<snip>
> I.e. we've independently arrived at the same place. Yay!
>
Yay!
> It just took me a lot longer... :-)
>
> >> On the bright side, if the 10^-9 formula (Chris #1, Chris #2, Sage)
> >> are anywhere near correct, they indicate with a block size of 4K and
> >> 32-bit checksum, you'd need to read 5 * 10^21 bits, or 0.5 ZB, to get
> >> to a 1% chance of seeing unflagged bad data, e.g.:
> >>
> >> Chris #2 @ U=10^-15, C=32, D=(4 * 1024), N=(5 * 8 * 10^21)
> >> = 1 - (((2^-32) * ((1-(10^-15))^(4 * 1024)-1)+1) ^ ((5 * 8 * 10^21) / (4 *
> 1024)))
> >> = 0.009269991978615439689381062400448881380202158231615685460
> >> = 0.92%
>
> Just for completeness, using the corrected D = (4 * 8 * 1024):
>
> P(bad data) @ U=10^-15, C=32, D=(4 * 8 * 1024), N=(5 * 8 * 10^21)
> = 1 - (2^-C * (1-U)^D - 2^-C + 1) ^ (N / D)
> = 1 - (2^-32 * (1-(10^-15))^(4 * 8 * 1024) - 2^-32 + 1) ^ ((5 * 8 * 10^21) / (4 *
> 8 * 1024))
> = 0.009269991978483162962573463579660791470065102520727107106
> = 0.92%
>
> I.e. the same as before, up to 10^-12. Not surprising as the previous D =
> (1024 * 1024) example demonstrated it's relatively insensitive to the block
> size.
Where does 10^21 come into the equation? I thought we were dealing with 5PB.
>
> Chris
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