@dshin said in Anti-Dredge Strategy - from a Dredge Expert:

A more rigorous take on this debate:

**Suppose you are allowed to run as many Leylines as you want.** Let p[n] be the probability of winning a tournament with optimal deck construction given the constraint that you must run exactly n Leyline’s.

For any integer n, let S(n) be the following assertion:

max(p[0], p[n]) > max(p[1], p[2], ..., p[n-1])

The “0 or 4” crowd is then asserting that S(4) is true and can be deduced from first principles. Let’s assume they are right.

Now if Leyline is playable, then S(75) is clearly false.

This implies that if you consider the statements S(4), S(5), S(6), ..., S(75), then there is some magic k for which S(k) can be proven to be true from first principles, while S(k+1) cannot.

What is this magic k, and what is so special about it, that allows you to make a from-first-principles argument for S(k) but not for S(k+1)?

If no such k can be identified, then the “0 or 4” crowd must be wrong. It may indeed be the case that 0 or 4 is better than 1, 2, or 3, but that fact cannot be deduced from first principles.

If your premise is

- you
**must** mulligan until you find a Leyline (even if you have zero Leylines in your deck) and you cannot attempt any other deck construction or mulligan strategy

then you and @ajfirecracker are entirely correct. The first Leyline most increases your chance of success, and each subsequent Leyline is weaker than the first (with 75 Leylines obviously useless, unless you're on the draw-go-until-they-deck-themselves plan).

I don't understand why people are so hung up on the (obviously absurd) premise I boxed above. The alternative to running 1 Leyline and mulliganing until you find it is **not** running 0 Leylines and mulliganing until you find it. It's building a completely different deck with a different strategy for defeating Dredge (such as cantripping into multiple copies of Priest or Rest in Peace.) In this sense a "Leyline mulligan" strategy with 4 Leylines can be more effective than a completely different "cantrip into hate" strategy with zero Leylines which is in turn more effective than a "Leyline mulligan" strategy with 3 Leylines.

This is the argument I and other "4 or 0" people are advancing. Not some straw-man (and incorrect) simplistic assertion about hypergeometric distributions. @dshin's calculations are completely valid but irrelevant to the argument I'm actually advancing.

I again suggest thinking of Workshops as a second view on the same phenomenon. A blue deck with zero non-Moxen artifacts is strictly worse if you add a random Workshop. Once you add the first Workshop and start adding expensive artifacts, you are pursuing a new strategy and now adding three more Workshops is optimal (with diminishing returns as you add more Workshops, with the tipping point well above the maximum allowed 4 copies). The vast majority of Vintage decks want 0 or 4 Workshops. This has nothing to do with p[n].