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
Re^3: [OT] The statistics of hashing.
by BrowserUk (Pope) on Apr 01, 2012 at 09:04 UTC

Thanks syphilis. Your calculations make sense to me. But I'm not sure that it gels with the actual data?
Assuming I've coded your formula correctly (maybe not!), then using 10 hashes & vectors, I get the odds of having seen a dup after 1e9 inserts as (1  ((4294967295/4294967296)**1e9) ) **10 := 0.00000014949378123.
By that point I had actually seen 13 collisions:
And looking at the figure for 4e9 := 0.00667569553892502, by which time the 10 vectors will be almost fully populated, it looks way too low to me?
I would have expected that calculation (for N=4e9) to have yielded odds of almost 1?
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I get the odds of having seen a dup after 1e9 inserts as (1  ((4294967295/4294967296)**1e9) ) **10 := 0.00000014949378123
That's not the probability of "having seen a dup", but the probability that the 1000000001st random selection of 10 numbers would be reported as a dup (ie the probability that each of the relevant bits in all 10 bit vectors was already set for that 1000000001st random selection of the 10 numbers).
If I get a chance I'll try to work out the probability of "having seen a dup" in the first 1e9 iterations. (But, judging by some of the figures being bandied about, it probably has little bearing on this actual case where we're looking at MD5 hashes instead of random selections.)
Cheers, Rob
 [reply] 

If I get a chance I'll try to work out the probability of "having seen a dup" in the first 1e9 iterations.
Thank you if you do find that time at some point. If you could also show your workings, I might be able to wrap my muddled brain around it and stand a chance of reapplying your derivation.
it probably has little bearing on this actual case where we're looking at MD5 hashes instead of random selections.
Whilst the MD5 hash is known to be imperfect, it has been well analysed and has been demonstrated to produce a close to perfect random distribution of bits from the input data by several practical measures.
Eg. If you take any single input text, and produce its MD5; and then vary a single bit in the input and produce a new MD5, then  on average  half of the bits in the new MD5 will have changed relative to the original.
And if you repeat that process  varying a single bit in the input and then compare the original and new MD5s  the average number of bits changed in the outputs will tend towards 1/2. That is about as good a measure of randomness as you can hope for from a deterministic process.
I am aware of the limitations on the distribution of the hashes when derived from a nonfull spectrum of inputs; but given that 2**32 (the maximum capacity of the vectors), represent such a minuscule proportion of the 1e44 possible inputs, I'd have to be extremely unlucky in my random selection from the total inputs for the hashing bias to actually have a measurable affect upon the probabilities of false positives.
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