# 3.2: Odds

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

When we talk about probability of getting tails on a coin, we generally think of the ratio of the one outcome in the event "tails" to the two outcomes in the sample space. So, we say the probability of getting tails is $$\frac{1}{2}$$ or 50%.

However, you also may have heard phrases such as “fifty-fifty” or “1 to 1” to describe the chance of getting tails. These phrases mean that the coin is as likely to come up tails as heads (each event occurring 50% of the time). This latter way to describe the chance of getting tails is what we refer to as the odds.

The odds of an event is a ratio that compares two sets of outcomes – those in the event and those not in the event. The odds in favor of event $$E$$ is the ratio of the number of outcomes in event $$E$$ to the number of outcomes that are not in the event in event $$E$$. Recall that we call this second set the complement of $$E$$. So, when we speak of odds in favor of $$E$$, we are giving the ratio $$n(E)$$ to $$n(E^{\prime})$$.

For example, suppose we roll a regular six-sided die and define event $$E$$ to be "roll the number six." There is only one way to roll a six, and there are five ways not to roll a six. This ratio is 1 to 5. So, the odds in favor of rolling the number six is $$1:5$$.

There are two types of odds, odds in favor and odds against, so it is always important that we are careful to distinguish between them. Casinos, race tracks and other types of gambling usually state the odds against an event happening. If the odds in favor of rolling a six on a die is $$1:5$$, then the odds against rolling a six on a die is the reciprocal ratio $$5:1$$.

## Finding Odds from the Sample Space and from Probability

The formulas for odds in favor and odds against are shown below in two different ways. One way shows how to find odds if you know the number of outcomes in event $$E$$ and in the event $$E^{\prime}$$. The second way shows how to find odds if instead you know $$P(E)$$ and $$P(E \text{ does not occur})$$.

##### Odds

The odds in favor of event $$E$$ is given by

$$\text{odds in favor of } E = \dfrac{\text{number of ways for E to occur}}{\text{number of ways for E to not occur}}$$ OR $$\dfrac{P(E)}{P(E\text{ does not occur})}$$

Also, the odds against event $$E$$ is given by

$$\text{odds against } E = \dfrac{\text{number of ways for E to not occur}}{\text{number of ways for E to occur}}$$ OR $$\dfrac{P(E \text{ does not occur})}{P(E)}$$

It should be noted that odds always use whole numbers. You never include decimals or percents with odds. Also, we should always write odds in lowest terms using the greatest common divisor just as we would write a fraction. Let's explore odds using some examples.

##### Example 1

A single card is drawn from a well-shuffled deck of 52 cards. Find the odds in favor of "drawing a red eight."

### Solution

There are two red eights in the deck -- which means there are $$52-2=50$$ cards that are not red eights. We can find the odds in favor of this event by forming the ratio

$$\text{odds in favor of a red eight}= \dfrac{\text{number of ways to get a red eight}}{\text{number of ways not to get a red eight}} = \dfrac{2}{50} = \dfrac{1}{25}$$.

The odds in favor of getting a red eight is 1 to 25, or $$1:25$$.

## Try it Now 1

A bag of 30 candies contain 12 that are peppermint. What are the odds in favor of selecting peppermint?

$$2:3$$
##### Example 2

Many roulette wheels have slots numbered 0, 00, and 1 through 36. The slots numbered 0 and 00 are green. The even numbered slots are red and the odd numbered slots are black. The game is played by spinning the wheel one direction and rolling a marble around the outer edge the other direction. Players bet on which slot the marble will fall into. What are the odds in favor of the marble landing in a red slot?

### Solution

There are 38 slots in all. The slots 2, 4, 6, …, 36 are red so there are 18 red slots. The other 20 slots are not red.

We can find the odds in favor of the marble landing on red by forming the ratio

$$\text{odds in favor of red} = \dfrac{\text{number of ways to land on red}}{\text{number of ways not to land on red}} = \dfrac{18}{20} = \dfrac{9}{10}$$.

The odds in favor of the marble landing on red is 9 to 10, or $$9:10$$.

## Try it Now 2

On a roulette wheel, what are the odds in favor of

1. the marble landing on green?
2. the marble landing on a slot numbered with two digits?
1. $$1:18$$
2. $$14:5$$

In the previous two examples, it was easy to count the outcomes in the event and its complement to form the odds. Sometimes, you may not know the number of outcomes in the event but you know the probability instead. This is where the second formula is useful.

##### Example 3

The probability of an event is $$\frac{3}{8}$$. What are the odds in favor of this event?

### Solution

Here we know $$P(E)=\frac{3}{8}$$, so $$P(E \text{ does not occur}) = 1 - \frac{3}{8} =\frac{5}{8}$$. We can find the odds in favor of the event by forming the ratio

$$\text{odds in favor of E} = \dfrac{P(E)}{P(E\text{ does not occur})} = \dfrac{\frac{3}{8}}{\frac{5}{8}} = \frac{3}{\cancel{8}} \times \frac{\cancel{8}}{5} = \dfrac{3}{5}$$.

The odds in favor of the event is 3 to 5, or $$3:5$$.

##### Example 4

The probability that it snows tomorrow is 60%. What are the odds that it snows tomorrow?

### Solution

Here we know $$P(\text{snow})=60\%$$ or $$\frac{60}{100}$$. This means $$P(\text{no snow})= 1 - \frac{60}{100} =\frac{40}{100}$$. We can find the odds in favor of the event by forming the ratio

$$\text{odds in favor of snow}= \dfrac{P(\text{snow})}{P(\text{no snow})} = \dfrac{\frac{60}{100}}{\frac{40}{100}} = \frac{60}{\cancel{100}} \times \frac{\cancel{100}}{40} = \dfrac{60}{40} = \dfrac{3}{2}$$.

The odds in favor of snow is 3 to 2, or $$3:2$$.

## Try it Now 3

The probability that someone who takes a flu shot stills gets the flu is 12%. What are the odds in favor of getting the flu if you get the shot?

$$3:22$$

##### Example 5

Two fair dice are tossed and the sum is recorded. Find the odds against rolling a sum of nine.

### Solution

The event "roll a sum of nine" contains 4 outcomes: $$E = \{(3, 6), (4, 5), (5, 4), (6, 3)\}$$.

There are $$6 \times 6 =36$$ ways to roll two dice and four ways to roll a sum of nine. That means there are 32 ways to roll a sum that is not nine.

$$\text{odds against rolling a sum of nine} = \dfrac{\text{number of ways not to roll a sum of nine}}{\text{number of ways to roll a sum of nine}} = \dfrac{32}{4} = \dfrac{8}{1}$$

The odds against rolling a sum of nine is 8 to 1, or $$8:1$$.

## Try it Now 4

A bag contains 6 red marbles, 3 blue marbles, and 1 yellow marble.

1. What are the odds in favor of choosing a red marble?
2. What are the odds against choosing a blue marble?
1. $$3:2$$
2. $$7:3$$
##### Example 6

The probability that Stephan wins his race is $$\frac{2}{15}$$. What are the odds against Stephan winning his race?

### Solution

Here we know $$P(\text{Stephan wins})=\frac{2}{15}$$, so $$P(\text{Stephan doesn't win}) = 1 - \frac{2}{15} =\frac{13}{15}$$. We can find the odds against the event by forming the ratio

$$\text{odds against Stephan wining} = \dfrac{P(\text{Stephan doesn't win})}{P(\text{Stephan wins})}= \dfrac{\frac{13}{15}}{\frac{2}{15}} = \frac{13}{\cancel{15}} \times \frac{\cancel{15}}{2} = \dfrac{13}{2}$$.

The odds against Stephan winning his race is 13 to 2, or $$13:2$$.

## Try it Now 5

The probability winning a prize in a game is 0.15. What are the odds against winning a prize?

$$17:3$$

## Finding Probabilities from Odds

Sometimes we may want to express the probability of an event happening based on the odds for the event. Saying that the odds in favor of an event are 3 to 5 means that the event happens three times for every five times it does not happen. If we add up the possibilities of both we get a sum of $$3+5=8$$. So the event happens about three out of every eight times. We would say the probability is $$3/8$$.

In other words, the sum of the odds gives the denominator of the probability. The value used in the numerator of the probability fraction depends on whether you are finding the probability of the event happening or the probability of the event not happening.

This relationship is summarized in the box.

##### Going from Odds to Probability

If the odds in favor of event $$E$$ are $$a:b$$, then

$$P(E) = \dfrac{a}{a+b}$$ and $$P(E \text{ does not occur}) = \dfrac{b}{a+b}$$.

##### Example 7

A local little league baseball team is going to a tournament. The odds of the team winning the tournament are 3 to 7. Find the probability of the team winning the tournament.

### Solution

$$P(\text{winning}) = \dfrac{3}{3+7} = \dfrac{3}{10} = 0.3$$, or 30%.

##### Example 8

The odds against Mathemagic winning the horse race derby are $$5:9$$. What is the probability that Mathemagic wins the race?

### Solution

Be cautious. The odds are given as odds against winning, but the question asks for the probability of winning.

$$P(\text{winning}) = \dfrac{9}{5+9} = \dfrac{9}{14} \approx 0.6429$$, or 64.29%.

## Try it Now 5

Jenny estimates her odds of earning an A on the next test as $$7:18$$. What is Jenny's probability of earning an A?

$$\frac{7}{25} = 0.28$$, or 28%.