Texas Hold'em - Bad Beat

I think you need to formulate the question more precisely for it to be answered. If the question is: what is the chance I win with AA against KdJd, then you're about 80% to win before the flop, on the flop your opponent has only 4 card that help (the four 10s) and you can still make a full house or four-of-a-kind, so you are about 88% at that stage; after the turn, you need to help your hand, and only 9 cards give you a full house, so you have a 9/44 = 20% chance to win.

If instead your question is: "I have AA here, and my opponent has two completely random cards, what is the probability I lose when the river looks like this?" then we know 7 cards, so there are 45 left. Your opponent's first card must be Kd or Jd, which is a 2/45 chance, and the remaining card needs to be exactly the other royal flush card, so a 1/44 chance. So there's a (2/45)(1/44) = 1/990 chance you lose to two random cards on the river.

That said, people don't normally play just any two random cards in poker, unless they don't care about money very much. Well, on Zynga they might, I guess. But because someone is more likely to play Kd Jd than they are to play 8c 3h, it's a bit more likely than 1/990 that you lose here. Of course it's still extremely improbable.

And if you're asking "what's the probability someone makes a Royal Flush and someone else makes four-of-a-kind on the same hand", that's a pretty complicated calculation, but the short answer is "very very very unlikely". You'd probably have a better chance of winning the lottery than of seeing that if you play just one hand of poker.