Using Quanta Index to determine the optimal number of Pillars in a deck

[Zuphix]

Yes a devourer, especially a burrowed one, definately counts as a pillar for all intensive purposes.

You need to come up with something better than that.

Reason why a Devourer will never produce as much quanta as a Pillar is because Devourer is more likely to be removed from the game. EQ is only real anti-Pillar card, but there are tons of anti-Devourer cards. If you have two decks, one with 18 Pillars and one with 12 Pillars and 6 Devourers, the first one will produce more quanta on average.

I'm not saying counting Devourers as Pillars give totally wrong results. I'm just saying that it's not a "perfect" solution.

They would produce the same ammount on average (except for the fact you have to pay to summon a devourer, but that has already been taken into account)

[wei2912]

The mark should be caculated together with pillars.

[Drusyc]

Wouldn't it make sense to consider the HP of the quanta generating creatures as well? for example:

Devourer costs 2  , has 2  , and can generate 1 per turn, including the summoned turn.
Even if Devourer is destroyed on the next turn, it has generated 1  to offset its cost.
Devourer is now 1 for the QI.

Being generous and assuming a non-rush deck will contain, at any given point in time, 4-6 creature control cards, a Devourer will in most casts survive for 1-3 turns.

The following will assume a 40 card deck with 5 creature control cards.
r is the number of CC remaining in the deck.
n is the number of cards drawn.
So, that means at the start of any game, we have a complicated probability formula that I'll "simplify" (incorrectly for simplicity's sake) to (d/40)+(1/40-n) x 100% chance to draw these cards if none have been drawn, since any possible creature control card can occupy any particular spot in the deck.
Your first CC has a 14.75% base chance to be drawn.
Your second card, 12.56%.
Your third, 10.13%.
Your fourth, 7.70%.
Your last, 5.78%.
The probability of drawing all five in the starting hand is something like 1%, and it would be easier to show if I really wanted to plug in the numbers, but I don't know fractional exponents in my head. anyway;

The probability of drawing a card that can kill a devourer goes up as the game goes on, and the chances of having at least 2 CC by turn, say, 3, is rather high (60%+).
This means a Devourer has a lower chance of surviving, reducing the amount of quanta it can generate.
Let's assume we multiply the next set of values by (hp/3)+1, where hp is, well, the hp of the critter. I assume /3 because I'm assuming, through "efficient" play, that fire bolt and rain of fire are the baseline spells for dealing damage to critters, and that no one is hoarding quanta to pump the damage.

So, for the case of QI, include the cost of the creatures. In ScaredGirl's original example,

,
Let the base QI be (93-12)/12 = 6.75 as she calculated.
For the first two Devourers, subtract 1 per devourer.
For the next two, subtract 2 per devourer.
For and after that (I'm looking at FFQ right now), subtract 3.
In this case, we get 1+1+2+2+3+3, for a total of 12. If we apply this to our value, we get this:
(93-12-12)/12 = 5.75, which makes a ton of sense. If they were instead Pests, we would multiply 12 by (4/3)+1 rounded down to 2, estimating our QI to be  (93-12-24)/12 = 4.75, which might be a bit of a stretch.

In the special case of FFQ, she costs 7  , her ability costs 2 to generate 1 or  ; and can generate 2 fireflies per queen. so, if you were to have 2 Queens, that would be the equivalent of having 4 0 cost fireflies with HPs of 2, cutting the deck QI down by (x-6-towers)/towers.
At 3 queens, the third queen by assumption of these calculations would amount to (x-12+n-towers)/towers, where n = the number of fireflies you assault your opponent with.

...That's a lot of numbers.
It's also kinda confusing.
I'm gonna paraphrase myself:

Up to 2 Critters that generate quanta are worth (cost-(hp/3)-1)), rounded down.
The next two critters that generate quanta are worth (cost-(hp/3)+2)), rounded down.
Any critters after 4 that generate quanta are worth (cost-(hp/3)+3)), rounded down.

If someone manages to make sense of all of that, I hope it makes sense XD

[teffy]

So, that means at the start of any game, we have a complicated probability formula that I'll "simplify" (incorrectly for simplicity's sake) to (d/40)+(1/40-n) x 100% chance to draw these cards if none have been drawn, since any possible creature control card can occupy any particular spot in the deck.

Can´t understand your formulas and your percentages. I know Hypergeometric Distribution and understand the formulas of it, but this doesn´t look like sth I know.


It makes sense to see the HP of creatures, the more HP a creature has, the longer it will survive.
But you can´t use the HP for the QI !

The worth of a hit point for the player is dependent on the cards we have in Elements.
The difference between a 3 HP creature and a 4 HP creature is big ( RoF, Fire Bolt, Owl´s Eye).
The difference between 1 HP and 2 HP normally not ("only" 2 shields kill the creature one turn later).

Devourers survive longer with an Earth Mark. And who kills devourers with spells except mass creature control ?.

[TheForbiddenOracle]

Gah, too much formulating...

[jallenw]

Gah, too much formulating...

Cards in deck = D
Quantum Pillars in deck = Q
Percent of Qpillars in deck = Q/D = Q%
Average number of pillars in starting hand = 7*Q% = Qs
Average number of turns you can survive without playing most expensive cost card= T
Number of pillars drawn and played during a game after first turn = (Q - Qs)(Q%-(Qs/D-7)T = Qp
Quanta generated by pillars in starting hand = 3Qs*(T+1)  (Remove the +1 for non-upgraded pillars)
Quanta generated by pillars drawn = 3Qp*(Q%*T)  (Use T-1 for non upgraded pillars)
Total Quanta Generated = (3Qs*(T+1))+3Qp*(Q%*T) = QT
Average Quanta of each element generated = QT/12
Cost of the most expensive card you have to play nearly every game in order to win.

Assuming miracle is your most expensive card, and you're not using supernovas or mark of life, then you need 12 light quanta before you die.  Really, you have many other cards that will save you, but those are less expensive, so you are only concerned with producing enough for the most expensive one before you die.

I have 10 pillars in a 40 card deck
D=40, Q=10, Q%=.25,Qs=1.75,Qp =T*1.62525
QT = 3*1.75*(T+1)+3*T*1.62525*.25*T

(5.25T+5.25+1.2189375 *T^2)/12=12

5.25t+5.25+1.2189375*T^2 = 12*12=144
5.25+1.2189375*t = 138.75/t
t = 138.75/1.2189375t - 5.25/1.2189375
t = 138.75/1.2189375t - 4.30703
t = 8.7307060839

So with 10 pillars and a need to cast miracle every game to win, you would have to survive 9 turns before casting miracle.  Of course, lets assume that 6 of your cards are supernovas.

Each supernova you draw effectively reduces the cost of the miracle from quantum generated by 2.  Your average chance to draw a supernova in those turns is 7+T)/33-T.  This means that in just 3 turns you have a 30% chance to have 2 supernovas, so we can safely assume that 2 supernovas is likely in every game.  Now we just reduce the cost of our target card by 4.

With a cost of 8 we come to 5.25t+5.25+1.2189375*t^2=12*8=96

This resuts in t = 6.7396123218, so you need to survive for 7 turns, but in 6 turns, you actually have a %50 or better chance of drawing a 3rd supernova, so again the cost is reduced.

This gives us a result of t = 5.5535229926.

This means that with 10 quantum towers and 6 supernovas in a 40 card deck, you should have enough quanta to drop anything you need by the 6th turn. 

Now go out there an make your algebra teachers proud by building rainbow decks.

Jallen

[Drusyc]

Can´t understand your formulas and your percentages. I know Hypergeometric Distribution and understand the formulas of it, but this doesn´t look like sth I know.

I noted that's incorrectly simplified , but it was the easiest way to describe increasing chances of probability with a linear decrease in available options. The formula I'd love to use would be Xⁿ= (1-1/qⁿ)ⁿ, but plugging in the numbers would determine the chances to not draw the card, rather than the other way around. The below quote, while ridiculously accurate, is an algebraic interpretation of the above formula... but it's algebra, and algebra is limited.

Cards in deck = D
Quantum Pillars in deck = Q
Percent of Qpillars in deck = Q/D = Q%
Average number of pillars in starting hand = 7*Q% = Qs

is, again, simplification of reality; the percentage of pillars in the deck increases as deck size decreases, and the percentage of pillars in the deck decreases as the deck increases. Calculus is required to calculate the exact average of drawn pillars; but of course that makes a difference of maybe a percentage or two.

[jallenw]

Yes, I use algebra because it comes easier to me at 2am than calculus.  I'm too lazy to get out my graphing calculator and plot curves and ratios and geometric progressions just for a slight increase in accuracy, especially when you're dealing with items that can't be divided into fractions, such as the number of cards in a deck. 

Sure, you're method will bring a more accurate result, but there is no difference between 5.4XXX and 5.6XXX when dealing with indivisible integers.  The practical result is 6.

Also, people here are more likely to be able to understand the algebraic equations more readily than the more complex calculus and such. 

I suppose I could write a qbasic program right quick that asks you the number of cards in your deck and number of pillars, towers, q pillars, q towers, novas, supernovas, and quanta generating creatures.  It would then run all the numbers through the equations automatically and report the average quanta production each turn and the total quanta production through the game, as well as running 1000 sample games with randomization in order to get a quick sample of brute force test data. 

Then I could compile the program to an .exe file and provide it for download.

I could do all this, I mean, the qbasic shortcut is right on my toolbar...  just take a few clicks, a little typing, a little math, a few test cycles...

Okay, after an hour or so I have the skeleton program.  I'm going to play elements for a while.  I'll finish the program at some point.

[BluePriest]

HP shouldnt be calculated AT ALL. QI doesnt take into account things getting destroyed. If it did then pillars wouldnt count for a full  because they can be destroyed too. Youre overcomplicating something that doesnt need overcomplicating.

[EvaRia]

I personally don't believe in QI.

I find the most important thing in determining how much quanta you need is basically "How long am I planning on making the game last" and "Am I going to get all the Quanta I need by that time?"

As such, stall decks can generally afford to have a much higher QI because by the time they stall out, they have the quanta you need anyways.

In rush decks, the QI should be to the lower side because rushes need more quanta faster.

This is just my opinion though.

[jmdt]

I personally don't believe in QI.

I find the most important thing in determining how much quanta you need is basically "How long am I planning on making the game last" and "Am I going to get all the Quanta I need by that time?"

As such, stall decks can generally afford to have a much higher QI because by the time they stall out, they have the quanta you need anyways.

In rush decks, the QI should be to the lower side because rushes need more quanta faster.

This is just my opinion though.

I agree with you in part, but mostly disagree.  QI is VERY important, but at the same time the target QI is also dependant on what the deck is doing.  I have effective decks with QI ranging anywhere from 3.5 to 8.  Does this mean QI is broken...no.

When I first started deck building, I sucked (yeah I said it); solid deckbuilding is a learned skill.  Studying QI really helped me to better understand and hence better build solid decks.  When I first discovered QI, I develped a spreadsheet that allowed me to 'virtually' build decks.  For a period there, I literally built every deck in a spreadsheet, forcing the quanta to be in the ~5.0 range.  Regardless of everything else, a deck with a QI of ~5.0 will function well mechanically.  It may suck in practice due to the strategy used, but you will generally find no excess quanata or shortage of quanta hindering the deck.

Over time studying popular decks that worked, I learned that rushes favor lower QI's (increasingly lower as the cards get more expensive) and that stalls can often fully function with only minimal quanta generation.  I look at the target QI as a necessity factor with 5.0 being the average value for an average deck.  Rushes need extra quanta early to get a field of damage out as quick as possible so you lower the target QI to compensate, and since a stall such as mono aether only needs to play an expensive shield every 3 turns (and even then you usually wait for the opponent to get out damage) as opposed to quickly after drawn the necessity of quanta generation is lowered and the target QI can be raised to compensate.Eventually I learned enough that I rarely use the spreadsheet now (except for upped pendulums) and can generally ballpark in initial design (its also quicker to just pull out the TI-83 to check).

Regardless of target QI value, only testing will reveal the ideal value for the particular deck.  If I test a deck and I find I have too much quanta sitting around most of the time I either add some larger hitters for smaller ones or remove a pillar for something cheap.  Conversely if my hand is sitting full most of the time, a dragon or 2 may need to go.  To further hammer this point, someone (I believe kevkev) did a ttw of of 3 unupped mono with varying QI but essentially the same deck.  The study showed that the deck with ttw closest to 5.0 had the fastest ttw.  Again, with a high QI, critters could not come out quick enough and with too low a QI tere was not enough damage to go around.

So yeah I definately believe in QI, but there a number of myths that have developed about QI that need debunking.

(I should expand that to a wiki article, lol)

[Vyrys_Complex]

Small necro to throw in a small idea.

We were having an issue where larger decks had unusually high QI ratings, right?

Divide [QI formula] by deck size and multiply by 30? QI remains the same for thirty-card decks but is "normalized" as the deck grows larger. (Or would this make it too complicated? After all, draw cards will throw it off.)