Chapter 4.1, Problem 47EC

Odds Odds are used in gambling games to make them fair. For example, if you rolled a die and won every time you rolled a 6, then you would win on average once every 6 times. So that the game is fair, the odds of 5 to 1 are given. This means that if you bet $1 and won, you could win $5. On average, you would win $5 once in 6 rolls and lose $1 on the other 5 rolls—hence the term fair game.

In most gambling games, the odds given are not fair. For example, if the odds of winning are really 20 to 1, the house might offer 15 to 1 in order to make a profit.

Odds can be expressed as a fraction or as a ratio, such as $\frac{5}{1}$, 5:1, or 5 to 1. Odds are computed in favor of the event or against the event. The formulas for odds are

Odds in favor = $\frac{P(E)}{1-P(E)}$

Odds against = $\frac{P(E)}{1-P(E)}$

In the die example,

Odds in favor of a 6 = $1/6/5/6=1/5$ or 1:5

Odds against a 6 = $5/6/1/6=5/1$ or 5:1

Find the odds in favor of and against each event.

a. Rolling a die and getting a 2

b. Rolling a die and getting an even number

c. Drawing a card from a deck and getting a spade

d. Drawing a card and getting a red card

e. Drawing a card and getting a queen

f. Tossing two coins and getting two tails

g. Tossing two coins and getting exactly one tail

To determine

To obtain: The odds in favour of and against of the event rolling a die and getting a 2.

Answer to Problem 47EC

The odds in favour of and against of the event rolling a die and getting a 2 are 1:5 and 5:1

Explanation of Solution

Given info:

A single dice is rolled.

Calculation:

On rolling a single die has 6 different outcomes. That is, the outcomes are ‘1, 2, 3, 4, 5, and 6’.

The formula to obtain odds in favour of an event rolling a die and getting a 2 is,

$\text{Odds in favor}=\frac{P\left(E\right)}{1-P\left(E\right)}$

Here, the event E is defined as rolling a die and getting 2. Hence, the favourable outcomes for getting 2 is ‘2’. That is, there is 1 outcome for event E.

The probability of event E is,

$P\left(E\right)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes in the sample space}}$

Substitute 1 for ‘Number of favourable outcomes in E’ and 6 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{1}{6}$

Substitute $\frac{1}{6}$ for $P\left(E\right)$ in Odds in favor,

$\begin{array}{c}\text{Odds in favor}=\frac{\frac{1}{6}}{1-\frac{1}{6}}\\ =\frac{\frac{1}{6}}{\frac{5}{6}}\\ =\frac{1}{5}\end{array}$

Thus, the odds in favour of an event rolling a die and getting a 2 is1:5.

The formula to obtain odds against of an event rolling a die and getting a 2 is,

$\text{Odds against}=\frac{P\left(E\right)}{1-P\left(E\right)}$

Here, the event E is defined as rolling a die and not getting 2. Hence, the favourable outcomes for not getting 2 is ‘1, 3, 4, 5, 6’. That is, there are 5 outcomes for event E.

The probability of event E is,

$P\left(E\right)=\frac{\text{Number of not favourable outcomes}}{\text{Total number of outcomes in the sample space}}$

Substitute 5 for ‘Number of favourable outcomes in E’ and 6 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{5}{6}$

Substitute $\frac{5}{6}$ for $P\left(E\right)$ in Odds against,

$\begin{array}{c}\text{Odds against}=\frac{\frac{5}{6}}{1-\frac{5}{6}}\\ =\frac{\frac{5}{6}}{\frac{1}{6}}\\ =\frac{5}{1}\end{array}$

Thus, the odds against of an event rolling a die and getting a 2 is $5:1$.

To determine

To obtain: The odds in favour of and against of the event rolling a die and getting an even number.

Answer to Problem 47EC

The odds in favour of and against of the event rolling a die and getting an even number are 1:1 and 1:1

Explanation of Solution

Calculation:

On rolling a single die has 6 different outcomes. That is, the outcomes are ‘1, 2, 3, 4, 5, and 6’.

The formula to obtain odds in favour of an event rolling a die and getting an even number is,

$\text{Odds in favor}=\frac{P\left(E\right)}{1-P\left(E\right)}$

Here, the event E is defined as rolling a die and getting an even number. Hence, the favourable outcomes for getting even numbers are ‘0, 2, 6’. That is, there are 3 favourable outcomes for event E.

The probability of event E is,

$P\left(E\right)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes in the sample space}}$

Substitute 3 for ‘Number of favourable outcomes in E’ and 6 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{3}{6}$

Substitute $\frac{3}{6}$ for $P\left(E\right)$ in Odds in favor,

$\begin{array}{c}\text{Odds in favor}=\frac{\frac{3}{6}}{1-\frac{3}{6}}\\ =\frac{\frac{3}{6}}{\frac{3}{6}}\\ =\frac{1}{1}\end{array}$

Thus, the odds in favour of an event rolling a die and getting an even number is1:1.

The formula to obtain odds against of an event rolling a die and getting an even number is,

$\text{Odds against}=\frac{P\left(E\right)}{1-P\left(E\right)}$

Here, the event E is defined as rolling a die and not getting an even number. Hence, the favourable outcomes for not getting an even number are ‘1, 3, 5’. That is, there are 3 outcomes for event E.

The probability of event E is,

$P\left(E\right)=\frac{\text{Number of not favourable outcomes}}{\text{Total number of outcomes in the sample space}}$

Substitute 3 for ‘Number of favourable outcomes in E’ and 6 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{3}{6}$

Substitute $\frac{3}{6}$ for $P\left(E\right)$ in Odds against,

$\begin{array}{c}\text{Odds against}=\frac{\frac{3}{6}}{1-\frac{3}{6}}\\ =\frac{\frac{3}{6}}{\frac{3}{6}}\\ =\frac{1}{1}\end{array}$

Thus, the odds against of an event rolling a die and getting an even number is 1:1.

To determine

To obtain: The odds in favour of and against of the event getting spade.

Answer to Problem 47EC

The odds in favour of and against of the event getting spade are 1:3 and 3:1

Explanation of Solution

Calculation:

In ordinary deck of cards there are 4 suits. They are hearts, clubs, diamonds and spades. In each suite there are 13 cards. In 13 cards, nine cards are numbers from 2 to 10 and remaining four cards are king, queen, ace and jack cards.

The total number of outcomes is 52.

The formula to obtain odds in favour of an event getting spade is,

$\text{Odds in favor}=\frac{P\left(E\right)}{1-P\left(E\right)}$

Here, the event E is defined as drawing a card from a deck and getting spade. Hence, the favourable outcomes for getting spade cards is ‘13’. That is, there is 13 outcomes for event E.

The probability of event E is,

$P\left(E\right)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes in the sample space}}$

Substitute 13 for ‘Number of favourable outcomes in E’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{13}{52}$

Substitute $\frac{13}{52}$ for $P\left(E\right)$ in Odds in favor,

$\begin{array}{c}\text{Odds in favor}=\frac{\frac{13}{52}}{1-\frac{13}{52}}\\ =\frac{\frac{13}{52}}{\frac{39}{52}}\\ =\frac{1}{3}\end{array}$

Thus, the odds in favour of an event getting a spade card is 1:3.

The formula to obtain odds against of an event getting spade is,

$\text{Odds against}=\frac{P\left(E\right)}{1-P\left(E\right)}$

Here, the event E is defined as not getting a spade. Hence, the favourable number of outcomes for not getting a spade is ‘39’. That is, there are 39 outcomes for event E.

The probability of event E is,

$P\left(E\right)=\frac{\text{Number of not favourable outcomes}}{\text{Total number of outcomes in the sample space}}$

Substitute 39 for ‘Number of favourable outcomes in E’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{39}{52}$

Substitute $\frac{39}{52}$ for $P\left(E\right)$ in Odds against,

$\begin{array}{c}\text{Odds against}=\frac{\frac{39}{52}}{1-\frac{39}{52}}\\ =\frac{\frac{39}{52}}{\frac{13}{52}}\\ =\frac{3}{1}\end{array}$

Thus, the odds against of an event getting a spade is 1:3.

To determine

To obtain: The odds in favour of and against of the event getting a red card.

Answer to Problem 47EC

The odds in favour of and against of the event getting red card are  1:1 and 1:1

Explanation of Solution

Calculation:

In ordinary deck of cards there are 4 suits. They are hearts, clubs, diamonds and spades. In each suite there are 13 cards. In 13 cards, nine cards are numbers from 2 to 10 and remaining four cards are king, queen, ace and jack cards.

The total number of outcomes is 52.

The formula to obtain odds in favour of an event getting a red card is,

$\text{Odds in favor}=\frac{P\left(E\right)}{1-P\left(E\right)}$

Here, the event E is defined as drawing a card from a deck and getting a red card. Hence, the favourable outcomes for getting red cards is ‘26’. That is, there are 26 outcomes for event E.

The probability of event E is,

$P\left(E\right)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes in the sample space}}$

Substitute 26 for ‘Number of favourable outcomes in E’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{26}{52}$

Substitute $\frac{26}{52}$ for $P\left(E\right)$ in Odds in favor,

$\begin{array}{c}\text{Odds in favor}=\frac{\frac{26}{52}}{1-\frac{26}{52}}\\ =\frac{\frac{26}{52}}{\frac{26}{52}}\\ =\frac{1}{1}\end{array}$

Thus, the odds in favour of an event getting a red card is 1:1.

The formula to obtain odds against of an event getting a red card is,

$\text{Odds against}=\frac{P\left(E\right)}{1-P\left(E\right)}$

Here, the event E is defined as not getting a red card. Hence, the favourable number of outcomes for not getting a red card is ‘26’. That is, there are 26 outcomes for event E.

The probability of event E is,

$P\left(E\right)=\frac{\text{Number of not favourable outcomes}}{\text{Total number of outcomes in the sample space}}$

Substitute 26 for ‘Number of favourable outcomes in E’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{26}{52}$

Substitute $\frac{26}{52}$ for $P\left(E\right)$ in Odds against,

$\begin{array}{c}\text{Odds against}=\frac{\frac{26}{52}}{1-\frac{26}{52}}\\ =\frac{\frac{26}{52}}{\frac{26}{52}}\\ =\frac{1}{1}\end{array}$

Thus, the odds against of an event getting a red card is 1:1.

To determine

To obtain: The odds in favour of and against of the event getting a queen.

Answer to Problem 47EC

The odds in favour of and against of the event getting a queen are 1:12 and 12:1

Explanation of Solution

Calculation:

In ordinary deck of cards there are 4 suits. They are hearts, clubs, diamonds and spades. In each suite there are 13 cards. In 13 cards, nine cards are numbers from 2 to 10 and remaining four cards are king, queen, ace and jack cards.

The total number of outcomes is 52.

The formula to obtain odds in favour of an event getting a queen is,

$\text{Odds in favor}=\frac{P\left(E\right)}{1-P\left(E\right)}$

Here, the event E is defined as drawing a card from a deck and getting a queen. Hence, the favourable outcomes for getting queen cards is ‘4’. That is, there are 4 outcomes for event E.

The probability of event E is,

$P\left(E\right)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes in the sample space}}$

Substitute 4 for ‘Number of favourable outcomes in E’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{4}{52}$

Substitute $\frac{4}{52}$ for $P\left(E\right)$ in Odds in favor,

$\begin{array}{c}\text{Odds in favor}=\frac{\frac{4}{52}}{1-\frac{4}{52}}\\ =\frac{\frac{4}{52}}{\frac{48}{52}}\\ =\frac{1}{12}\end{array}$

Thus, the odds in favour of an event getting a queen is 1:12.

The formula to obtain odds against of an event getting a queen is,

$\text{Odds against}=\frac{P\left(E\right)}{1-P\left(E\right)}$

Here, the event E is defined as not getting a queen. Hence, the favourable number of outcomes for not getting a queen is ‘48’. That is, there are 48 outcomes for event E.

The probability of event E is,

$P\left(E\right)=\frac{\text{Number of not favourable outcomes}}{\text{Total number of outcomes in the sample space}}$

Substitute 48 for ‘Number of favourable outcomes in E’ and 52 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{48}{52}$

Substitute $\frac{48}{52}$ for $P\left(E\right)$ in Odds against,

$\begin{array}{c}\text{Odds against}=\frac{\frac{48}{52}}{1-\frac{48}{52}}\\ =\frac{\frac{48}{52}}{\frac{4}{52}}\\ =\frac{12}{1}\end{array}$

Thus, the odds against of an event getting a queen is 12:1.

To determine

To obtain: The odds in favour of and against of the event tossing two coins and getting two tails.

Answer to Problem 47EC

Then odds in favour of and against of the event tossing two coins and getting two tails are 1:3 and 3:1

Explanation of Solution

Given info:

Tossing two coins.

Calculation:

The possible outcomes for tossing two coins is,

$\left\{HH,HT,TH,TT\right\}$

The total number of outcomes is 4.

The formula to obtain odds in favour of an event tossing two coins and getting two tails is,

$\text{Odds in favor}=\frac{P\left(E\right)}{1-P\left(E\right)}$

Here, the event E is defined as event tossing two coins and getting two tails. Hence, the favourable outcomes for getting two tails is ‘1’. That is, there is 1 outcomes for event E.

The probability of event E is,

$P\left(E\right)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes in the sample space}}$

Substitute 1 for ‘Number of favourable outcomes in E’ and 4 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{1}{4}$

Substitute $\frac{1}{4}$ for $P\left(E\right)$ in Odds in favor,

$\begin{array}{c}\text{Odds in favor}=\frac{\frac{1}{4}}{1-\frac{1}{4}}\\ =\frac{\frac{1}{4}}{\frac{3}{4}}\\ =\frac{1}{3}\end{array}$

Thus, the odds in favour of an event tossing two coins and getting two tails is 1:3.

The formula to obtain odds against of an event tossing two coins and getting two tails is,

$\text{Odds against}=\frac{P\left(E\right)}{1-P\left(E\right)}$

Here, the event E is defined as event tossing two coins and not getting two tails. Hence, the favourable number of outcomes for not getting a two tails is ‘3’. That is, there are 3 outcomes for event E.

The probability of event E is,

$P\left(E\right)=\frac{\text{Number of not favourable outcomes}}{\text{Total number of outcomes in the sample space}}$

Substitute 3 for ‘Number of favourable outcomes in E’ and 4 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{3}{4}$

Substitute $\frac{3}{4}$ for $P\left(E\right)$ in Odds against,

$\begin{array}{c}\text{Odds against}=\frac{\frac{3}{4}}{1-\frac{3}{4}}\\ =\frac{\frac{3}{4}}{\frac{1}{4}}\\ =\frac{3}{1}\end{array}$

Thus, the odds against of an event tossing two coins and getting two tails is 3:1.

To determine

To obtain: The odds in favour of and against of the event tossing two coins and getting exactly one tail.

Answer to Problem 47EC

Then odds in favour of and against of the event tossing two coins and getting exactly one tail are 1:1 and 1:1

Explanation of Solution

Given info:

Tossing two coins.

Calculation:

The possible outcomes for tossing two coins is,

$\left\{HH,HT,TH,TT\right\}$

The total number of outcomes is 4.

The formula to obtain odds in favour of an event tossing two coins and getting two tails is,

$\text{Odds in favor}=\frac{P\left(E\right)}{1-P\left(E\right)}$

Here, the event E is defined as event tossing two coins and getting exactly one tail. Hence, the favourable outcomes for getting exactly one tail is ‘2’. That is, there are 2 outcomes for event E.

The probability of event E is,

$P\left(E\right)=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes in the sample space}}$

Substitute 2 for ‘Number of favourable outcomes in E’ and 4 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{2}{4}$

Substitute $\frac{2}{4}$ for $P\left(E\right)$ in Odds in favor,

$\begin{array}{c}\text{Odds in favor}=\frac{\frac{2}{4}}{1-\frac{2}{4}}\\ =\frac{\frac{2}{4}}{\frac{2}{4}}\\ =\frac{1}{1}\end{array}$

Thus, the odds in favour of an event tossing two coins and getting exactly one tail is 1:1.

The formula to obtain odds against of an event tossing two coins and getting exactly one tail is,

$\text{Odds against}=\frac{P\left(E\right)}{1-P\left(E\right)}$

Here, the event E is defined as event tossing two coins and not getting exactly one tail. Hence, the favourable number of outcomes for not getting exactly one tail is ‘2’. That is, there are 2 outcomes for event E.

The probability of event E is,

$P\left(E\right)=\frac{\text{Number of not favourable outcomes}}{\text{Total number of outcomes in the sample space}}$

Substitute 2 for ‘Number of favourable outcomes in E’ and 4 for ‘Total number of outcomes in the sample space’,

$P\left(E\right)=\frac{2}{4}$

Substitute $\frac{2}{4}$ for $P\left(E\right)$ in Odds against,

$\begin{array}{c}\text{Odds against}=\frac{\frac{2}{4}}{1-\frac{2}{4}}\\ =\frac{\frac{2}{4}}{\frac{2}{4}}\\ =\frac{1}{1}\end{array}$

Thus, the odds against of an event tossing two coins and getting exactly 2 tails is 1:1.