I'm interested in the math, if you care. I thought that it wouldn't change the likelihood for potions, but if i'm wrong i'd like to know the right answer.

If you have all sets, but no promos, then there are 200 kingdom cards, of which 10 have potion in its cost. If we ignore the bane (which will hardly matter for computations), there are 10 kingdom cards in each game. If we choose the kingdom cards (uniformly) at random, every card shows up on average in 1 in 20 games, i.e. with probability 0.05.

Now let's compute how often potion cost cards appear in your setup rule. If we use standard notation for probabilities (P[A] is probability of A and P[A|B] is probability of A given B)

P[no potion cost card in in the first eight cards] =

P[first card has no potion cost]*P[second card has no potion cost | first card has no potion cost]*P[third card has no potion cost | first two cards have no potion cost]*... =

(190/200)*(189/199)*(188/198)*...*(183/193) = 0.658.

So the probability that there is at least one potion cost card is in the first eight kingdom cards is 0.342. According to the setup rules, in all such games there will be at least 3 potion cost cards, so there will be on average more than 3*0.342=1.02 potion cost cards in each game, which means that each potion cost card appears with probability at least 0.102, more than twice as much as what would happen when you choose cards randomly. The fact that I ignored in this computation the possibility that there are more than 3 cards with potion cost in the game doesn't matter, because that probability is very very small (I think it starts to matter in the 5th digit after the decimal point).

Similarly one can compute that the probability of a potion cost card in the first 4 cards is 0.187, which means that each potion cost card has a probability of 0.0560 of appearing in each game using "my" setup rule, which is just slightly more than the 0.05 when choosing randomly.