Reasons behind QI Definition

[OldTrees]

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.

In the absence of removal, a card's benefit is proportional to Cost x Turns remaining.
The value of playing all your cards before the turn you kill your opponent is proportional to what fraction of your TTW remains.
The value of playing all your cards is reduced by the cost of the quanta inefficiency (including more quanta producers and less quanta consumption).

[summerz88]

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.

I was going to try and post about this but then i got carried away, back when you suggested the 2 different decks with frogs cockatrice and dragons, I ran them both, and almost independently of the draw they seemed to both kill their opponent 2 turns after the expected T value, this was regardless of whether or not my hand was emptied that turn or not, therefore the dragon deck was slower to kill its opponents (ai3) and the frog deck was significantly faster, there was a slight +ive correlation between when my hand emptied and winning earlier, but only slight.

Therefore given the very strong correlation, an easy solution to my question about what turn can i kill my opponent, is T +2, obviously there has to be some lower limit on this, and this only applies to opponents with 100 HP (as quanta to damage ratio is roughly constant), but this could be at least an indication that we can link T and expected turn damage output.

There's also some sort of assumption in there that its a reasonably balanced deck, for instance a deck with 4 dragons and 26 pillars may have decent damage potential, but its pretty likely that you're going to be waiting 10 turns for your second dragon, so T will be ~ 3 and the finishing turn would be more like 18.  I don't really know how to handle that concept without actually looking at the number of creatures in your deck and the individual cost/damage potential

[summerz88]

I also have some ideas about the frequency of ability usage, for instance If I play a chrysaora on turn 1 and I always have death quanta, I'm going to use that ability every turn (assuming that the creature doesn't die), if I draw it on the last turn then I only play it once, therefore you can use the number of chrysaora's in the deck of N cards to indicate the expected death quanta usage. This deviates away from scared's method of saying X number of times on average, and more toward how many times are you likely to have the opportunity.

For things like Maxwell's and say BE (butterfly effect) you could have such a program allow you to enter an expected style of opponent, with X number of creatures and Y number of permanents, with things like shards and hourglasses taking priority (im highly likely to want to destroy/steal these) to pillars being say (if i have lots of opportunities for PC) i have a likely hood of wanting to destroy z pillars.

For example if you play a shard of focus on the first turn of a match, you're almost always going to destroy a pillar, because it hampers their quanta generation, and SoF have lots of PC, but if i've only got one SoF in my deck, chances are im not going to get it/be able to play it on the first turn.

[farrugiamaths]

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.

I was going to try and post about this but then i got carried away, back when you suggested the 2 different decks with frogs cockatrice and dragons, I ran them both, and almost independently of the draw they seemed to both kill their opponent 2 turns after the expected T value, this was regardless of whether or not my hand was emptied that turn or not, therefore the dragon deck was slower to kill its opponents (ai3) and the frog deck was significantly faster, there was a slight +ive correlation between when my hand emptied and winning earlier, but only slight.

Therefore given the very strong correlation, an easy solution to my question about what turn can i kill my opponent, is T +2, obviously there has to be some lower limit on this, and this only applies to opponents with 100 HP (as quanta to damage ratio is roughly constant), but this could be at least an indication that we can link T and expected turn damage output.

I had actually found an expected amount of damage dealt by the end of the t^th turn formula two days ago, but then I dismissed it. Seems like I need to dig it up again.

It's a pretty complex formula: first it starts as a cubic in t and then it degenerates to a quadratic when t exceeds T_d, which is the same as T but with c replaced by just the total cost of the damage dealing cards only (instead of the total cost of every card). I had tried it out on an Excel sheet and I found out that the damage usually becomes 100 at around the T value (T here is the usual T), 150 at around the T+2 value and 200 at around the T+3 value. I still believe your T+2 value though as my damage formula assumes that no hindrances whatsoever are done by the opponent.

[10 men]

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

Sorry, what I meant here was whether you won the coin toss (are on the play) or lost it (are on the draw). In the latter case you have one more card in your opening hand which affects your quanta production.

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.

The problem is still that I can't expect to play all my Chrysaoras on turn one. Sometimes I will get a draw with 0 Novas and I'm just stuck with them for a couple turns...

[farrugiamaths]

Sorry, what I meant here was whether you won the coin toss (are on the play) or lost it (are on the draw). In the latter case you have one more card in your opening hand which affects your quanta production.

I had said this in an earlier post in this thread - I guess you missed it, which is understandable considering the amount of text I posted. If you know that you're starting first, replace the number 14 in the formula by 13. If you know you're starting second, replace it by 15. If you don't know who's starting first, leave the formula as is.

The problem is still that I can't expect to play all my Chrysaoras on turn one. Sometimes I will get a draw with 0 Novas and I'm just stuck with them for a couple turns...

Yes, to be honest, the case when Q=0 and m=0 has me a bit confused. You don't have continuous quanta generation in such a case. Let me try to reason this case out again from scratch.

Suppose I have 6 Nova and 1 Blue Crawler in my deck (requiring 3 Water Quanta to be cast), and nothing else in my deck can generate Water Quanta. Now in such a case, obviously T is not 1 because T must be at least equal to the average amount of turns I need to draw 3 Novas, which is the solution of the equation 6(2t+13)/(2n) = 3, hence t = (n-13)/2 which is larger than 1 even for n=30, the smallest possible n. Now suppose I have 6 Nova and 2 Chrysaora in my deck. T would be 1 now because the Chrysaora requires only 1 Water Quantum to be cast and you should be able to get 1 Nova in your starting hand more often than not (6(2t+13)/(2n)=1 has the solution t=-1.5 if n=30, meaning you should get a Nova among the first 5 or 6 cards of the deck on average), so from this point onwards, we are already generating at least as much Water Quanta as required. So, judging by this reasoning, T in such cases would be the solution of the equation U(2t+13)/(2n) = c / d, limiting the minimum value of the solution to 1, where d is the number of cards requiring casts of the type we're considering. So let me try

T = max(1, (2nc - 13dU) / (2dU)), if Q=0 and m=0, where max(a,b) = a if a>b and max(a,b) = b if a<=b.

Now, to answer your question, yes you should expect to play all your Chrysaoras from your hand on turn 1, because you are expected to have only 1 Chrysaora in your hand on your first turn. On average, the number of Nova in the first i cards of a shuffled deck must be greater than the number of Chrysaora in the same i cards. This doesn't always happen in practice, I know, but we're talking about the average here.