It has 53 bits of precision because there's an implied leading 1 bit that's not stored.

log10(2^53) ~ 15.95, not quite 16. So not exactly 16 decimal digits of precision, but damn close. What's important here is that it's more than 15.

$ perl -e'
   printf "%.16f\n", 0.1234567890123456;
   printf "%.16f\n", 0.1234567890123457;
'
0.1234567890123456
0.1234567890123457
[download]