6 months is 26 weeks, or 52 draws.
There are 49 balls, unless I'm mistaken, and each draw includes one bonus ball. All draws are unrelated. I'm going to assume all 6 random numbers are different and the order doesn't matter - because I assume that's the case for a valid ticket? I don't do lottery tickets.
The probability that your first number never appears is (48/49)^52
The probability that your first number did show at some point then is is (1-(48/49)^52)
Given that your first number has shown up, the probability you never see your second is (48/49)^51. Similarly the probability you do is (1-(48/49)^51).
The probability you see both is then (1-(48/49)^52)*(1-(48/49)^51).
You can see then that the probability of getting all six is:
Product from n=47 to 52 of (1-(48/49)^n)
Which is about 1 in 15 .
No doubt someone with a better grasp of probability will be along shortly to point out the errors in my maths...