1
00:00:01,240 --> 00:00:04,270
We will now apply our
multinomial formula for

2
00:00:04,270 --> 00:00:07,270
counting the number of
partitions to solve the

3
00:00:07,270 --> 00:00:09,510
following probability problem.

4
00:00:09,510 --> 00:00:12,690
We have a standard 52-card
deck, which we

5
00:00:12,690 --> 00:00:14,190
deal to four persons.

6
00:00:14,190 --> 00:00:18,890
Each person gets 13 cards as,
for example, in bridge.

7
00:00:18,890 --> 00:00:21,360
What is the probability
that each person

8
00:00:21,360 --> 00:00:24,710
gets exactly one ace?

9
00:00:24,710 --> 00:00:27,500
Well, before we start, as
always we will need a

10
00:00:27,500 --> 00:00:29,170
probability model.

11
00:00:29,170 --> 00:00:34,700
We deal the cards fairly, and
this is going to be our model.

12
00:00:34,700 --> 00:00:37,760
But we still need to interpret
our statement.

13
00:00:37,760 --> 00:00:41,200
To give this interpretation,
let us first think of the

14
00:00:41,200 --> 00:00:42,640
outcomes of the experiment.

15
00:00:42,640 --> 00:00:44,250
What are the possible
outcomes?

16
00:00:44,250 --> 00:00:49,440
An outcome of this experiment is
a partition of the 52 cards

17
00:00:49,440 --> 00:00:52,720
into the four persons
so that each person

18
00:00:52,720 --> 00:00:55,360
gets exactly 13 cards.

19
00:00:55,360 --> 00:01:00,050
Our statement about dealing the
cards fairly will be an

20
00:01:00,050 --> 00:01:03,680
assumption that all partitions
are equally likely.

21
00:01:12,180 --> 00:01:15,280
So since all partitions, all
outcomes of the experiment,

22
00:01:15,280 --> 00:01:18,030
are equally likely, this means
that we can solve a

23
00:01:18,030 --> 00:01:20,750
probability question
by counting.

24
00:01:20,750 --> 00:01:24,070
We need to count the number of
elements of our sample space,

25
00:01:24,070 --> 00:01:27,200
the number of possible outcomes,
and then count the

26
00:01:27,200 --> 00:01:29,820
number of outcomes that
make the event

27
00:01:29,820 --> 00:01:32,370
of interest to occur.

28
00:01:32,370 --> 00:01:34,360
Let us start with the
number of elements

29
00:01:34,360 --> 00:01:36,680
of the sample space.

30
00:01:36,680 --> 00:01:39,810
This is the problem that
we just dealt with a

31
00:01:39,810 --> 00:01:41,020
little while ago--

32
00:01:41,020 --> 00:01:45,250
the number of outcomes, the
number of partitions of 52

33
00:01:45,250 --> 00:01:52,015
items into four persons, where
we give 13 cards to person

34
00:01:52,015 --> 00:01:58,150
one, 13 cards to person two, 13
cards to person three, and

35
00:01:58,150 --> 00:02:00,700
13 cards to person four.

36
00:02:00,700 --> 00:02:04,700
The number of possible ways of
doing this is equal to this

37
00:02:04,700 --> 00:02:07,160
multinomial coefficient.

38
00:02:07,160 --> 00:02:10,360
So now let us count the number
of outcomes that belong to the

39
00:02:10,360 --> 00:02:13,980
event of interest, namely the
outcomes where each person

40
00:02:13,980 --> 00:02:16,250
gets an ace.

41
00:02:16,250 --> 00:02:19,400
We think of the process of
constructing such an outcome

42
00:02:19,400 --> 00:02:20,870
as a multi-stage process.

43
00:02:20,870 --> 00:02:22,740
And we count the number
of choices that we

44
00:02:22,740 --> 00:02:24,579
have at each stage.

45
00:02:24,579 --> 00:02:26,890
The process is as follows.

46
00:02:26,890 --> 00:02:30,390
We first distribute
the four aces.

47
00:02:30,390 --> 00:02:33,820
We take the ace of spades and
give it to one person.

48
00:02:33,820 --> 00:02:35,620
In how many ways can we do it?

49
00:02:35,620 --> 00:02:38,690
We can do it in four ways.

50
00:02:38,690 --> 00:02:40,660
Then we take the next ace.

51
00:02:40,660 --> 00:02:43,170
The next ace must be given
to a different person.

52
00:02:43,170 --> 00:02:46,530
And so at that stage, we have
three different choices about

53
00:02:46,530 --> 00:02:48,750
who to give that ace to.

54
00:02:48,750 --> 00:02:51,200
Then we consider the next ace.

55
00:02:51,200 --> 00:02:54,220
At this point, two persons
already have aces.

56
00:02:54,220 --> 00:02:57,610
So we have two available
choices for who can

57
00:02:57,610 --> 00:02:59,760
get the next ace.

58
00:02:59,760 --> 00:03:02,660
And finally for the last ace,
we do not have any choice.

59
00:03:02,660 --> 00:03:05,370
We give it to the only remaining
person who doesn't

60
00:03:05,370 --> 00:03:08,510
yet have an ace.

61
00:03:08,510 --> 00:03:13,990
Having distributed the four
aces, then we need to somehow

62
00:03:13,990 --> 00:03:19,590
distribute the remaining 48
cards to the four people.

63
00:03:19,590 --> 00:03:22,050
But we can do that in
any way we want.

64
00:03:22,050 --> 00:03:25,950
So all we need to do is to just
partition the 48 cards

65
00:03:25,950 --> 00:03:28,950
into four subsets of given
cardinalities.

66
00:03:28,950 --> 00:03:34,680
And this can be done by a number
of ways, which is the

67
00:03:34,680 --> 00:03:36,190
number of such partitions.

68
00:03:36,190 --> 00:03:39,350
We have already found
what that number is.

69
00:03:39,350 --> 00:03:43,135
And it is this particular
multinomial coefficient.

70
00:03:46,880 --> 00:03:50,440
So the number of ways that we
can distribute the cards so

71
00:03:50,440 --> 00:03:53,940
that each person gets an ace,
according to the counting

72
00:03:53,940 --> 00:03:56,870
principle, is going to be the
number of ways that we can

73
00:03:56,870 --> 00:04:00,150
distribute the aces times the
number of ways that we can

74
00:04:00,150 --> 00:04:02,060
distribute the remaining
cards.

75
00:04:02,060 --> 00:04:05,860
The product of this number gives
us the count, gives us

76
00:04:05,860 --> 00:04:09,510
the cardinality, of the
event of interest.

77
00:04:09,510 --> 00:04:12,860
We also have the cardinality
of the sample space.

78
00:04:12,860 --> 00:04:15,680
So the desired probability
can be found by

79
00:04:15,680 --> 00:04:18,000
dividing these two numbers.

80
00:04:18,000 --> 00:04:21,230
And the final answer
takes this form.

81
00:04:24,810 --> 00:04:27,300
Let us now look at the
same problem but

82
00:04:27,300 --> 00:04:28,910
in a different way.

83
00:04:28,910 --> 00:04:32,560
Probability problems can often
be solved in multiple ways.

84
00:04:32,560 --> 00:04:35,120
And some can be faster
than others.

85
00:04:35,120 --> 00:04:38,690
So we want to look for a smarter
solution that perhaps

86
00:04:38,690 --> 00:04:42,970
will get us in a faster way
to the desired answer.

87
00:04:42,970 --> 00:04:45,750
We will use the following
trick.

88
00:04:45,750 --> 00:04:48,990
We will think about a very
specific way of dealing the

89
00:04:48,990 --> 00:04:51,230
cards which is the following.

90
00:04:51,230 --> 00:04:56,210
We take the 52 cards, the card
deck, and stack it so that the

91
00:04:56,210 --> 00:04:59,180
four aces are at the top.

92
00:04:59,180 --> 00:05:01,200
So they are first.

93
00:05:01,200 --> 00:05:05,970
And then we deal those cards
to the players as follows.

94
00:05:05,970 --> 00:05:12,480
We think of each player having
13 slots of his own.

95
00:05:12,480 --> 00:05:16,630
And the cards will be
placed randomly into

96
00:05:16,630 --> 00:05:19,290
the different slots.

97
00:05:19,290 --> 00:05:24,060
So we can do this one card at a
time, starting from the top.

98
00:05:24,060 --> 00:05:31,380
We take the first ace and send
it to a random location.

99
00:05:31,380 --> 00:05:34,050
Then we will take the second
ace, send it to a random

100
00:05:34,050 --> 00:05:35,930
location, and so on.

101
00:05:35,930 --> 00:05:39,780
What we want to calculate is the
probability that the four

102
00:05:39,780 --> 00:05:45,150
aces will end up in locations
or in slots that are

103
00:05:45,150 --> 00:05:48,320
associated with different
persons.

104
00:05:48,320 --> 00:05:50,610
So let us calculate
this probability.

105
00:05:50,610 --> 00:05:52,290
The first ace can go anywhere.

106
00:05:52,290 --> 00:05:53,710
It doesn't matter.

107
00:05:53,710 --> 00:05:59,720
For the second ace, it has
51 slots to choose from.

108
00:05:59,720 --> 00:06:03,120
It's 51 because we started
with 52, but one slot has

109
00:06:03,120 --> 00:06:06,130
already been taken by
that particular ace.

110
00:06:06,130 --> 00:06:10,430
So for the ace of hearts,
we have 51 slots

111
00:06:10,430 --> 00:06:12,910
that it can go to.

112
00:06:12,910 --> 00:06:18,500
And out of those 51, we have
39 of them that belong to

113
00:06:18,500 --> 00:06:20,970
people who do not
yet have an ace.

114
00:06:20,970 --> 00:06:24,820
So this is the probability that
the ace of hearts gets

115
00:06:24,820 --> 00:06:29,110
placed into a slot that belongs
to a person who is

116
00:06:29,110 --> 00:06:34,180
different than the person
who got the first ace.

117
00:06:34,180 --> 00:06:36,570
Now let us consider this ace.

118
00:06:36,570 --> 00:06:40,890
What is the probability that
this ace will get into a slot

119
00:06:40,890 --> 00:06:44,620
which belongs to either this
person or that person?

120
00:06:44,620 --> 00:06:50,330
It has 26 slots in which
this desired

121
00:06:50,330 --> 00:06:52,220
event is going to happen.

122
00:06:52,220 --> 00:06:57,280
And it's 26 out of the
50 available slots.

123
00:06:57,280 --> 00:07:00,300
Finally, let us consider
this ace.

124
00:07:00,300 --> 00:07:03,410
So having placed that ace and
assuming that it got to a

125
00:07:03,410 --> 00:07:08,060
different person, what is the
probability now that this ace

126
00:07:08,060 --> 00:07:12,760
is going to go to this person
who doesn't yet have one?

127
00:07:12,760 --> 00:07:15,050
The probability of this
happening is the number of

128
00:07:15,050 --> 00:07:20,060
slots associated with that
person, which is equal to 13

129
00:07:20,060 --> 00:07:25,580
divided by the number
of slots that this

130
00:07:25,580 --> 00:07:27,320
card can choose from.

131
00:07:27,320 --> 00:07:31,710
And the number of slots is 52
minus the 3 slots that have

132
00:07:31,710 --> 00:07:36,310
already been taken,
so it's 49.

133
00:07:36,310 --> 00:07:40,280
And so this is the answer
to our problem.

134
00:07:40,280 --> 00:07:42,909
This expression looks very
different from the expression

135
00:07:42,909 --> 00:07:45,520
that we derived a
little earlier.

136
00:07:45,520 --> 00:07:49,210
But you can do the algebra, the
arithmetic, simplify the

137
00:07:49,210 --> 00:07:52,560
answer, and you will verify that
indeed it's exactly the

138
00:07:52,560 --> 00:07:53,860
same answer.

139
00:07:53,860 --> 00:07:58,520
And in case you're curious, the
numerical value turns out

140
00:07:58,520 --> 00:08:01,130
to be 0.105.

141
00:08:01,130 --> 00:08:02,880
So there's about 10% [chance]

142
00:08:02,880 --> 00:08:05,820
that when you deal the cards
in bridge, each one of the

143
00:08:05,820 --> 00:08:12,100
players is going to end up
having exactly one ace.

144
00:08:12,100 --> 00:08:16,230
So this was a faster way of
getting to the answer to our

145
00:08:16,230 --> 00:08:18,860
problem, compared to
the previous one.

146
00:08:18,860 --> 00:08:21,750
But it raises a legitimate
question.

147
00:08:21,750 --> 00:08:25,560
Is the way that we dealt the
cards by putting the aces on

148
00:08:25,560 --> 00:08:29,530
top and then dealing them,
is that way a fair way

149
00:08:29,530 --> 00:08:31,340
of dealing the cards?

150
00:08:31,340 --> 00:08:34,990
Is it true that with this way
of dealing the cards all

151
00:08:34,990 --> 00:08:38,090
partitions are equally likely?

152
00:08:38,090 --> 00:08:41,120
It turns out that this
is indeed the case.

153
00:08:41,120 --> 00:08:45,210
But it does require
a bit of thinking.

154
00:08:45,210 --> 00:08:47,530
Maybe you can see
it intuitively

155
00:08:47,530 --> 00:08:49,020
that this is the case.

156
00:08:49,020 --> 00:08:53,290
But if not, then it is something
that one can prove.

157
00:08:53,290 --> 00:08:55,850
It can be proved formally
as follows.

158
00:08:55,850 --> 00:09:01,360
One first needs to check that
all permutations, that is all

159
00:09:01,360 --> 00:09:03,940
possible allocations
of cards into

160
00:09:03,940 --> 00:09:07,270
slots, are equally likely.

161
00:09:07,270 --> 00:09:12,420
And because of this, one can
then argue that any possible

162
00:09:12,420 --> 00:09:16,250
partition into subsets of [13]

163
00:09:16,250 --> 00:09:18,280
is also equally likely.

164
00:09:18,280 --> 00:09:22,160
So this is an equivalent way
of dealing the cards to the

165
00:09:22,160 --> 00:09:25,330
one that we considered earlier,
which was that every

166
00:09:25,330 --> 00:09:27,710
partition is equally likely.

167
00:09:27,710 --> 00:09:31,350
Therefore, we did indeed solve
the same problem, and so this

168
00:09:31,350 --> 00:09:35,650
is a legitimate alternative way
of getting to the answer.

169
00:09:35,650 --> 00:09:39,140
And of course, it's reassuring
to check that this numerical

170
00:09:39,140 --> 00:09:41,940
expression agrees with the
numerical expression we had

171
00:09:41,940 --> 00:09:43,190
derived earlier.