In no way do I actually believe this deck to be competitive, but this is my best attempt at making this work. The deck is premised on casting Fluctuator, cycling a bunch of cards and then playing Shadow of the Grave, getting them back and repeating until you win via Labratory Maniac.
4 Cloud of Faeries
1 Laboratory Maniac
4 Fluctuator
4 Enlightened Tutor
4 Shadow of the Grave
4 Lotus Petal
4 Dark Ritual
4 Forsake the Wordly
4 Miscalculation
4 Lingering Mirage
2 Rune of Protection: White
2 Rune of Protection: Red
2 Unearth
4 Blasted Landscape
4 Plains
4 Irrigated Farmland
2 Drifting Meadow
2 Remote Island
1 Polluted Mire
The best case scenario of this deck is:
Turn 1 - Land (6 cards in hand)
Turn 2 - Land, cast Fluctuator (5 cards in hand)
Turn 3 - Cycle 3 cards, should get about 2 extra cyclers. Cycle them too. Should get at least one more cycler. Cycle it. You should have cycled about 6 cyclers and have 4 non-cycling cards in hand. Cast Shadow of the Grave. Cycle the 6 cards you discarded. You should get about 3 extra cyclers. Cycle them. Maybe you'll get another cycler. Cycle it. You should have 7 non-cycling cards in your hand at this point and no mana open. Cast a lotus petal into a dark ritual into another Shadow of the grave that you should have drawn at this point. You at this point cycled 10 cards, cycle them all again. You should draw about 6 other cyclers, cycle them and get 3 more cylcers and then another. Cycle them. You should have about 15 non-cycling cards in your hand at this point and 1 mana open. Repeat the process and draw most of your deck. Using the extra mana from the dark rituals, cast another lotus petal into lab maniac and cycle to win. If you need more mana, you can cast unearth into cloud of faeries to generate net 1 mana.
It seems like you're a little overcommitted on using cycling as a discard outlet. How about Careful Study or Zombie Infestation?
I mean, those may be other ways to use Shadow of the Grave. . . but either of them is pointless in a Fluctuator deck. Really the only strength of this list is that a Fluctuator is now consistent. Before, cycling could lead to dead ends. Now, with Shadow, that doesn't happen as often.
While building this deck for the fastest speed might be a good deck-building exercise, the deck fizzles on itself still entirely too much. Perhaps having an approach that wins consistently on turn 4/5 or so might be better?
How could we approach this from a turn 4/5 point of view instead of a turn 2? It would obviously need some disruption in the maindeck, but I think this could be fun to have a "competitive" tier 2.5 deck. It could also be used to generate a lot of card advantage instead of going for the instant win. Maybe running 3-4 loam, +3-4 of this new card, +4 fluctuator.
Zvi's deck that won the PT back in Urza's block didn't really go off until turn 4, and that was the most successful fluctuator list to date.
-rob
Rune of Protection is rather nice. I would also add the blue one to get rid of delver, true name nemesis and leovold damage
Some cards that could be useful are Thran Turbine and Braid of Fire to help generate some mana for free cycling when you don't draw fluctuator. Not sure how good they would be, I like the first one as it helps a bit.
-rob
Isn't this an update to "Mr. Toad's Wild Ride"? Is the name change from the removal of Bloated Toad?
Never heard of it, but it's not that this deck is that original of an idea. It's a fluctuator deck, with shadow of the grave added.
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@ bruizar: I like the idea of adding more runes just to be annoying until you can combo out.
@ mistercakes: Currently, it is more of a turn 3 combo than a turn 2. But I agree, this deck can afford to slow down and include more protection.
Honestly, it's more of a mathematical problem than a deck.
Let's say we aim for a turn 4 win. That would mean that you're casting Fluctuator turn 3 and then turn 4 cycling into Shadow, casting Shadow, and then hopefully cycling into mana and another shadow, repeat until you win. How many cyclers do you need?
On average, you'd need to dig through 60/4=15 cards to find the next shadow, right?
Now, if r=(cyclers)/(total cards) then then you want 15/11 <= r /(1-r) for the 11 cards you start with to lead to 15 cyclers.
So r>=15/26 which works out to 35 or more cyclers.
That's a bit low, since you really expect to have to get to card 30 in order to hit the second shadow, then
r>=19/30 so 38 or more cyclers.
You can probably improve that a little with tutor effects or something like Noxious Revival.
Lets say you need to get to card 30 to get the 2nd Shadow, and you drew 10 cards, by turn 4, you'll need 20 cyclers. To get to 20 cyclers, that would be 2(X+Y)+Z = 20, where X is how many cycler's in your hand prior to starting cycling, Y is how many cyclers you drew from the X cyclers, X and Y are doubled due to Shadow, and Z is how many cyclers you drew from the X+Y cards after they were returned from being cycled. Each of them are based on percentages right. I.e, each of they is based on Cylers in the deck/Cards in the Deck. With A equaling Cyclers in the Deck, and without fancy geometric math in the above equation: X=10A/60, Y=XA/60 or A^2/360, and Z= 2(X+Y))A/60) or A((A/6)+(A^2/360))/30. All the way down it would look like: 2((10A/60)+(A^2/360))+ (A((A/6)+(A^2/360))/30)=20. Solving for A, I got that we only need about 27.934 cyclers.
What do you think? That checks out?
Ah, right, forgot about doubling up with the first shadow.
OK, so supposing you don't dig until the fluctuator turn, and you see D cards initially, and you don't play any of the cyclers.
On the first run you get to card D/(1-r) and want to go at least 15 deep to find the first shadow, so
D(1/1-r)>15
For D=10 that works out to r>1/3 so 20 cyclers. That's pretty reasonable.
The second run gets you D(1/1-r) + Dr(1/1-r)^2 deep and you want the total be bigger than 30.
D/(1-r) + Dr/(1-r)^2 > 30
...
r > 1 - sqrt(D/30)
For D=10 that's around 25.35.
That's surprisingly low, but agrees with your numbers. Those are both only ~0.65 chances so you really want the number of cyclers to be bigger though.
I agree it is way lower than I thought it would be, and while it's not 100% it does leave us wiggle room. For example, lets say assuming that we want the deck to go off turn 4, the following configuration should work:
20 Land Cyclers
4 Blasted Landscape
4 Cycling Cards
4 Fluctuator
4 Shadow of the Grave
3 Lotus Petal
3 Enlightened tutor
3 Dark Ritual
1 Laboratory Maniac
Taken into account is the fact that by turn 3 we either want a Fluctuator in hand or Enlightened in hand turn 2, and assuming 4 mana open turn 4 for the combo to work we need to be able to generate an additional 4BBU (2BB for the rest of the shadow of the graves and 2U for the Laboratory Maniac. 3 Petals and 3 Dark Rituals should be enough for that (with an additional mana in case we only had 3 mana open to start with)).
That leaves us with 14 open spots. We can either include more cycling for more consistency or start including in discard and other protection cards.
I think protection is mandatory. If fluctuator doesn't stick around or hit the table the deck is pretty hosed.
I'm curious about how good the land cyclers like Sanctum Plowbeast are in a deck like this, now that they can fetch cycling lands. Similarly, Summoner's Pact can fetch desirable stuff, though it does tend to make the combo turn an 'all-in' proposition.
Good call on the land cyclers. Getting two cyclers for the cost of one. I'd love to see Eternal Dragon make a comeback! I don't see how this deck can do anything without Fluctuator on the table, so I'd agree that it needs much more protection.
A list for the lazy: http://gatherer.wizards.com/Pages/Se...;swampcycling]
Landcycling sounds like a great way to increase consistency since they can fetch you a guaranteed cycler.
With the high amount of CITP lands and due the synergy with Shadow of the Grave, maybe Mox Diamond might be an option?
Land Cyclers are a great idea - they help, but not that much, they will only reduce variance in the combo the same amount as a fetchland in a regular combo deck (actually twice as much due to being returned due to SOTG).
Additionally, a slight addendum to my math. Warning, I had fun just writing variables and stuff: to get to 20 cyclers, that would be Q+2(X+Y)+Z = 20, where Q is how many cyclers you used prior to the turn you decided to go off X is how many cycler's in your hand prior to starting cycling, Y is how many cyclers you drew from the X cyclers, X and Y are doubled due to Shadow, Z is how many cyclers you drew from the X+Y cards after they were returned from being cycled. Q would be 9A/60 (based on my previous post, and assuming you use each such cycler), and X would be (((10-Q)A/60)+(QA/60)) which should miraculously simply to A/6. So the final equaltion should be: (9A/60)+2((A/6)+(A^2/360))+((A^2/200)+(A^3/10800))=20. Solving for A would lead to a final answer of about 24.892. Which is even lower than the previous answer!
Again, check my math and assumptions, but it seems that you can get away with cycling prior to going off, and that doing so actually helps you get to card 30? Edit: I feel that one of my assumptions MUST be wrong, since this doesn't really make sense intuitively.
If the above checks out, that allows us to free up 3 more spots in the deck and/or replace some cycle lands with other lands and run more landcyclers to reduce variance.
Serum Powder may be a wise choice for a deck like this. The main engine is Fluctuator itself, so you need to be able to find it first without destroying your keeps. Bazaar in Vintage - same thing.
That to me would be more important than cycling X number of cards down and drawing one.
This one doesn't make sense to me. I don't understand what "to get to 20 cards" means or what Q is supposed to represent.
As it stands, I'm sure the cycler counts are too low because we're not accounting for the fact that the lands are cyclers that get played out and that E-tutor cuts into the card count.
For example, let's say that you go off on turn 3 on the play with 2 lands in play having cast E-tutor once. Then the expected number of starting cyclers in hand is
(9 (total cards drawn) -1 (e-tutor))*r - 2 = (8r-2)
Now, you want to start with (8r-2) cyclers and draw at least 6 more cards to get to 15 and a decent chance at seeing a shadows. (E-tutor enriches the deck a little, but not significantly at this point)
(8r-2)/(1-r) > 6
..
r> 8 / 14
60*r>34.28...
and to get to 40 you'll want to get 21 cards deeper
(8r-2)/((1-r)^2) > 21
...
r > 0.623...
60*r>37.38...
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