No, you would get it right once with probability approaching 1-1/e, or
about 63.212%. The probability of success on one try is 1/N, where N is
the number of possibilities, 2^32 in this case; the probability of failure
on one try is 1-1/N; the probability of failure on N tries is (1-1/N)^N,
which approaches 1/e as N approaches infinity, which means the probability
of success on N tries approaches 1-1/e. It's really quite a good
approximation, in this case, good to about ten digits, I think.
I just tried this in GNU bc:
scale=100
onetry=(2^32-1)/2^32
half=onetry^(2^16)
half^(2^16)
The result is the probability of failure.
Kragen