Probability when selecting from 2 populations with different medians Given differing median values for two populations, what is the probability that a randomly selected person from the higher-median population will have a higher value than someone from the lower-median population? (I think I need to assume the populations are of similar size)
 A: The median is not enough information. It could even be more likely for the woman's income to be higher.
Consider the following example.

This figure is an example histogram for the income of the two populations "men" (blue) and "women" (red). Clearly, the median of men is higher than the one of women, yet if you randomly choose a (man, woman) pair, it is more likely that the woman is having a higher income.
A: Your problem does not contain enough information.
In general, if you have 2 random variables $X$ and $Y$, corresponding to 2 independent observations of your distributions, you should compute $\mathbb{P}\left(X>Y\right)$. This depends on the distributions of $X$ and $Y$. In order to do compute that, you need to have either:

*

*Assumptions about these distributions. Ex: Pareto distribution with parameters $p_X$ and $p_Y$ for instance.

*Empirical distributions/samples for your populations.

Depending on what you have on hand, you can compute a simple analytical formula to solve your problem (only with "nice" distributions), or use simulations to estimate your probabilities.
EDIT following your comments and your new question:
I do not think that you will be able to get any reasonable estimate with only 2 medians. If you have $m_Y>m_X$ (where $m$ represents the median, the only estimate you will be able to get is $\mathbb{P}(Y>X)\geq\frac{1}{2}$ in my opinion. (for some distributions, @frank was able to get a counterexample) Take these 2 extreme examples:

*

*$X=m_X, Y=m_Y$, i.e., there is no variations, every man has the same salary, same for women. Then you will have $\mathbb{P}(Y>X)=1$.

*The standard deviations of the distributions are huge compared to the median, like 100 times the median (for example). Then, these variations in the individual distributions will overweight the difference between the medians, and you might get close to $\mathbb{P}(Y>X)=\frac{1}{2}$.

In general, remember that $\mathbb{P}(Y>X)=\mathbb{P}(Y-X>0)$. So you are interested in the distribution of $Y-X$. It is impossible to get information about this distribution without information about the individual distributions of $X$, $Y$, and their dependencies (no dependency in your case). Some tricks might exist with some assumptions/estimates of the standard deviation of the distributions, but that's it. You cannot build a good estimate without good input data. In your case, your data is insufficient in my opinion.
