Reasons behind QI Definition

[summerz88]

Sure, if you can show me how to do it, actually yes, for this situation we dont care the order that the photon or the cremation are drawn and therefore it just degenerates down to simple probability, but ive been using hypergeometrics for my analysis of win rates for the instosis deck, i think teffy asked you to have a look at it, and id appreciate it.

[TribalTrouble]

Sure, if you can show me how to do it, actually yes, for this situation we dont care the order that the photon or the cremation are drawn and therefore it just degenerates down to simple probability, but ive been using hypergeometrics for my analysis of win rates for the instosis deck, i think teffy asked you to have a look at it, and id appreciate it.

No not me, I don't know the first thing about that! 

[farrugiamaths]

To summerz, about the damage thing you said, I'm not fully agreeing with your argument. A QI index should be an index that tells you whether your quanta generation is efficient or not, not whether you are dealing efficient damage. For all I know, someone's deck might have only 2 cards that deal damage and the rest are stalling cards. For such a deck, it is still important to have quanta being generated efficiently to be able to feed the stalling tactics. Some decks win by decking the opponent, in fact.

A few examples are in order:

Example 1: Suppose deck is: 12 Emerald Pillar, 6 Horned Frog, 6 Cockatrice, 6 Adrenaline with Mark of Life.

We have q=12, p=r=0, so Q=12x4 = 48.
n=30, m=1 and c=6x2+6x3+6x4 = 54. So K = 54 - 1x30 = 24.

Thus T = sqrt((7Q-4K)^2+60Qc)-(6Q-4K))/Q = (sqrt((7x48-4x24)^2 + 60x48x54) - (6x48 - 4x24)) / 48 = (sqrt(240^2 + 155520) - 192) / 48 = (sqrt(213120) - 192) / 48 = (461.65 - 192) / 48 = 5.62.

Example 2: Same as Example 1 but we change the 6 Cockatrice to 6 Emerald Dragon.

Q, n and m are still the same. c is now 6x2 + 6x10 + 6x4 = 96. So K = 96 - 1x30 = 66.

T would now be (sqrt((7x48-4x66)^2 + 60x48x96) - (6x48 - 4x66)) / 48 = (sqrt(72^2 + 276480) - 24) / 48 = (sqrt(281664) - 24) / 48 = (530.72 - 24) / 48 = 10.56.

EXample 3: Same as Example 1 but we change the Mark to something other than Life (for example Gravity).

We first calculate T for Life. Q, n and c are the same as in Example 1. m=0 this time. Thus K = 54 - 0x30 = 54.

T would now be (sqrt((7x48-4x54)^2 + 60x48x54) - (6x48 - 4x54)) / 48 = (sqrt(120^2 + 155520) - 72) / 48 = (sqrt(169920) - 72) / 48 = (412.21 - 72) / 48 = 7.09.

We now calculate T for Gravity. Q=0 and m=1 in this case. c=0 as well. So K=0-30(1) = -30.

T would be 1 - (15c / 2K) = 1 - (15x0 / 2x(-30)) = 1.

This shows that changing the mark of a deck affects its quanta generation by quite a lot.

From the examples above, it looks like Example 1 generates slightly too much Life quanta, Example 2 generates too little Life quanta, while Example 3 seems to generate a fine amount of Life quanta but then generates way too much Gravity quanta (obviously).

Theoretically, T is the average amount of turns where you start generating at least as much quanta as the amount required to cast all the cards in your hand (of the type being analysed). Thus, a small T means that you are generating too much quanta and a large T means that you are generating too little quanta. We simply need to define what 'too much' and 'too little' mean.

I'm currently leaning on a deck having an efficient quanta generation if T is somewhere between 6 and 7, but I think this depends on the deck. Some decks require speed, so they can get away with a value of T less than 6 (though not too smaller than 6). Other decks, like stall decks, can allow T to be larger than 7.

[farrugiamaths]

Update on the formula:

Now supports the following quanta-generation cards:

- Pillars
- Quantum Pillars
- Pendula
- Nova (new)
- Supernova (new)
- Towers (new)
- Quantum Towers (new)
- Upgraded Pendula (new)

Does not support (yet):

- Skills requiring quanta
- Skills generating quanta

The formulae are already becoming unwieldy, so I guess a program that generates these automatically given a deck (or a deck code) would be handy.

Suppose:
p is the number of Pendula and Upgraded Pendula generating same type as that being analysed
q is the number of Pillars and Towers (except Quantum Pillars and Quantum Towers) generating same type as that being analysed
r is the number of Quantum Pillars and Quantum Towers
s is the number of Supernova in your deck if type being analysed is not Entropy, and 0 if type being analysed is Entropy
u is the number of Upgraded Pendula generating same type as that being analysed
v is the number of Nova in your deck
w is the number of Towers (except Quantum Towers) generating same type as that being analysed
y is the number of Quantum Towers generating same type as that being analysed
n is the number of cards in the deck
m is 1 if deck Mark is the same as type being analysed, and 0 if it is not
c is the total casting costs of all the cards in your deck (excluding all skills)
Q is 4q + 2p + r
K is c - mn
U is 8s + 4u + 4v + 4w + y

Then the expected number of quanta generated at the end of the t^th turn is

GQ(t) = 0, if t=0
GQ(t) = (Qt^2 + (14Q + 8mn + 2U)t + 13U) / (8n), if t>0

The expected number of turns required to start generating at least as much quanta as can be cast from your hand is

T = (sqrt((7Q - 4K + U)^2 + Q(60c - 13U)) - (6Q - 4K + U)) / Q, if Q>0
T = (60c - 8K - 11U) / (2(U - 4K)), if Q=0 and U > 4K and 60c > 8K + 11U
T is undefined otherwise.

Example: We saw in the last post that the deck consisting of 12 Emerald Pillar, 6 Frog, 6 Cockatrice and 6 Adrenaline has T = 5.62. Suppose we replace all the Emerald Pillars with Emerald Towers.

We have q=12, p=r=0, so Q=12x4 = 48.
n=30, m=1 and c=6x2+6x3+6x4 = 54. So K = 54 - 1x30 = 24.
w=12, s=u=v=y=0. Thus U = 4x12 = 48.

T = (sqrt((7Q - 4K + U)^2 + Q(60c - 13U)) - (6Q - 4K + U)) / Q
T = (sqrt((7x48 - 4x24 + 48)^2 + 48(60x54 - 13x48)) - (6x48 - 4x24 + 48)) / 48 = (sqrt(288^2 + 125568) - 240) / 48 = (sqrt(208512) - 240) / 48 = (456.63 - 240) / 48 = 4.51

showing that the quanta generation became even faster (probably much too fast) by replacing Pillars with Towers.

[Chapuz]

I like thinking minds, awesome!

well, previous QI wasn't so wrong 

[bucky1andonly]

I've always been confused by the QI, not the formula though, but why anyone would use it anyway.  There's no need to use any simple or complex mathematical formula to determine how many pillars you should put into your deck.  There's only 2 things you need to do.

1) Trial and error, you draw too much all the time, take some out, not enough, put some in.
2) If your deck has a lot of high cost cards, you will most likely need a bit more, whereas if you have only low cost cards slightly less, and average amount for average costs.

If I were to play a light/water deck, I would add up all the light and water required for the entire deck, and that would be my ratio for light/water pillars, the amount exactly is unknown, but easily determined by using the 2 step process above.

[Xenocidius]

Interesting stuff. Nice job.

The formulae are already becoming unwieldy, so I guess a program that generates these automatically given a deck (or a deck code) would be handy.

I know a certain someone who may be interested in making such a thing once the formula includes skills and the like. Combined with the game simulator to determine average TTW, you've got yourself a ridiculously accurate improved quantum index calculator.

[Chapuz]

Interesting stuff. Nice job.

The formulae are already becoming unwieldy, so I guess a program that generates these automatically given a deck (or a deck code) would be handy.

I know a certain someone who may be interested in making such a thing once the formula includes skills and the like. Combined with the game simulator to determine average TTW, you've got yourself a ridiculously accurate improved quantum index calculator.

In "draw and play" decks, yes, yes.

[Absol]

Interesting stuff. Nice job.

The formulae are already becoming unwieldy, so I guess a program that generates these automatically given a deck (or a deck code) would be handy.

I know a certain someone who may be interested in making such a thing once the formula includes skills and the like. Combined with the game simulator to determine average TTW, you've got yourself a ridiculously accurate improved quantum index calculator.

Please make it. Thank you.

[10 men]

GQ(t) = (Qt^2 + (14Q + 8mn + 2U)t + 13U) / (8n)

I'm missing a term whether you're on the draw/on the play there? Right now looks as if you're playing all the time.

The expected number of turns required to start generating at least as much quanta as can be cast from your hand is

T = (sqrt((7Q - 4K + U)^2 + Q(60c - 13U)) - (6Q - 4K + U)) / Q, if Q>0
T = (60c - 8K - 11U) / (2(U - 4K)), if Q=0 and U > 4K and 60c > 8K + 11U
T is undefined otherwise.

Ok let's say I have a deck with 5 Chrysaoras and I plan on paying for them with 6 Novas only in 30 cards (I've done worse ). So T in my case would be -0,5. No wait, it's undefined actually. But why? I can, in fact pay for all my 1,17 Crysaoras with my 1,4 Novas right on the first turn! And I even have a couple quanta left, so it makes sense that T would be negative.
I think here we're starting to see a problem of the approach to not use hypergeometric probability (which would be too complicated, granted) and instead just take the averages. Why undefined? Some decks (pillarless Immo and/or Nova decks) can indeed get a T = 0 or T < 0 here. This is because all their cards can be played from their Immolations and Novas only.
Unfortunately, they can't expect to play all their cards on turn 1 always (because draws are inhomogenous).

Theoretically, T is the average amount of turns where you start generating at least as much quanta as the amount required to cast all the cards in your hand (of the type being analysed). Thus, a small T means that you are generating too much quanta and a large T means that you are generating too little quanta. We simply need to define what 'too much' and 'too little' mean.

You generally want to have played all your cards a little before you kill your opponent. (A little before because cards that enter the game early have a much bigger impact on the game than cards that enter late.) I think that would be the best starting point, become aware of by which turn you want to kill your opponent and figure out how many quanta you need to play all your cards before that.

[farrugiamaths]

I'm thinking about Skills costs now (not Skills that produce quanta, for now). Obviously adding the total Skill costs to the total number of casting costs works if the skill can be used just once.

However, there are skills which can be used more than once, in which case we would need the average number of times that the skill would need to be used during the game. Now this obviously varies, not just due to the nature of the card, but also due to the player's playing style (in my opinion, at least). In the original QI definition, ScaredGirl used the following system:

- if the ability is non-situational as in player will play it if he has quanta (for example Chrysaora, Hourglass) it gets counted twice
- if the ability is situational (for example Maxwell's Demon) it gets counted once

If you want to use ScaredGirl's system here as well, go ahead. If not, I am not going to suggest a better system, because there simply isn't any; the choice depends too much on the card and on the player.

The theoretical system would be multiplying each skill cost by the average number of times that it is used per game. Sometimes, this depends on how the game evolves, like for Maxwell's Demon mentioned above, and sometimes the game does not influence the usage of the skill too much, like for Rustler's Photosynthesis.

I guess that if you don't know this average, then using ScaredGirl's rough guide would be a good start, and then update your pillars accordingly if you see that you are using a particular skill more or less than you expected. So ScaredGirl's system assumes that you use situational skills once per game on average, and non-situational ones twice per game on average. Note that this is per card, so if you have 1 Pegasus in play, ScaredGirl's system assumes that you use its Dive skill twice per game, but if you have 2 Pegasus in play, you are assumed to be using its Dive skill twice per game per card, so 4 times on average.

So let x be a_1x_1 + a_2x_2 + ... + a_dx_d, where x_1 is the cost of using the skill of card 1 and a_1 is the (empirical or ScaredGirl) average number of times that this skill is expected to be used per game (this can be a decimal if you so desire). Obviously we're only counting the skills needing the type that we are analysing. Then let c be equal to the total costs of all the cards of your deck plus x.

Now simply apply the formulae in my last post with this new modified value of c.

To 10 men, I'll answer you later.

[farrugiamaths]

I'm missing a term whether you're on the draw/on the play there? Right now looks as if you're playing all the time.

That formula is the expected number of quanta you would have generated by the end of your t^th turn. No quanta generation (or destruction) is assumed to happen during your opponent's turn; if it happens, I have no way of predicting it from your deck!

Ok let's say I have a deck with 5 Chrysaoras and I plan on paying for them with 6 Novas only in 30 cards (I've done worse ). So T in my case would be -0,5. No wait, it's undefined actually. But why? I can, in fact pay for all my 1,17 Crysaoras with my 1,4 Novas right on the first turn! And I even have a couple quanta left, so it makes sense that T would be negative.

Fair point. I did not consider the case when Q=0 and m=0, which is the case in your example. Sorry about that. The formula for T should read the following:

T = (sqrt((7Q - 4K + U)^2 + Q(60c - 13U)) - (6Q - 4K + U)) / Q, if Q>0
T = (60c - 8K - 11U) / (2(U - 4K)), if Q=0, m=1 and U>4K
T = 1, if Q=0, m=0 and U>4K
T is undefined otherwise

I think this should address the problem you mentioned. I haven't yet considered cards like Immolation.

You generally want to have played all your cards a little before you kill your opponent. (A little before because cards that enter the game early have a much bigger impact on the game than cards that enter late.) I think that would be the best starting point, become aware of by which turn you want to kill your opponent and figure out how many quanta you need to play all your cards before that.

This is interesting. I'm still a pretty new player here, so this is an interesting suggestion for me.