> an algorithm that takes 1/(randFloat()) time to insert an element to the data structure where randFloat() returns any possible floating point number with equal distribution, is still O(1) algorithm, even though there is no upper limit on how long it can take to execute.
According to the definition, if 1/(randFloat()) = O(1) then there must be a constant M that satisfies 1/(randFloat()) <= M * 1. But according to your own words there's no upper limit to 1/(randFloat()), therefore there's no such constant M, therefore it's not O(1).
(In practice on most systems there would be an upper limit since a float can't be infinitely close to zero, but let's act as if that wasn't the case.)
