For a Bernoulli distribution with parameter p, the number of trials with a 50% probability of at least one success is about (1/p) * ln(2). But the expected value of the corresponding geometric distribution is 1/p. Intuitively I would expect them to be the same (1/p). What am I missing?
The mean tends to exceed the median when the distribution is right-skewed.
Indeed, mean-median is (when normalized appropriately) a common measure of skewness.