Correlation of handedness to sexual orientation? There have been studies done that report that heterosexual individuals are somewhat more likely to be right-handed than homosexual individuals.  Does it necessarily follow that left-handed individuals are more likely to be homosexual than right-handed individuals?  I feel like this is a mathematical concept I can't place my finger on right now, and could be solved with a formal proof.
Wikipedia article on this topic
Edit: I think this is might be the method.
Assume individuals can only be either heterosexual (group $A$) or homosexual (group $B$), and can be either right-handed (group $R$) or left-handed (group $L$).  Given, 
$P(A,R) > P(B,R)$
$\because P(A,R)+P(A,L)=1, P(B,R)+P(B,L)=1$
Combine the two equations,
$P(A,R)+P(A,L)=P(B,R)+P(B,L)$
move terms,
$P(A,R)-P(B,R)=P(B,L)-P(A,L)>0$
$\therefore P(B,L)>P(A,L)$
Therefore, a person who is left-handed is more likely to be homosexual than (fixed) a right-handed person is.  
Is this correct, based on the (obviously) simplified assumptions?
 A: Your notation is quite nonstandard, but your idea is correct as long as you don't mean the last statement as written. The correct conclusion is that "a person who is left-handed is more likely to be homosexual than a right-handed person is," not that a majority of people who are left-handed are homosexual. You are writing $P(A,R)$ for "the probability of $R$ given $A$." Usually that is written $P(R|A)$. The usual interpretation of the symbols $P(A,R)$ would be the probability that both $A$ and $R$ occur.
Another way to derive the same implication is Bayes' Theorem. This says $P(R|A) = \frac{P(A|R)P(R)}{P(A)}$, which is equivalent to $\frac{P(R|A)}{P(R)} = \frac{P(A|R)}{P(A)}$. That the left hand side is greater than $1$ means $P(R|A) \gt P(R)$. In words, if you are given $A$ then $R$ is more likely. The left hand side is greater than $1$ means that the right hand side is greater than $1$, which means that if you are given $R$ then $A$ is more likely than without that information.
To connect this with $P(A|L)$, note that $P(A)$ is a weighted average of $P(A|L)$ and $P(A|R)$. $P(A) = P(A \cap L) + P(A \cap R) =P(A|L)P(L) + P(A|R)P(R)$. So, if $P(A|R) \gt P(A)$, then $P(A|L) \lt P(A)$, and $P(A|L) \lt P(A|R)$.
A: Given the enormity of the mistake, to understand better where it lies, I think it will be easier to use figures instead of formulae. 
Imagine a population of 100 individuals, with 2 left-handed homosexual, 8 right-handed homosexual, 9 left-handed heterosexuals, 81 right-handed heterosexuals.
There are 20% of left-handed individuals among homosexuals, and 10% only among heterosexual, so "heterosexual individuals are somewhat more likely to be right-handed than homosexual individuals". However pick a random left-handed individual : the probability of this individual to be homosexual is 2/11, that is 18%. He is more likely to be heterosexual than homosexual.
Additionally, there are 10% of homosexual in this population, and 9% in the right-handed subpopulation. This is all you can get: more homosexuals among left-handed persons that among right-handed persons. Cf Douglas’ answer for a formalization of this – or try it with other figure to get a better understanding of what is going on.
