CHAPTER
5
Odds and Expectation
Introduction
In this chapter you will learn about two concepts that are often used in
conjunction with probability. They are odds and expectation. Odds are used
most often in gambling games at casinos and racetracks, and in sports betting
and lotteries. Odds make it easier than probabilities to determine payoffs.
Mathematical expectation can be thought of more or less as an average over
the long run. In other words, if you would perform a probability experiment
many times, the expectation would be an average of the outcomes. Also,
expectation can be used to determine the average payoff per game in a gamb-
ling game.
77
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Odds
Odds are used by casinos, racetracks, and other gambling establishments to
determine payoffs when bets are made. For example, at a race, the odds that
a horse wins the race may be 4 to 1. In this case, if you bet $1 and the horse
wins, you get $4. If you bet $2 and the horse wins, you get $8, and so on.
Odds are computed from probabilities. For example, suppose you roll a
die and if you roll a three, you win. If you roll any other number, you lose.
Furthermore, if you bet one dollar and win, what would the payoff be if you
win? In this case, there are six outcomes, and you have one chance (outcome)
of winning, so the probability that you win is
1
6
. That means on average you
win once in every six rolls. So if you lose on the first five rolls and win on the
sixth, you have lost $5 and therefore, you should get $5 if you win on the
sixth roll. So if you bet $1 and win $5, the odds are 1 to 5. Of course, there is

no guarantee that you will win on the sixth roll. You may win on the first roll
or any roll, but on average for every six rolls, you will win one time over the
long run.
In gambling games, the odds are expressed backwards. For example,
if there is one chance in six that you will win, the odds are 1 to 5, but in
general, the odds would be given as 5 to 1. In gambling, the house (the people
running the game) will offer lower odds, say 4 to 1, in order to make a profit.
In this case, then, the player wins on average one time in every 6 rolls and
spends on average $5, but when the player wins, he gets only $4. So the house
wins on average $1 for every six rolls of the player.
Odds can be expressed as a fraction,
1
5
, or as a ratio, 1 : 5. If the odds of
winning the game are 1 : 5, then the odds of losing are 5 : 1. The odds of
winning the game can also be called the odds ‘‘in favor’’ of the event
occurring. The odds of losing can also be called ‘‘the odds against’’ the event
occurring.
The formulas for odds are
odds in favor ¼
PðEÞ
1 À PðEÞ
odds against ¼
PðEÞ
1 À Pð
EÞ
where P(E) is the probability that the event E occurs and Pð
EÞ is the
probability that the event does not occur.
CHAPTER 5 Odds and Expectation

78
EXAMPLE: Two coins are tossed; find the odds in favor of getting two heads
SOLUTION:
When two coins are tossed, there are four outcomes and PðHHÞ¼
1
4
: PðEÞ¼1
À
1
4
¼
3
4
; hence,
odds in favor of two heads ¼
PðE Þ
1 À PðE Þ
¼
1
4
1 À
1
4
¼
1
4
3
4
¼
1

4
Ä
3
4
¼
1
4
1
Á
4
1
3
¼
1
3
The odds are 1 : 3.
EXAMPLE: Two dice are rolled; find the odds against getting a sum of 9.
SOLUTION:
There are 36 outcomes in the sample space and four ways to get a sum of 9.
Pðsum of 9Þ¼
4
36
¼
1
9
, PðEÞ¼1 À
1
9
¼
8

9
: Hence,
odds of not getting a sum of 9 ¼
PðE Þ
1 À Pð
E Þ
¼
8
9
1 À
8
9
¼
8
9
1
9
¼
8
9
Ä
1
9
¼
8
9
1
Á
9
1

1
¼
8
1
The odds are 8 : 1.
If the odds in favor of an event occurring are A : B, then the odds against
the event occurring are B : A. For example, if the odds are 1 : 15 that an event
will occur, then the odds against the event occurring are 15 : 1.
Odds can also be expressed as
odds in favor ¼
number of outcomes in favor of the event
number of outcomes not in favor of the event
For example, if two coins are tossed, the odds in favor of getting two heads
were computed previously as 1 : 3. Notice that there is only one way to get
two heads (HH) and three ways of not getting two heads (HT, TH, TT);
hence the odds are 1 : 3.
CHAPTER 5 Odds and Expectation
79
When the probability of an event occurring is
1
2
, then the odds are 1 : 1.
In the realm of gambling, we say the odds are ‘‘even’’ and the chance of the
event is ‘‘fifty–fifty.’’ The game is said to be fair. Odds can be other numbers,
such as 2 : 5, 7 : 4, etc.
PRACTICE
1. When two dice are rolled, find the odds in favor of getting a sum
of 12.
2. When a single card is drawn from a deck of 52 cards, find the odds
against getting a diamond.

3. When three coins are tossed, find the odds in favor of getting two tails
and a head in any order.
4. When a single die is rolled, find the odds in favor of getting an even
number.
5. When two dice are rolled, find the odds against getting a sum of 7.
ANSWERS
1. There is only one way to get a sum of 12, and that is (6, 6). There are
36 outcomes in the sample space. Hence, P(sum of 12) ¼
1
36
: The
odds in favor are
1
36
1 À
1
36
¼
1
36
35
36
¼
1
36
Ä
35
36
¼
1

36
1
Á
36
1
35
¼
1
35
The odds are 1 : 35.
2. There are 13 diamonds in 52 cards; hence, Pð
^
) ¼
13
52
¼
1
4
:
Pð
^
Þ¼1 À
1
4
¼
3
4
: The odds against getting a diamond
3
4

1 À
3
4
¼
3
4
1
4
¼
3
4
Ä
1
4
¼
3
4
1
Á
4
1
1
¼
3
1
The odds are 3 : 1.
CHAPTER 5 Odds and Expectation
80
3. When three coins are tossed, there are three ways to get two tails and
a head. They are (TTH, THT, HTT), and there are eight outcomes in

the sample space. The odds in favor of getting two tails and a head are
3
8
1 À
3
8
¼
3
8
5
8
¼
3
8
Ä
5
8
¼
3
8
1
Á
8
1
5
¼
3
5
The odds are 3 : 5.
4. There are 3 even numbers out of 6 outcomes; hence, PðevenÞ¼

3
6
¼
1
2
:
The odds in favor of an even number are
1
2
1 À
1
2
¼
1
2
1
2
¼
1
2
Ä
1
2
¼
1
2
1
Á
2
1

1
¼
1
1
The odds are 1:1.
5. There are six ways to get a sum of 7 and 36 outcomes in the sample
space. Hence, P(sum of 7) ¼
6
36
¼
1
6
and P(not getting a sum of 7) ¼
1 À
1
6
¼
5
6
: The odds against getting a sum of 7 are
5
6
1 À
5
6
¼
5
6
1
6

¼
5
6
Ä
1
6
¼
5
6
1
Á
6
1
1
¼
5
1
The odds are 5 : 1.
Previously it was shown that given the probability of an event, the odds in
favor of the event occurring or the odds against the event occurring can be
found. The opposite is also true. If you know the odds in favor of an event
occurring or the odds against an event occurring, you can find the probability
of the event occurring. If the odds in favor of an event occurring are A : B,
then the probability that the event will occur is PðE Þ¼
A
AþB
:
If the odds against the event occurring are B : A, the probability that the
event will not occur is Pð
EÞ¼

B
BþA
:
Note: Recall that Pð
EÞ is the probability that the event will not occur or
the probability of the complement of event E.
CHAPTER 5 Odds and Expectation
81
EXAMPLE: If the odds that an event will occur are 5 : 7, find the probability
that the event will occur.
SOLUTION:
In this case, A ¼ 5 and B ¼ 7; hence, PðE Þ¼
A
AþB
¼
5
5 þ 7
¼
5
12
: Hence, the
probability the event will occur is
5
12
:
EXAMPLE: If the odds in favor of an event are 2 : 9, find the probability that
the event will not occur.
SOLUTION:
In this case, A ¼ 2 and B ¼ 9; hence, the probability that the event will not
occur is

Pð
EÞ¼
B
B þ A
¼
9
9 þ 2
¼
9
11
:
PRACTICE
1. Find the probability that an event E will occur if the odds are 5:2 in
favor of E.
2. Find the probability that an event E will not occur if the odds against
the event E are 4 : 1.
3. Find the probability that an event E will occur if the odds in favor of
the event are 2 : 3.
4. When two dice are rolled, the odds in favor of getting a sum of 8 are
5 : 31; find the probability of getting a sum of 8.
5. When a single card is drawn from a deck of 52 cards, the odds against
getting a face card are 10 : 3, find the probability of selecting a face
card.
ANSWERS
1. Let A ¼ 5 and B ¼ 2; then PðE Þ¼
5
5 þ 2
¼
5
7

:
2. Let B ¼ 4 and A ¼ 1; then Pð
EÞ¼
4
4 þ 1
¼
4
5
:
3. Let A ¼ 2 and B ¼ 3; then PðE Þ¼
2
2 þ 3
¼
2
5
:
CHAPTER 5 Odds and Expectation
82