Probabilty calculation of having 3 of a kind by river holding poker pairs?

Hi Everyone,
I need some help about calculating the probability of basic things in poker.
I have Poker Propheiser which is a free program to calculate certain probabilities with given hands and I also have and old version of Texas Calculatem which also works the same way.
In this program I gave pocket pairs (like QQ) and no flop. They calculate the probability of having a three of a kind by river as 11.75%. But when I want to calculate it by hand I find much higher value like %20.
My calculation is [2*C(4,49)]/C(5,50). Without multipling by two, 10% is also not the correct answer. Why do I do mistake? Why my approach is wrong? Can someone explain me how to calculate it the correct way?
Thanks in advance

Both are right because they are giving you answers to two different questions. The programs saying 11.75% is not a set by the river, it is the probability of flopping a set (8.5 to 1). The odds of having a set by the river when you start with a pocket pair is about 4 to 1. Actually it's just under 20% but that's close enough.

Do the math and see:
Flop Set = 1 - (48/50)*(47/49)*(46/48) = 11.76%
Set by river = 1 - (48/50)*(47/49)*(46/48)*(45/47)*(44*46) = 19.22%

I don't know the math but I just read this in Harrinton volume one. Suppose you have a pocket pair that misses the flop. With two more cards to come, your odds of hitting the set are 8.4%.

True. I didn't illustrate it above, but it works the same just that we're now saying what if we did see a flop, missed, and want a set by the river.

1 - (45/47)*(44*46) = 8.4%