14.2 Odds
Consider again the small kidney stone data (Table 14.1).
For Method A, the sample contains 81 successes and 6 failures. Apart from proportions and percentages, another way to numerically summarise this information is to see that there are \(81\div 6 = 13.5\) times as many successes than failures in the sample.
In other words, for small kidney stones, the odds of success for Method A is 13.5 (in the sample). The sample odds is a statistic, and the population odds is a parameter.
Definition 14.3 (Odds) The odds are the proportion (or percentage, or number) of times that an event happens, divided by the proportion (or percentage, or number) of times that the event does not happen:
\[ \text{Odds} = \frac{\text{Proportion of times that something happens}} {\text{Proportion of times that something doesn't happen}}, \] or (equivalently)
\[ \text{Odds} = \frac{\text{Number of times that something happens}}{\text{Number of times that something doesn't happen}}. \] The odds show how many times an event happens compared to not happening. Alternatively, it is how many times the event happens for every 100 times that it does not happen.Notice that, when computing odds, we divide the relevant number by the remaining number, which is different than how percentages are computed.
percentages and proportions, we divide the relevant number by the total number relevant to the context.
Software usually works with odds rather than percentages (for good reasons that we will not delve into). However, understanding how software computes the odds is important.
Software usually computes odds as comparing either
- Row 1 to Row 2; or
- Column 1 to Column 2.
Example 14.1 (Interpreting odds) For the small kidney stone data, the odds of a success for Method A is \(81\div6 = 13.5\) (in the sample). This can be interpreted as:
- There are \(13.5\) times as many successes as failures (in the sample);
- There are \(13.5\times 100 = 1350\) successes for every 100 failures (in the sample).
Example 14.2 (Odds) Suppose that about 67% of students at a particular university were female. The population odds of finding a female is about \(67 / (100 - 67) = 2.03\): about twice as many females are students as non-females.
Suppose one tutorials had 18 females and 5 non-females. The sample odds of finding a female in this class is \(18/5 = 3.60\). Another classes had 16 females and 9 non-females. The sample odds of finding a female in this class is \(16/9 = 1.79\).Example 14.3 (Computing odds) Consider again the small kidney stone data (Table 14.1). The odds of a success using Method B can also be found (Table 14.1):
\[\begin{align*} &\text{Odds}(\text{Success with Method B})\\ = &\frac{\text{Number of successes for Method B}}{\text{Number of failures for Method B}} =\frac{234}{36} = 6.52. \end{align*}\] Working with the proportions (or percentages) (Table 14.2) rather than the numbers, the same value results:
\[\begin{align*} & \text{Odds}(\text{Success with Method B})\\ = &\frac{\text{Percentage of successes for Method B}}{\text{Percentage of failures for Method B}} =\frac{86.7}{13.3} = 6.52. \end{align*}\]When interpreting odds:
- When the odds are greater than one: the event is more likely to happen than to not happen.
- When the odds are equal to one: the event is just as likely to happen as it is to not happen.
- When the odds are less than one: the event is less likely to happen than to not happen.