00:07:15,430
It's not a bijection,
because there's

141
00:07:15,430 --> 00:07:18,440
no difference between the
first pair and the second pair.

142
00:07:18,440 --> 00:07:21,290
There's just a couple of pair.

143
00:07:21,290 --> 00:07:24,460
If I do that, then I
can fix this formula.

144
00:07:24,460 --> 00:07:26,740
Now that I realize that
the mapping from these six

145
00:07:26,740 --> 00:07:28,660
tuples, which I've
counted correctly

146
00:07:28,660 --> 00:07:31,390
to the things I want to count--
namely, the two-pair hands-- is

147
00:07:31,390 --> 00:07:34,420
two to one, then, by the
generalized product rule,

148
00:07:34,420 --> 00:07:36,470
or by the division
rule, all I need to do

149
00:07:36,470 --> 00:07:39,410
is divide this number by a half.

150
00:07:39,410 --> 00:07:44,943
And that is really the answer
of the number of two-pair hands.