Showing the calculation you described: If p is the value drawn for A, then each time B is drawn, the probability that B>A is (1-p), So, the chance that B is d

Showing the calculation you described: If p is the value drawn for A, then each time B is drawn, the probability that B>A is (1-p), So, the chance that B is drawn n times before being less than or equal to A is, p^(n-1) (1-p) (a geometric distribution). The expected number of draws is then (1/p) . Then, E[draws] = E[E[draws|A=p]] = \int_0^1 E[draws|A=p] dp = \int_0^1 (1/p) dp, which diverges to infinity (as you said). (I wasn’t doubting you, I just wanted to see the calculation.)