Quote:
Originally Posted by yusunchina
P(#heads_min = i) = 1000 * P( one gets i intersect the others get >= i)

This formula assumes that the events "one gets i intersect the others get >= i" (with the "one" fixed at one coin at a time) are disjoint, otherwise this would be an overcount. These events are in fact overlapping. For instance, the case where all the coins get exactly i (which is admittedly an extremely unlikely event but chosen here as a simple illustration) would be counted in all 1000 cases, although it happens only "once."