4 ms·
Not necessarily - it could also mean 2,776 database searches returning just over an average 20 communications per violation. (Correct me if my math is wrong)
by csandreasen 13y ago
Not necessarily - it could also mean 2,776 database searches returning just over an average 20 communications per violation. (Correct me if my math is wrong)
(2,776 violations per year) / (.00001156666667 violations per search) = 240,000,000 total searches per year
(56,000 communications) / (2,776 searches) = average 20.172910663 records returned per search
(240,000,000 totals searches) * (20.172910663 records per search) = 4,841,498,559 total results per year
This ignores things like duplicate results and upper/lower bounds on the number of search results. Theoretically, the duplicate results would only factor into the non-violating result set, since any two queries that returned a US communication would both be a violation (although that might inflate the upper bound on search result set). That's also ignoring the possibility that a violating search could return both violating and non-violating results; in that case the average number of records per search would be higher, but not the total number of American communications, resulting in a larger set of total results.
Assuming that those number are somewhere in the ballpark, 56 thousand records of American communications out of at least 4.8 billion total records is a far cry from the 64% error rate that you proposed.