At some point, I really think you'd be better off just doing the calculations for any specific hand in your head. It's really not that difficult, but it requires some skill in analyzing the "texture" of the board (if you're talking about hold 'em, which you mostly likely do).

You hold [Js] [10s] on a flop of

[Kh] [10d] [7s]

You suspect that your opponent has AK. What's the odds of you beating him?

The first step to calculating this is **counting your outs**. Let's start with the easy ones:

Any ten or jack will give you a stronger hand than he has. There are 5 of those cards in the deck left (2 tens, 3 jacks), so you have five outs there. Furthermore, you have a backdoor flush and two separate backdoor straight draws. There are mnemonic estimation tricks for these kind of draws, and they're usually calculated to be worth roughly one out each (or slightly more).

Edit: ... accidently hit "post" button, didn't mean to. Continuing:

So, if we estimate your outs to be 7 (5 for the two pair/trip draw, and another two for the two backdoor draws) that means that you expect that 7 cards of the remaining 47 (or 45, but let's not go into that now, it doesn't really make a big difference) will win you the hand. So what then?

Your odds are 40-7. 40 of the cards will lose you the pot, 7 of them will win it for you. 40-7 is roughly 6-1, so you need pot odds better than 6-1 to continue.

If you want percentages specifically, you calculate 7/47 = 14.8%. Picking up the calculator is usually not fast enough, so the mnemonic trick is to think of the deck as containing 50 cards (close enough) and then every out will increase your chance of winning by 2% (since 1 card is 2% of 50). 7 outs = 7*2 = 14%.

If this seems like a lot to calculate, don't worry: It doesn't take long to get a hang of. Once you start just looking at each out as 2% of a chance to improve, the difficulty is just properly finding your outs. This is the harder part of it, but wasn't really what you asked. We could talk about that too, though, if you want.