4
CHAPTER
The Multiplication
Rules
Introduction
The previous chapter showed how the addition rules could be used to solve
problems in probability. This chapter will show you how to use the
multiplication rules to solve many problems in probability. In addition, the
concept of independent and dependent events will be introduced.
56
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CHAPTER 4 The Multiplication Rules
57
Independent and Dependent Events
The multiplication rules can be used to find the probability of two or more
events that occur in sequence. For example, we can find the probability of
selecting three jacks from a deck of cards on three sequential draws. Before
explaining the rules, it is necessary to differentiate between independent and
dependent events.
Two events, A and B, are said to be independent if the fact that event A
occurs does not affect the probability that event B occurs. For example, if a
coin is tossed and then a die is rolled, the outcome of the coin in no way
affects or changes the probability of the outcome of the die. Another
example would be selecting a card from a deck, replacing it, and
then selecting a second card from a deck. The outcome of the first card, as
long as it is replaced, has no effect on the probability of the outcome of the
second card.
On the other hand, when the occurrence of the first event in some way
changes the probability of the occurrence of the second event, the two events
are said to be dependent. For example, suppose a card is selected from a deck
and not replaced, and a second card is selected. In this case, the probability of
1
, but since this card is not
selecting any specific card on the first draw is 52
replaced, the probability of selecting any other specific card on the second
1
, since there are only 51 cards left.
draw is 51
Another example would be parking in a no parking zone and getting a
parking ticket. Again, if you are legally parked, the chances of getting a
parking ticket are pretty close to zero (as long as the meter does not run out).
However, if you are illegally parked, your chances of getting a parking ticket
dramatically increase.
PRACTICE
Determine whether the two events are independent or dependent.
1. Tossing a coin and selecting a card from a deck
2. Driving on ice and having an accident
3. Drawing a ball from an urn, not replacing it, and then drawing a
second ball
4. Having a high I.Q. and having a large hat size
5. Tossing one coin and then tossing a second coin
CHAPTER 4 The Multiplication Rules
58
ANSWERS
1. Independent. Tossing a coin has no effect on drawing a card.
2. Dependent. In most cases, driving on ice will increase the probability
of having an accident.
3. Dependent. Since the first ball is not replaced before the second ball is
selected, it will change the probability of a specific second ball being
selected.
4. Independent. To the best of the author’s knowledge, no studies have
been done showing any relationship between hat size and I.Q.
5. Independent. The outcome of the first coin does not influence the
outcome of the second coin.
Multiplication Rule I
Before explaining the first multiplication rule, consider the example of tossing
two coins. The sample space is HH, HT, TH, TT. From classical probability
theory, it can be determined that the probability of getting two heads is 14,
since there is only one way to get two heads and there are four outcomes in
the sample space. However, there is another way to determine the probability
of getting two heads. In this case, the probability of getting a head on the first
toss is 12, and the probability of getting a head on the second toss is also 12.
So the probability of getting two heads can be determined by multiplying
1 1
1
2 ! 2 ¼ 4 This example illustrates the first multiplication rule.
ONLY for
INDEPENDENT
events
Multiplication Rule I: For two independent events A and B, PðA and BÞ ¼
PðAÞ ! PðBÞ.
In other words, when two independent events occur in sequence, the probability that both events will occur can be found by multiplying the probabilities of each individual event.
The word and is the key word and means that both events occur in
sequence and to multiply.
EXAMPLE: A coin is tossed and a die is rolled. Find the probability of
getting a tail on the coin and a 5 on the die.
SOLUTION:
1
.
Since PðtailÞ ¼ 12 and Pð5Þ ¼ 16 ; Pðtail and 5Þ ¼ PðtailÞ ! Pð5Þ ¼ 12 ! 16 ¼ 12
Note that the events are independent.
CHAPTER 4 The Multiplication Rules
The previous example can also be solved using classical probability. Recall
that the sample space for tossing a coin and rolling a die is
H1, H2, H3, H4, H5, H6
T1, T2, T3, T4, T5, T6
Notice that there are 12 outcomes in the sample space and only one
1
.
outcome is a tail and a 5; hence, P(tail and 5) ¼ 12
EXAMPLE: An urn contains 2 red balls, 3 green balls, and 5 blue balls.
A ball is selected at random and its color is noted. Then it is replaced and
another ball is selected and its color is noted. Find the probability of each of
these:
a. Selecting 2 blue balls
b. Selecting a blue ball and then a red ball
c. Selecting a green ball and then a blue ball
SOLUTION:
Since the first ball is being replaced before the second ball is selected, the
events are independent.
a. There are 5 blue balls and a total of 10 balls; therefore, the probability
of selecting two blue balls with replacement is
P(blue and blue) ¼ PðblueÞ $ PðblueÞ
5 5
$
10 10
25
1
¼
¼
100 4
¼
b. There are 5 blue balls and 2 red balls, so the probability of selecting a
blue ball and then a red ball with replacement is
Pðblue and redÞ ¼ PðblueÞ $ PðredÞ
5 2
$
10 10
10
1
¼
¼
100 10
¼
59
CHAPTER 4 The Multiplication Rules
60
c. There are 3 green balls and 5 blue balls, so the probability of selecting
a green ball and then a blue ball with replacement is
Pðgreen and blueÞ ¼ PðgreenÞ $ PðblueÞ
3 5
¼ $
10 10
15
3
¼
¼
100 20
The multiplication rule can be extended to 3 or more events that
occur in sequence, as shown in the next example.
EXAMPLE: A die is tossed 3 times. Find the probability of getting three 6s.
SOLUTION:
When a die is tossed, the probability of getting a six is 16; hence, the probability of getting three 6s is
Pð6 and 6 and 6Þ ¼ Pð6Þ $ Pð6Þ $ Pð6Þ
1 1 1
¼ $ $
6 6 6
1
¼
216
Another situation occurs in probability when subjects are selected from a
large population. Even though the subjects are not replaced, the probability
changes only slightly, so the change can be ignored. Consider the next
example.
EXAMPLE: It is known that 66% of the students at a large college favor
building a new fitness center. If two students are selected at random, find the
probability that all of them favor the building of a new fitness center.
SOLUTION:
Since the student population at the college is large, selecting a student
does not change the 66% probability that the next student selected will
favor the building of a new fitness center; hence, the probability of selecting
two students who both favor the building of a new fitness center is
(0.66)(0.66) ¼ 0.4356 or 43.56%.
CHAPTER 4 The Multiplication Rules
61
PRACTICE
1. A card is drawn from a deck, then replaced, and a second card is
drawn. Find the probability that two kings are selected.
2. If 12% of adults are left-handed, find the probability that if three
adults are selected at random, all three will be left-handed.
3. If two people are selected at random, find the probability that both
were born in August.
4. A coin is tossed 4 times. Find the probability of getting 4 heads.
5. A die is rolled and a card is selected at random from a deck of 52
cards. Find the probability of getting an odd number on the die and a
club on the card.
ANSWERS
1. The probability that 2 kings are selected is
Pðking and kingÞ ¼ PðkingÞ $ PðkingÞ ¼
41
41
1
$
¼
13
13
169
52
52
2. The probability of selecting 3 adults who are left-handed is
(0.12)(0.12)(0.12) ¼ 0.001728.
3. Each person has approximately 1 chance in 12 of being born in
August; hence, the probability that both are born in August is
1 1
1
$ ¼
:
12 12 144
1 1 1 1
1
4. The probability of getting 4 heads is $ $ $ ¼ .
2 2 2 2 16
5. The probability of getting an odd number on the die is 36 ¼ 12, and the
1
probability of getting a club is 13
52 ¼ 4; hence, the P(odd and
1 1
1
club) ¼ PðoddÞ $ PðclubÞ ¼ 2 $ 4 ¼ 8.
Important
Multiplication Rule II
When two sequential events are dependent, a slight variation of the
multiplication rule is used to find the probability of both events occurring.
For example, when a card is selected from an ordinary deck of 52 cards the
CHAPTER 4 The Multiplication Rules
62
1
probability of getting a specific card is 52
, but the probability of getting a
1
specific card on the second draw is 51 since 51 cards remain.
EXAMPLE: Two cards are selected from a deck and the first card is not
replaced. Find the probability of getting two kings.
SOLUTION:
4
and the probability
The probability of getting a king on the first draw is 52
3
of getting a king on the second draw is 51, since there are 3 kings left and
51 cards left. Hence the probability of getting 2 kings when the first card is
4
3
12
1
! 51
¼ 2652
¼ 221
.
not replaced is 52
When the two events A and B are dependent, the probability that the
second event B occurs after the first event A has already occurred is written as
P(B | A). This does not mean that B is divided by A; rather, it means and is
read as ‘‘the probability that event B occurs given that event A has already
occurred.’’ P(B | A) also means the conditional probability that event B occurs
given event A has occurred. The second multiplication rule follows.
Multiplication Rule II: When two events are dependent, the probability of
both events occurring is PðA and BÞ ¼ PðAÞ ! PðB j AÞ
EXAMPLE: A box contains 24 toasters, 3 of which are defective. If
two toasters are selected and tested, find the probability that both are
defective.
SOLUTION:
Since there are 3 defective toasters out of 24, the probability that the first
3
¼ 18. Since the second toaster is selected from the
toaster is defective is 24
remaining 23 and there are two defective toasters left, the probability that
2
. Hence, the probability that both toasters are defective is
it is defective is 23
PðD1 and D2 Þ ¼ PðD1 Þ ! PðD2 jD1 Þ ¼
31
24 8
4
!
21
1
¼
23 92
CHAPTER 4 The Multiplication Rules
EXAMPLE: Two cards are drawn without replacement from a deck of
52 cards. Find the probability that both are queens.
SOLUTION:
PðQ and QÞ ¼ PðQÞ $ PðQjQÞ
4 3
¼ $
52 51
1
¼
221
This multiplication rule can be extended to include three or more events, as
shown in the next example.
EXAMPLE: A box contains 3 orange balls, 3 yellow balls, and 2 white balls.
Three balls are selected without replacement. Find the probability of
selecting 2 yellow balls and a white ball.
SOLUTION:
3 2 2
Pðyellow and yellow and whiteÞ ¼ $ $
8 7 6
31 21 21
¼ 4$ $ 1
8 7 6
1
¼
28
Remember that the key word for the multiplication rule is and. It means to
multiply.
When two events are dependent, the probability that the second event
occurs must be adjusted for the occurrence of the first event. For the
mathematical purist, only one multiplication rule is necessary for two events,
and that is
PðA and BÞ ¼ PðAÞ $ PðB j AÞ:
The reason is that when the events are independent PðBjAÞ ¼ PðBÞ since
the occurrence of the first event A has no effect on the occurrence of the
second event B.
63
CHAPTER 4 The Multiplication Rules
64
PRACTICE
1. In a study, there are 8 guinea pigs; 5 are black and 3 are white. If
2 pigs are selected without replacement, find the probability that both
are white.
2. In a classroom there are 8 freshmen and 6 sophomores. If three
students are selected at random for a class project, find the probability that all 3 are freshmen.
3. Three cards are drawn from a deck of 52 cards without replacement.
Find the probability of getting 3 diamonds.
4. A box contains 12 calculators of which 5 are defective. If two calculators are selected without replacement, find the probability that both
are good.
5. A large flashlight has 6 batteries. Three are dead. If two batteries are
selected at random and tested, find the probability that both are dead.
ANSWERS
3 21
3
" ¼
4 7
28
8
1
1
8 7 6
2
freshmen) ¼ 2 " " 2 ¼
13
14 13 12
1
13 12 11
11
diamonds) ¼ 4 " " ¼
51
50
850
52
7 61
7
good) ¼ 2 " ¼
12 11 22
3 21
3
1
batteries dead) ¼ 3 " ¼
¼
15 5
6 5
1. P(white and white) ¼
2. P(3
3. P(3
4. P(2
5. P(2
Conditional Probability
Previously, conditional probability was used to find the probability of
sequential events occurring when they were dependent. Recall that P(B|A)
means the probability of event B occurring given that event A has already
occurred. Another situation where conditional probability can be used is
when additional information about an event is known. Sometimes it might be
CHAPTER 4 The Multiplication Rules
known that some outcomes in the sample space have occurred or that some
outcomes cannot occur. When conditions are imposed or known on events,
there is a possibility that the probability of the certain event occurring may
change. For example, suppose you want to determine the probability that a
house will be destroyed by a hurricane. If you used all houses in the United
States as the sample space, the probability would be very small. However, if
you used only the houses in the states that border the Atlantic Ocean as the
sample space, the probability would be much higher. Consider the following
examples.
EXAMPLE: A die is rolled; find the probability of getting a 4 if it is known
that an even number occurred when the die was rolled.
SOLUTION:
If it is known that an even number has occurred, the sample space is reduced
to 2, 4, or 6. Hence the probability of getting a 4 is 13 since there is one chance
in three of getting a 4 if it is known that the result was an even number.
EXAMPLE: Two dice are rolled. Find the probability of getting a sum of 3 if
it is known that the sum of the spots on the dice was less than six.
SOLUTION:
There are 2 ways to get a sum of 3. They are (1, 2) and (2, 1), and there are
10 ways to get a sum less than six. They are (1, 1), (1, 2), (2, 1), (3, 1), (2, 2),
(1, 3), (1, 4), (2, 3), (3, 2), and (4, 1); hence, P(sum of 3|sum less than 6) ¼
2
1
10 ¼ 5.
The two previous examples of conditional probability were solved using
classical probability and reduced sample spaces; however, they can be solved
by using the following formula for conditional probability.
The conditional probability of two events A and B is
PðAjBÞ ¼
PðA and BÞ
:
PðBÞ
P(A and B) means the probability of the outcomes that events A and B
have in common. The two previous examples will now be solved using the
formula for conditional probability.
EXAMPLE: A die is rolled; find the probability of getting a 4, if it is known
that an even number occurred when the die was rolled.
65
CHAPTER 4 The Multiplication Rules
66
SOLUTION:
P(A and B) is the probability of getting a 4 and an even number at the same
time. Notice that there is only one way to get a 4 and an even number—the
outcome 4. Hence P(A and B) ¼ 16. Also P(B) is the probability of getting an
even number which is 36 ¼ 12. Now
PðA and BÞ
PðBÞ
1
¼ 6
1
2
1 1
¼ $
6 2
PðAjBÞ ¼
1 21
%
63 1
1
¼
3
Notice that the answer is the same as the answer obtained when classical
probability was used.
¼
EXAMPLE: Two dice are rolled. Find the probability of getting a sum of 3 if
it is known that the sum of the spots on the dice was less than 6.
SOLUTION:
P(A and B) means the probability of getting a sum of 3 and a sum less than 6.
2
Hence P(A and B) ¼ 36
: PðBÞ means getting a sum less than 6 and is 10
36.
Hence,
PðA and BÞ
PðAjBÞ ¼
¼
PðBÞ
2
36
10
36
2 10
$
36 36
21 36
¼ % 5
36 10
1
¼
5
¼
CHAPTER 4 The Multiplication Rules
EXAMPLE: When two dice were rolled, it is known that the sum was an even
number. In this case, find the probability that the sum was 8.
SOLUTION:
When rolling two dice, there are 18 outcomes in which the sum is an even
number. There are 5 ways to get a sum of 8; hence, P(sum of 8|sum is
8
or 49.
even) is 18
EXAMPLE: In a large housing plan, 35% of the homes have a deck and
a two-car garage, and 80% of the houses have a two-car garage. Find the
probability that a house has a deck given that it has a garage.
SOLUTION:
Pðdeck and two-car garageÞ ¼ 0:35 and Pðtwo-car garageÞ ¼ 0:80; hence;
(deck and two-car garage)
P(deckjtwo-car garage) ¼
P(two-car garage)
0:35
¼
0:80
7
or 0:4375
¼
16
PRACTICE
1. When two dice are rolled, find the probability that one die is a 6 given
that the sum of the spots is 8.
2. Two coins are tossed. Find the probability of getting two tails if it is
known that one of the coins is a tail.
3. A card is selected from a deck. Find the probability that it is an ace
given that it is a black card.
4. The probability that a family visits the Sand Crab Water Park and
the Rainbow Gardens Amusement Park is 0.20. The probability
that a family visits the Rainbow Gardens Amusement Park is 0.80.
Find the probability that a family visits the Sand Crab Water Park
if it is known that they have already visited Rainbow Gardens
Amusement Park.
5. Three dice are rolled. Find the probability of getting three twos if it is
known that the sum of the spots of the three dice was six.
67
CHAPTER 4 The Multiplication Rules
68
ANSWERS
1. There are five ways to get a sum of 8. They are (6, 2), (2, 6), (3, 5), (5, 3),
and (4, 4). There are two ways to get a sum of 8 when one die is a 6.
Hence,
P(one die is a sixjsum of 8) ¼
¼
¼
P(one die is a six and sum of 8)
P(sum of 8)
2
36
5
36
2
5
2
361
" ¼ 1#
36 36 36
5
2
5
The problem can also be solved by looking at the reduced sample
space. There are two possible ways one die is a 6 and the sum of the
dice is 8. There are five ways to get a sum of 8. Hence, the probability
is 25.
¼
2. There are 3 ways to get one tail: HT, TH, and TT. There is one way to
get two tails; hence, the probability of getting two tails given that one
of the coins is a tail is 13. The problem can also be solved using the
formula for conditional probability:
1
4
3
4
P(two tailsjgiven one coin was a tail) ¼
1 3
¼ "
4 4
¼
1
41
"
3
41
¼
1
3
CHAPTER 4 The Multiplication Rules
3. There are 2 black aces and 26 black cards; hence, P(ace|black
2
1
or 13
or
card) ¼ 26
P(ace and black card)
P(black card)
2
¼ 52
26
52
2 26
¼
#
52 52
21 521
¼ 1 $ 13
52 26
1
¼
13
P(acejblack cardÞ ¼
4. P (Sand CrabjRainbow Gardens)
P(Sand Crab and Rainbow Gardens)
¼
P(Rainbow Gardens)
0:20
¼
¼ 0:25
0:80
5. There is one way to get 3 twos—(2, 2, 2) and there are 10 ways to get
a sum of six—(2, 2, 2), (3, 2, 1), (1, 2, 3), (2, 1, 3), (2, 3, 1), (3, 1, 2),
(1, 3, 2), (4, 1, 1), (1, 4, 1,) and (1, 1, 4). Hence,
P(3 twos and sum of 6)
P(sum of 6)
1
¼ 36
10
36
1 10
#
¼
36 36
1 361
¼ 1$
36 10
1
¼
10
P(3 twosjsum of 6) ¼
69
CHAPTER 4 The Multiplication Rules
70
Summary
When two events occur in sequence, the probability that both events occur
can be found by using one of the multiplication rules. When two events
are independent, the probability that the first event occurs does not affect
or change the probability of the second event occurring. If the events are
independent, multiplication rule I is used. When the two events are
dependent, the probability of the second event occurring is changed after
the first event occurs. If the events are dependent, multiplication rule II is
used. The key word for using the multiplication rule is ‘‘and.’’ Conditional
probability is used when additional information is known about the
probability of an event.
CHAPTER QUIZ
1. Which of the following events are dependent?
a.
b.
c.
d.
Tossing a coin and rolling a die
Rolling a die and then rolling a second die
Sitting in the sun all day and getting sunburned
Drawing a card from a deck and rolling a die
2. Three dice are rolled. What is the probability of getting three 4s?
1
216
1
b.
6
1
c.
36
1
d.
18
a.
CHAPTER 4 The Multiplication Rules
3. What is the probability of selecting 4 spades from a deck of 52 cards if
each card is replaced before the next one is selected?
4
a.
13
1
b.
2561
1
c.
52
1
d.
13
4. A die is rolled five times. What is the probability of getting 5 twos?
1
a.
8
1
b.
4
1
c.
6
1
d.
32
5. A coin is tossed four times; what is the probability of getting 4 heads?
1
a.
2
1
b.
16
c. 1
d.
1
4
6. If 25% of U.S. prison inmates are not U.S. citizens, what is the
probability of randomly selecting three inmates who will not be
U.S. citizens?
a.
b.
c.
d.
0.75
0.421875
0.015625
0.225
71
CHAPTER 4 The Multiplication Rules
72
7. If three people are randomly selected, find the probability that they
will all have birthdays in June.
1
1728
1
b.
36
1
c.
12
1
d.
4
a.
8. In a sample of 10 telephones, 4 are defective. If 3 are selected
at random and tested, what is the probability that all will be nondefective?
1
30
8
b.
125
1
c.
6
27
d.
125
a.
9. A bag contains 4 blue marbles and 5 red marbles. If 2 marbles are
selected at random without replacement, what is the probability that
both will be blue?
16
81
1
b.
6
1
c.
2
1
d.
15
a.
CHAPTER 4 The Multiplication Rules
10. If two people are selected at random from the phone book of a large
city, find the probability that they were both born on a Sunday.
1
7
1
b.
49
2
c.
365
2
d.
7
a.
11. The numbers 1 to 12 are placed in a hat, and a number is selected.
What is the probability that the number is 4 given that it is known to
be an even number?
2
3
3
b.
4
1
c.
6
1
d.
8
a.
12. Three coins are tossed; what is the probability of getting 3 heads if it
is known that at least two heads were obtained?
1
4
2
b.
3
1
c.
2
3
d.
8
a.
73
CHAPTER 4 The Multiplication Rules
74
13. In a certain group of people, it is known that 40% of the people take
Vitamins C and E on a daily basis. It is known that 75% take Vitamin
C on a daily basis. If a person is selected at random, what is the
probability that the person takes Vitamin E given that the person
takes Vitamin C?
5
13
8
b.
15
3
c.
11
7
d.
12
a.
14. Two dice are tossed; what is the probability that the numbers are the
same on both dice if it is known that the sum of the spots is 6?
2
3
1
b.
6
4
c.
5
1
d.
5
a.
Probability Sidelight
THE LAW OF AVERAGES
Suppose I asked you that if you tossed a coin 9 times and got 9 heads, what
would you bet that you would get on the tenth toss, heads or tails? Most
people would bet on a tail. When asked why they would select a tail, they
would probably respond that a tail was ‘‘due’’ according to the ‘‘law of
averages.’’ In reality, however, the probability of getting a head on the
CHAPTER 4 The Multiplication Rules
tenth toss is 12, and the probability of getting a tail on the tenth toss is 12, so
it doesn’t really matter since the probabilities are the same. A coin is an
inanimate object. It does not have a memory. It doesn’t know that in the
long run, the number of heads and the number of tails should balance out.
So does that make the law of averages wrong? No. You see, there’s a big
difference between asking the question, ‘‘What is the probability of getting
10 heads if I toss a coin ten times?’’ and ‘‘If I get 9 heads in a row, what is
the probability of getting a head on the tenth toss?’’ The answer to the first
1
, that is about 1 chance in 1000, and the answer to the
question is 2110 ¼ 1024
second question is 1.
2
This reasoning can be applied to many situations. For example, suppose
that a prize is offered for tossing a coin and getting 10 heads in a row. If you
played the game, you would have only one chance in 1024 of winning, but if
1024 people played the game, there is a pretty good chance that somebody
would win the prize. If 2028 people played the game, there would be a good
chance that two people might win. So what does this mean? It means that the
probability of winning big in a lottery or on a slot machine is very small, but
since there are many, many people playing, somebody will probably win;
however, your chances of winning big are very small.
A similar situation occurs when couples have children. Suppose a husband
and wife have four boys and would like to have a girl. It is incorrect to reason
1
, so it is more likely that the
that the chance of having a family of 5 boys is 32
next child will be a girl. However, after each child is born, the probability that
the next child is a girl (or a boy for that matter) is about 12. The law of
averages is not appropriate here.
My wife’s aunt had seven girls before the first boy was born. Also, in the
Life Science Library’s book entitled Mathematics, there is a photograph of
the Landon family of Harrison, Tennessee, that shows Mr. and Mrs. Emery
Landon and their 13 boys!
Another area where people incorrectly apply the law of averages is in
attempting to apply a betting system to gambling games. One such system is
doubling your bet when you lose. Consider a game where a coin is tossed. If it
lands heads, you win what you bet. If it lands tails, you lose. Now if you bet
one dollar on the first toss and get a head, you win one dollar. If you get tails,
you lose one dollar and bet two dollars on the next toss. If you win, you are
one dollar ahead because you lost one dollar on the first bet but won two
dollars on the second bet. If you get a tail on the second toss, you bet four
dollars on the third toss. If you win, you start over with a one dollar bet, but
if you lose, you bet eight dollars on the next toss. With this system, you win
every time you get a head. Sounds pretty good, doesn’t it?
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76
CHAPTER 4 The Multiplication Rules
This strategy won’t work because if you play long enough, you will
eventually run out of money since if you get a series of tails, you must
increase your bet substantially each time. So if you lose five times in a row,
you have lost $1 þ $2 þ $4 þ $8 þ $16 or $31, and your next bet has to be $32.
So you are betting $63 to win $1. Runs do occur and when they do, hope that
they are in your favor.
Now let’s look at some unusual so-called ‘‘runs.’’
In 1950, a person won 28 straight times playing the game of craps (dice) at
the Desert Inn in Las Vegas. He lost on the twenty-ninth roll. He did not win
big though because after each win he stuffed some bills in his pocket. The
event took about one hour and twenty minutes.
In 1959 in a casino in Puerto Rico at a roulette game, the number 10
occurred six times in succession. There are 38 numbers on a roulette wheel.
At a casino in New York in 1943 the color red occurred in a roulette game
32 times in a row, and at a casino in Monte Carlo an even number occurred
in a roulette game 28 times in a row.
These incidents have been reported in two books, one entitled Scarne’s
Complete Guide to Gambling and the other entitled Lady Luck by Warren
Weaver.
So what can be concluded? First, rare events (events with a small
probability of occurring) can and do occur. Second, the more people who
play a game, the more likely someone will win. Finally, the law of averages
applies when there is a large number of independent outcomes in which the
probability of each outcome occurring does not change.