go ahead... be a heretic  
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IIUC it's just (N/4294967296)**4
I believe the figure I should have provided is: (1  ((4294967295/4294967296) ** N)) ** 4 And if that's not right, I give up. Update: The logic is simple. (Danger, Will Robinson ;) If you're picking numbers at random in the range (1 .. 4294967296), then the probability that the N+1th pick has already come up is: 1 minus the probability that it has *not* already come up and the probability that it has *not* already come up is (4294967295/4294967296) ** N So that gives us the 1  ((4294967295/4294967296) ** N) But because we're looking for the case where all *4* numbers have already come up, that probability needs to be raised to the 4th power ... which yields the final figure. Cheers, Rob In reply to Re^2: [OT] The statistics of hashing.
by syphilis

