# Probabilities vs. Odds Ratios

Suppose we know that the probability of a female getting into a program are $p=.7$ and $q=1-.7=.3$ for males. Then we know:

$$\mbox{odds}(\mbox{female}) = .7/.3 = 2.33333$$ $$\mbox{odds}(\mbox{male}) = .3/.7 = .42857$$

We could use this information to compute an odds ratio:

$$OR = 2.3333/.42857 = 5.44$$

Thus a female is 5.44 times more likely to get in.

But since the probability of an event is just $\frac{p}{p+q}$ the probability of a male getting in is 30%, while the probability of a female getting in is 70%. We might reason then that the probability of a female getting in is roughly double (2.333). What are we trying to accomplish with the odds ratio and why does it result in such a different answer from comparing probabilities?

• Why do the probably of a man getting in plus the probability of a woman getting in sum to one? – Neil G Jul 8 '15 at 4:15
• I am not sure if the OR should be interpreted as 'x times more/less likely of...' or as 'a x increase/decrease in the odds of...' – Enrique Jul 8 '15 at 5:15
• an odds ratio is literally a ratio of odds (for once a name of a statistic that makes sense...), so an odds ratio of 5.44 means that the odds of a female entering the program is 5.44 times the odds of a male entering that program, or equivalently, the odds of a female entering the program is $(5.44-1)\times100\%=444\%$ larger than the odds of a male entering the program. – Maarten Buis Jul 8 '15 at 7:12
• Suppose that a male and female always face off for a given spotc(unrealistic but that addresses your concern). – 114 Jul 8 '15 at 12:45

In an observational study, the odds ratio can be calculated either by conditioning on exposure ($E$ and its complement $E'$) or outcome ($C$ and its complement $C'$):

$\psi = \frac{P(C|E)/P(C'|E)}{P(C|E')/P(C'|E')} = \frac{P(E|C)/P(E'|C)}{P(E|C')/P(E'|C')}$

According to your observations, the binary outcome is whether subjects get into the program ($C$) or not ($C'$), and the exposure could be femininity ($E$) versus masculinity ($E'$).

The problem is that your data is ambiguous. You say:

Suppose we know that the probability of a female getting into a program are $p=0.7$

If this means $P(C|E)=0.7$ then clearly $P(C'|E)=1-P(C|E)=0.3$. But we have no information about $P(C|E')$ or $P(C'|E')=1-P(C|E')$. In other words, what's the "probability of a male getting into the program"? (It's not 0.3).

Likewise, if you interpret that as $P(E|C)=0.7$ then $P(E'|C)=1-P(E|C)=0.3$. But now you're missing $P(E|C')$ and $P(E'|C')=1-P(E|C')$.

So you need to ask yourself whether that $0.7$ number is the probability of getting into the program for females, or the probability of being a female for those who got into the program.

As for what the odds ratio is used for,

• like you imply, it measure relative odds in a single number that may be less than, equal to, or greater than 1 and so summarises the data well
• in can be calculated simply; for example, there's a simple formula for a $2\times2$ table of exposure counts versus outcome counts
• it is used in calculations when analysing matched pairs in observational studies
• it may be interpreted as the relative risk when certain outcomes (such as the incidence of lung cancer) are rare, and it approximates $\frac{P(C|E)}{P(C|E')}$ and there's probably some uses I haven't encountered yet.

I think your confusion here is that sex is one binary variable. It has two levels, male and female, but they are not separate variables. So there aren't two variables to calculate the odds ratio of.