AntiDredge Strategy  from a Dredge Expert

I agree that if your mulligan strategy involves finding Leyline of the Void in your opening hand you probably want to run as many as possible
On the other hand if your mulligan strategy involves finding some piece of Dredge hate and you have (for example) 4 Cage and 2 Leyline, then you have 6 cards that can count as a success. The fact that the Cages aren't literally Leylines doesn't matter if you're mulliganing for them like Leylines

@baishuu said in AntiDredge Strategy  from a Dredge Expert:
I noticed Pitch Dredge jumped back and forth between Leyline of the Void and Ravenous Trap for the same reasons listed by everyone else. Leyline of the Void is solid in Dredge because it greatly reduces the number of creatures the opponent can use to break Bridge from Below. However, there were lists for a while that used Ravenous Trap since you could still make use of it if you drew it for the turn or off of a Bazaar of Baghdad activation. Both have their merits, but it seems Leyline of the Void is winning out more often.
I personally think Ravenous Trap was always just wrong in that deck. You don't actually draw that many cards (because you are using all your draws to dredge), and it's much weaker in that matchup than Leyline.

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[n1])
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 fromfirstprinciples 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.

One consideration with respect to the possibility of the “magic” k is that there are diminishing returns in terms of the the probability of seeing a leyline in your opener when you add more leylines to your deck. The magic k may be the point at which the increasing likelihood of drawing 2 outweighs the increasing likelihood of getting 1 in your opener.

Returns diminish monotonically absent selfsynergy
Even if you accept that you should Mulligan until you hit a Leyline, the most valuable Leyline is the 1st one you put in your deck  each subsequent Leyline (starting already with the 2nd) suffers from diminishing returns

@ajfirecracker it seems like you clearly do not understand the basic concept of probability

@aelien said in AntiDredge Strategy  from a Dredge Expert:
@ajfirecracker it seems like you clearly do not understand the basic concept of probability
If you explain how you think I'm wrong it's a conversation
If you just tell me I don't understand it's an insult

@ajfirecracker said in AntiDredge Strategy  from a Dredge Expert:
Even if you accept that you should Mulligan until you hit a Leyline, the most valuable Leyline is the 1st one you put in your deck  each subsequent Leyline (starting already with the 2nd) suffers from diminishing returns
Each Leyline above the first ups the probability of you having it in your opening hand, making your strategy of mulliganing to the first one incredibly bad if you only play 1 and way better if you play the maximum amount of leylines.
I really dont get why this is even an argument at all... You play 4 bazaars, not because you always want a second bazaar but because you really want to have a bazaar in your opening hand, you should be very familiar with this concept. 
I used a hypergeometric calculator to put some actual numbers to it.
If you have X Leylines and mulligan until you have at least one in your opening hand, your odds of success are below:
 0%
 38.6%
 62.6%
 77.4%
 86.5%
So, yes, more Leylines does indeed translate into a higher chance of success.
However, the most impactful Leyline is the first one, and if we subtract the prior odds of success to see the marginal impact of adding 1 more Leyline, we see that returns diminish smoothly:
 38.6%
 24.0%
 14.8%
 9.1%
So the most valuable Leyline is the first one, and each one after that helps increase the probability of "success", but not as much as the one before.

This post is deleted! 
@dshin said in AntiDredge 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[n1])
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 fromfirstprinciples 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 drawgountiltheydeckthemselves 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 strawman (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 nonMoxen 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].

@evouga I think you make a very good point  if you are oriented around some particular strategy, the cards that support that strategy are good and should generally be maximized, but there might be some other strategy that is better or comparable, but you can't necessarily take half and half and get a good result
However, I think you are defining "strategy" in too narrow a way if your choices are some number of a given named card. "Mulligan until you hit a piece of graveyard hate" is a fine strategy for decks that can protect hate and/or have some way of winning fairly quickly without using a large number of cards. This strategy can be executed with 2 Leyline and 4 Grafdigger's Cage just as well as 4 Leyline 2 Cage. Typically we see Workshops decks lean on Cage and Leyline as their main hate cards, and I don't see why they shouldn't prioritize Cage from that selection in order to enhance other matchups like Oath of Druids.

@evouga said in AntiDredge Strategy  from a Dredge Expert:
I again suggest thinking of Workshops as a second view on the same phenomenon. A blue deck with zero nonMoxen 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].
Agreed, there are good generic arguments that for Magic card X you should run either 0 or the maximum legal number of copies, and only if you have some special reason should you run some number in between (the most common special reason being you actually tested it and the diminishing returns make 1 or 2 or whatever the right number).
With Workshops I would guess the optimal number is somewhere in the 1012 range, but of course there's no way to get there from first principles  in theory 3 could be the optimal number, but we all know from experience that 3 is silly and you'd prefer to have more of that particular effect. The same applies here  there's no way to get from first principles to the optimal number being 4 or greater.
Moreover, the discussion has all been in the context of a particular deck. Your 0 or 4 division is generated by having different decks with different strategies. I interpret the "0 or 4" claim as applied to a particular deck with a particular set of 71 other cards at its disposal. I agree that generally you would want 0 or 4 because either Leyline meshes well into your strategy or some other card does, and you want the better of the two. However, I don't see any reason at all why Leyline should be an exception to the same principles that frequently push people to play 1, 2, or 3 copies of a card.