The number of getting two winning tickets out of six is 6! / (2! x 4!) = 6x5/2 = 15.
Consider the probability of getting one of those 15 outcomes - say just the first two are winners, and the final four are losers. The chances of that happening - remember you're choosing WITHOUT replacement, so the odds of picking a winning ticket change every time - are (4/12) x (3/11) x (8/10) x (7/9) x (6/8) x (5/7) = (1/33).
Now consider the probability of getting a different combination of winners and losers in which there are still just two winners - say the 2nd and 4th one are the winners. The odds of THAT happening are (8/12) x (4/11) x (7/10) x (3/9) x (6/8) x (5/7). The denominators are exactly the same as before, and the numerators have just all been shuffled around - so the product of them all is the same as before, namely (1/33).
Try any other combination, and you'll see that exactly the same thing happens again. So the answer should be the total number of ways of getting two winners out of six times the probability of any one of them happening, namely 15 x (1/33). So I reckon the answer is (15/33).