This is an interesting game. I've started some calculations but it will be awhile before I can finish it.
This is what I have so far:
Probability of getting zero duplicates (all 18 numbers are distinct) = 4.6%. If this happens, the maximum you can win in the second tier is 6 of 18 for $7.
Probability of getting one pair = 19.1%. The maximum win in the second tier is 7 of 18 for $20.
Probability of getting two pairs = 30.1%. The maximum win in the second tier is 8 of 18 for $100.
The sum of the three above is 54%. The remaining calculations that make up the other 46% are difficult.
For the second tier prizes, I don't know why they call it "approximate probability". It's combinatorics. It's exact.
While it is true that your odds of winning a particular prize in the second tier depend on how many duplicates are on your ticket, the stated probability on their website is the one that applies before you buy your ticket. Once you buy your ticket, the probability of winning a particular prize in the second tier will depend on the number of duplicates.
Say you get two triples and three pairs (e.g. 1-2-3-4-5-6, 1-2-3-4-21-22, 1-2-3-5-31-32). I don't know what the odds are of that happening, but I suspect that your chances of winning a prize in the second tier will be much higher than the "before I bought a ticket" probability. If the winning numbers are 1,2,4,5,41,42, the main game wins 4 of 6 for $20 but the second tier wins 10 of 18 for $2,000. All the different "after I bought a ticket" probabilities will average out to the "before I bought a ticket" probability.
Think of it this way: suppose you have a game where you flip a fair coin in step one. If you get heads, you get to flip it again. A seconds heads is a win. Tails on the second flip or the first flip is a loser. Before the game starts, your chances of winning are 1 in 4. But after the first coin toss, your chances of winning change to either zero or 1/2, depending on the outcome of the first coin flip.
