Stupid question from a math illiterate

[Kevin J]

In hold'em when you're dealt two non-paired cards, you have a 32.43% chance of flopping at least a pair. But if you're against 1 opponent (assuming he is not holding one of your ranks), is there now a 64.86% chance that ONE of you will flop a pair? And if so, does this mean that if you do NOT flop a pair, the chances are better than 50/50 that your opponent DID flop a pair? Or is it that if you don't flop a pair, then the chances are still 32.43% that your opponent did?

I know this is easy stuff for you guys, I just really suck at math and probabilities. Is there a FAQ guide anywhere that explains how to calculate probabilities as they relate to hold'em? I'm lazy and don't want to go through the process of learning all about how to calculate combinational probabilities in general. I just want to know how to do it as it pertains to hold'em. Thanks.