19.3: Odds

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Learning Objectives

Compute odds.
Determine odds from probabilities.
Determine probabilities from odds.

A particular lottery instant-win game has 2 million tickets available. Of these, 500,000 win a prize. This means there are 1,500,000 losing tickets. When evaluating the risk associated with a game like this, it can be useful to compare the number of ways to win to the number of ways to lose. In this case, we compare the 500,000 winning tickets to the 1,500,000 losing tickets. In other words, there are three losing tickets for every winning ticket. Comparisons of this type are the focus of this section.

Defining and Computing Odds

Definition: Odds for the Event

The ratio of the number of equally likely outcomes in an event $E$ to the number of equally likely outcomes in the event $E^c$ is called the odds for (or odds in favor of) the event $E$.

Definition: Odds against the Event

The opposite ratio (the number of outcomes in the event $E^c$ to the number of outcomes in the event $E$) is called the odds against the event $E$.

Both odds and probabilities are calculated as ratios. To avoid confusion, we will always use fractions, decimals, or percents for probabilities, and we’ll use colons (as in, $A:B$), or the form "$A~to~B$" to indicate odds. The rules for simplifying fractions apply to odds as well. Thus, the odds of winning a prize in the game described in the section opener are 500,000 : 1,500,000 = 1 : 3, and the odds against winning a prize are 3 : 1. These are often described in words as "the odds of winning are one to three in favor" or "the odds of winning are three to one against."

Some sources, particularly state lottery websites, use the terms “odds” and “probability” interchangeably. Never assume that the word “odds” is being used correctly. Compute one of the odds or probabilities yourself to verify how the term is being used!

Note

As with many mathematics concepts, sometimes popular media will confuse odds with probabilities. A key distinction is that when they are used correctly:

Probabilities are always given as a single number between 0 and 1, e.g., $0.23$, $\frac{1}{2}$, $0.19\%$, or even "1 in 10," which translates to $\frac{1}{10}$.
Odds are always provided as a ratio of two numbers, e.g., "1:20" or "2 to 3."

Notice that, while probabilities must always be between zero and one inclusive, odds can include any non-negative number, as we will see in the next example.

Example $\PageIndex{1}$: Computing Odds

If you roll a fair six-sided die, what are the odds for rolling a 5 or higher?
If you roll two fair six-sided dice, what are the odds against rolling a sum of 7?
If you draw a card at random from a standard deck, what are the odds for drawing a $\heartsuit$?
If you draw two cards at random from a standard deck, what are the odds against them both being $\spadesuit$?

Solution

The sample space for this experiment is {1, 2, 3, 4, 5, 6}. Two of those outcomes are in the event “roll a 5 or higher,” while four are not. So, the odds for rolling a 5 or higher are 2 to 4, which is also 1 to 2.
The sample space for this experiment is shown in the table below (Figure $\PageIndex{1}$). There are six outcomes in the event “roll a sum of 7,” and there are 30 outcomes not in the event. So, the odds against rolling a sum of 7 are $ 30 : 6 = 5 : 1$.

$A table with 6 rows and 6 columns. The columns represent the first die and are titled, 1, 2, 3, 4, 5, and 6. The rows represent the second die and are titled, 1, 2, 3, 4, 5, and 6. The data is as follows: Row 1: 2, 3, 4, 5, 6, 7. Row 2: 3, 4, 5, 6, 7, 8. Row 3: 4, 5, 6, 7, 8, 9. Row 4: 5, 6, 7, 8, 9, 10. Row 5: 6, 7, 8, 9, 10, 11. Row 6: 7, 8, 9, 10, 11, 12.$

Figure $\PageIndex{1}$
There are 13 hearts in a standard deck, and $52-13=39$ others. So, the odds in favor of drawing a $\heartsuit$ are $13:39=1:3$.
There are $_{13}C_{2}=78$ ways to draw two spades, and $_{52}C_{2}−78=1,248$ ways to draw two cards that are not both $\spadesuit$. So, the odds against drawing two spades are $1,248:78 = 16:1$.

Try It $\PageIndex{1}$

You roll a pair of four-sided dice with faces labeled 1 through 4.

What are the odds for rolling a sum greater than 3?
What are the odds against both dice giving the same number?

Answer

$13 : 3$
$12:4=3:1$

Odds as a Ratio of Probabilities

We can also think of odds as a ratio of probabilities. Consider again the instant-win game from the section opener, with 500,000 winning tickets out of 2,000,000 total tickets. If a player buys one ticket, the probability of winning is $\displaystyle\frac{500,000}{2,000,000}=\frac{1}{4}$, and the probability of losing is $\displaystyle 1-\frac{1}{4}=\frac{3}{4}$. Notice that the ratio of the probability of winning to the probability of losing is $\displaystyle \frac{1}{4}:\frac{3}{4}=1:3$, which matches the odds in favor of winning.

FORMULA: Odds

For an event $E$, odds for $E\ = n(E):n(E^c) =P(E): P(E^c) = P(E):(1-P(E))$, and

odds against $E =n(E^c):n(E) = P(E^c):P(E) = (1-P(E)):P(E)$.

We can use these formulas to convert probabilities to odds, and vice versa.

Example $\PageIndex{2}$: Converting Probabilities to Odds

Given the following probabilities of an event, find the corresponding odds for and odds against that event.

$P(E) = 35\%$.
$P(E) = 17\%$.

Solution

Using the formula, the odds for $E = P(E):(1-P(E)) = 0.35:(1-0.35) = 0.35:0.65 = 7:13$, and the odds against $E = 13:7.$
Using the formula again, the odds for $E = P(E):(1-P(E)) = 0.17:(1-0.17) = 0.17:0.83 \approx 1:4.88$, and the odds against $E = 83:17 \approx 4.88:1$.

Try It $\PageIndex{2}$

If the probability of an event $E$ is 80%, find the odds for and the odds against $E$.

Answer: The odds for $E$ are $4:1$ and the odds against $E$ are $1:4$.

Now, let’s convert odds to probabilities. Let’s say the odds for an event are $A:B$. By the formula above, we have $ A:B=P(E):(1−P(E))$. Converting to fractions and solving for $P(E)$, we obtain:

$A:B = P(E): 1−P(E) $

$\displaystyle \frac{A}{B} =\frac{P(E)}{1−P(E)}$

$ B\cdot P(E) = A\cdot(1-P(E)) $

$BP(E) = A-AP(E)$

$(A+B)P(E) = A$

$\displaystyle P(E) = \frac{A}{A+B} $

Let’s put this result in a formula we can use.

FORMULA: Odds to Probability

\[\text{If the odds in favor of } E \text{ are } A:B, \text{ then } P(E) = \frac{A}{A+B}\label{1}\].

Example $\PageIndex{3}$: Converting Odds to Probabilities

Find $P(E)$ for each of the following:

The odds of $E$ are 2:1 in favor.
The odds of $E$ are 6:1 against.

Solution

In the formula \ref{1}, $A=2$ and $B=1$, so we have $\displaystyle P(E) =\frac{2}{2+1}=\frac{2}{3}$.
The odds against the event are 6:1, so the odds for the event are 1:6. Therefore, $\displaystyle P(E)=\frac{1}{1+6}=\frac{1}{7}$.

Exercise $\PageIndex{3}$

Find $P(E)$

if the odds against $E$ are $15:1$.
if the odds in favor of $E$ are $2.5 :1$.

Answer

$\displaystyle P(E)=\frac{1}{16}$
$\displaystyle P(E)=\frac{2.5}{3.5}=\frac{5}{7}$

Check Your Understanding

For the following exercises, you are rolling a six-sided die with three orange faces, two green faces, and one blue face.
1. What are the odds in favor of rolling a green face?
2. What are the odds against rolling a blue face?
3. What are the odds in favor of rolling an orange face?
What are the odds in favor of an event with probability $\displaystyle\frac{3}{8}$?
What are the odds against an event with probability $\displaystyle\frac{2}{13}$?
What is the probability of an event with odds $9:4$ against?
What is the probability of an event with odds $5:7$ in favor?

Display Answers

1. $1:2$
2. $5:1$
3. $1:1$
$3:5$
$11:2$
$\displaystyle\frac{4}{13}$
$\displaystyle\frac{5}{12}$

Exercises

For the following exercises, find the probabilities of events with the given odds in favor.
a. $9:4\quad$ b. $2:3\quad$ c. $5:4\quad$ d. $1:50\quad$ e. $7:5\quad$ f. $1:7\quad$ g. $10:9$
For the following exercises, find the probabilities of events with the given odds against.
a. $1:8\quad$ b. $2:3\quad$ c. $3:2\quad$ d. $5:4\quad$ e. $1:50\quad$ f. $7:5\quad$ g. $1:7\quad$ h. $10:9$
In the following exercises, find (i) the odds in favor of the given events and (ii) the odds against them, based on the provided probabilities. Give your answers as ratios of whole numbers. If neither of those two numbers is 1, also express the odds as a ratio involving 1 and a number greater than or equal to 1 (for example, the odds $5:2$ and $3:8$ can be rewritten as $2.5:1$ and $1:2.67$, respectively).
a. $\displaystyle\frac{2}{7}\quad$ b. $\displaystyle\frac{1}{217}\quad$ c. $\displaystyle\frac{8}{9}\quad$ d. $\displaystyle\frac{3}{8}\quad$ e. $\displaystyle\frac{9}{25}\quad$ f. $\displaystyle\frac{6}{7}\quad$ g. $\displaystyle\frac{10}{13}\quad$ h. $\displaystyle\frac{8}{15}$
In the following exercises, cards are drawn from a deck containing only these 10 cards: $A\heartsuit, A\spadesuit, A\clubsuit, A\diamondsuit, K\spadesuit, K\clubsuit, Q\heartsuit, Q\spadesuit, J\heartsuit, J\spadesuit.$ Find the probability of each event $E$, the odds in favor of $E$, and the odds against $E$.
1. Drawing an ace.
2. Drawing a heart.
3. Drawing two spades without replacement.

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Definition: Odds for the Event

Definition: Odds against the Event

Note

Example \(\PageIndex{1}\): Computing Odds

Try It \(\PageIndex{1}\)

FORMULA: Odds

Example \(\PageIndex{2}\): Converting Probabilities to Odds

Try It \(\PageIndex{2}\)

FORMULA: Odds to Probability

Example \(\PageIndex{3}\): Converting Odds to Probabilities

Exercise \(\PageIndex{3}\)