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What are the odds?

We often hear the words "odds," "probability," and "chance," used interchangeably. The truth is, though, that while probability and chance are similar in meaning, odds is its own concept. It is not quite the same as the other two words.

Probability and odds both describe the likeliness of an event. However, they look at likeliness through a different lens.

So, what exactly are odds?

Note how the probability, $\frac{5}{11},$ and odds, $\frac{5}{6},$ give quite different values. Mixing these terms up could give pretty misleading information.

Our numerators are the same for calculating probability and odds; it's the denominators that differ. For odds, the denominator is no longer a total of all of the outcomes.

As you can see from our example, odds are written as a ratio of the number of ways the desired event could occur to the number of ways it could not occur.

As a result, the denominators for the formulas for odds and probability will be different.

Let's try another example, this time just calculating odds.

Odds in favor vs. odds against

In the previous examples, when we ask, "What are the odds of picking green," we are asking what the odds are in favor of green.

Outside betting, finding odds in favor is basically the default. It's usually what you're being asked.

In gambling, however, you're more likely to be presented with odds against a particular outcome.

Notice how the odds against picking a diamond is the inverse (flipped over) version of the odds in favor of picking a diamond.

The probability that the side face up shows 1 dot when rolled is $\frac{1}{6}$. What are the odds in favor of this happening? The odds in favor of an event is

$$\frac{\text{probability of the event}}{\text{probability of NOT the event}}.$$

In other words, if there are $n$ possible outcomes and the favorable outcome occurs $t$ times out of $n$, then the odds in favor of the event equals

$$\frac{t}{n - t}.$$

Thus, the probability of rolling a $1$ is $\frac{1}{6}$. The probability of rolling otherwise is $\frac{5}{6}$ (we can roll a $2, 3, 4, 5,$ or $6$). Thus the odds in favor of one dot showing up is $\frac{1}{5}$. The odds against one dot showing up is the reciprocal of this, or more generally, $$\frac{\text{probability of NOT the event}}{\text{probability of the event}}$$ which is $5$, or in words "five to one".

Now I'm going to let you try a problem on your own.

(Video) Finding the odds in favor and against

by larryschmidt

In this video, we are finding the odds in favor or against an event happening.

$\large{\text{Summary}}$

Event ($E$) Amount of favorable outcome ($m$) Amount of unfavorable outcome ($n$)

Odds in favor of event $E =\frac{m}{n}$ $\\$ Odds against event $E = \frac{n}{m}$

Example:

Lena and Scott are at the grand opening celebration of a supermarket. They each spin a wheel with $10$ equal sized slices. When the wheel is spun the arrow stops on a random slice. (You can ignore the borders).

If the arrow stops on a black or white slice, Lena wins a gift card. What are the odds against Lena winning a gift card?

Odds against event $E = \frac{n}{m}$ $\\$ $n=3$$\\$ $m=7$ $\\$ so odds against Lena winning $E = \frac{3}{7}$$\\$