7 shuffles with 52 cards though, so I reckon anything less than 30 and your good with 4 or 5?
(In the same way it says that 2 decks would need to be shuffled 9 times)
If you're refer to the paper, I think 7 is too low, as this is only the 3/2log n-term, ignoring the \theta, which you also want to have to be greater than zero. And I don't think that number decreases much, as it's only logarithmically dependend on n, so 25 cards about one shuffle less than 50.
Table 3 gives numbers. What may rescue you with lower numbers is that we are not talking about 25 distinct cards, but usually many of them will be same.
Edit: If you think "real" shuffling will take to long and you want some preprocessing, do pile shuffling. That way you can't take advantage over "real" randomness because you can't* really influence the ....... wait ....
What about this: You play BMU, so every discard will consist of 6 cards. Now if your buy is always at the same position of this discarded cards, and you are pileshuffling on say 3 or 6 piles, you clumb all your buys together, which in the beginning will be the good cards (and in the end the bad ones). Assumming not perfect shuffling now this will result in a probably better deck than it would be without pile shuffling...