Probability of hash collision in the case of two parallel hashes
You can think of concatenating the two hash values and treating the result as a single hash.  Then you have $n$ records, each of which has a single hash.  The probability of a collision among $n$ hashes is roughly $n^2/2^{b+1}$, if the hash outputs a $b$-bit value.  In your case if each of the two individual hashes is 64 bits long, after concatenation you have a 128-bit hash for the record, so $b=128$.

How did I obtain the formula $n^2/2^{b+1}$?  It's an application of the birthday paradox, with $n$ people born on a planet that has $2^b$ days in the year (instead of 365 days).  Or, if you prefer an intuitive argument: think of all ways to pick a pair of $b$-bit hashes.  There are about $n^2/2$ pairs of hashes. For each one, there's a $1/2^b$ probability that those pair of hashes are equal (that they collide).  So the total probability that at least one of these collide is about $n^2/2$ times $1/2^b$.  This is a crude estimate that isn't totally accurate, but it turns out to be a reasonable approximation as long as $n^2/2$ is fairly small compared to $2^b$.