Paired door card in 7 card stud, 2 pair or trips?

[Precision1C]

In Roy West's "7 card stud" How to win at medium and lower limits" p 110, he states that when your opponent pairs his door card he is more likely to have 2 pair rather than trips. While I agree that at lower limits where people call the bring with complete junk that when someone pairs their door card they have 2 pair or less rather than trips but don't agree with this statement as written. Please check my math if you are into to probability.

Say your opponent has X X 7C 7D is he more likely to have exactly 2 pair or trips? My combo count for 2 pair is 6 for each pair other than 7's giving 72 combos and there are 48*2 combos giving trips so it looks like the odds of having trips is higher than 2 pair. However if you make the statement "When your opponent pairs his door card it is more likely to have 2 pair or less rather than trips or better it would be accurate even a tight game."

pair and 3-flush (in the suit of the door card) 66
pair and consecutive 3 straight 48
two pair 72
trips 96
quads 1

So there are 186 combos for 2 pair or less versus 97 for trips or better. The odds are even better at lower limits since people play an amazing assortment of crap hands : ). I am posting just to see if I am missing something and someone has a different angle making the statement by West correct.

[kurosh]

I don't understand your math but this is how I see it. There are 3 cards that can give you 2-pair. There are 2 cards that can give you trips. 3 > 2.