in reply to Re: [OT] The statistics of hashing.

in thread [OT] The statistics of hashing.

*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

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