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?
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3$\begingroup$ You are not missing anything; your intuition is leading your astray. The mean and the median are not necessarily the same for asymmetric distributions. See Problem 4.8 in Stirzaker's Elementary Probability which describes how the Royal Oak lottery of 18th century Britain exploited this fact. $\endgroup$– Dilip SarwateMar 28, 2012 at 2:22
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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.
