6 confusions: The bottom card must be a confusion, and I cannot at any point have 1 action, a watchtower in hand, and non-confusion cards left in my deck, nor can I have a hand without watchtowers and cards left in my deck.

Given 0 Watchtowers, the probability is 0.

Given 1 Watchtower, I must draw the watchtower and all necropolises in my starting hand. If you choose b elements from a set of size a, the probability that it contains c specific elements is (a - c) choose (b - c) divided by a choose b. a = 7 + n, b = 5, c = 1 + n. Now we have that 6 choose (4 - n) divided by (7 + n) choose 5 is the probability that we draw our deck with one watchtower, 6 confusions, and n necropolises.

Given 2 Watchtowers, there have to be 1 to 9 necropolises in my deck for there to be a positive probability, and well... this is not so simple. I'm not even sure what the optimal number of necros is, two? That's another interesting question. Gut says you want roughly the same number of necros as watchtowers, in general.

For 6 confusions, for the probability to be positive we have w - 1 <= n <= 5w - 1, w > 0.

Wow, this got off topic fast. I'll, uh, pass the problem back to heron now...