Odds of a Perfect Board

[KegNog]

OK, I'm not the best at this stuff but I want to know exactly how to produce this result.

Playing a tournament and the following hand comes up.

2 players see the following board - k-8-5-8-5
Player 1 shows K8
Player 2 shows 55

My question is what are the odds of this exact happening to occur? Here's what I have thus far, tell me how off I am or where I'm missing/adding values. Ignoring suits :

Odds of being dealt 55 = C(4,2)/C(52,2) = 1/221
Odds of being dealt K8 = [C(4,1)*C(4,1)/C(52,2)] = 8/663
Odds of having 2 players of 10 playing a hand = C(10,2) = 45
(Not really certain where this is used but I know it is)

Now to the board; To get a flop of K85:
Odds of hitting a 5 = C(2,1) = 2
Odds of hitting a K = C(3,1) = 3
Odds of hitting a 8 = C(3,1) = 3
Total board possibilities = C(48,3) = 17296
...Odds of K85 flop = (2*3*3)/17296 = 9/8648

Turning an 8 = 1/45
Rivering a 5 = 1/44

Therefore...am I correct to multiply all these values together and come up with the probability being:

(45)(1/221)(8/663)(9/8648)(1/45)(1/44) = 1.2913 E^-9...which is pretty much impossible! LOL

I'm really just trying to figure out where I'm wrong here because I know I am somewhere, and what the correct answer would be especially after the flop. I'm ignoring a LOT of possible outcomes (for example turning a K OR 8 OR 5, etc).

Thanks!