Probability and you: How to build the best deck

[agentflare]

So war's coming up, and I figured team  is forming. Since I'm not getting chosen for war, i might as well help Team Underworld win. So without further ado, I present: How to build the best deck.

Step 1: Strategy

This section is pretty easy. Decide what your deck is going to be like. Is it going to be a rush or a stall? Mono, duo, trio, rainbow? What's your key card? For the purposes of demonstration, I'm throwing together a deck.

The Example

Code: [Select]

5f0 5f0 5f0 5f0 5f0 5f0 5f0 5f0 5f0 5f0 5f1 5f1 5f3 5f3 5f4 5f4 5f4 5f6 5f6 5f6 5f7 5f8 5fa 5fa 5fa 5fa 5fa 5fa 5fb 5fb
So we can examine the strategy:
Deflags for Permanent Control
Rage Potions, Rain of fire and Firebolts for Creature control
Pillars for quanta *Extremely important*
Fahrenheit, Lavagolems, Ash Eaters and fire spirits for damage.

Alright so, onto step 2

Step 2: Quanta
It's at this point that differentiates pros at noobs. Noobs choose using gut instincts, which sometimes works. Pros should calculate.

We could take the QI, but that's not the best way. We need to take a look at the average number of pillars each turn, and in the starting hand.

We draw 7 cards in the starting hand. you draw an average of:
(1/3)*7 pillars per hand which equals ~2.3 pillars per hand

We now draw 1 card a turn. With 7.7 statistical pillars remaining, we know that we now gain ~7.7/23 pillars per hand which equals ~0.3 pillars for each subsequent turn. But we factor in our mark, which essentially is a pillar. 2.3+1=3.3

(Yes I know it's not the best way, but it's the simplest and gives a reasonably accurate answer.)

A better way is using the nCr function.

Step 3: Using quanta

Now calculate the average quanta cost of all cards. Here it's (5*6+1*2+2*2+3*3+2*3+3+7+3*2)/20=3.35

We can see that we produce 3.3+(0.3*n) quanta a turn, where n is the number of turns that has passed. We need 3.35 quanta a turn (parity between quanta and cost is best because that means we play each card we get the turn we get it). So obviously, we go back and increase the number of pillars. We achieve parity by turn 1. That's actually pretty good.

Step 4: Why doesn't that dang card show up

Alright, so, in this section, we're going to talk about the probability you get that key card by turn 'X'. So, this is going to be dependent probability. We're also going to calculate it through the indirect method.

So, consider this deck:

by agentflare

Hover over cards for details, click for permalink

Deck import code : [Select]

5bt 5bt 5bt 5bt 5bt 5bu 5bu 5bu 5bu 5bu 5bv 5bv 5bv 5bv 5c7 5c7 5c7 5c7 5c7 5c7 5l8 5l8 5l8 5l8 5l8 5l8 5l8 5l8 5l8 5l8 8pn

So as you can tell, the key card is RUSTLER! Who doesn't love rustlers. Turning 1 quanta into 2. We need to get that rustler out ASAP though. So, using probability, we'll find which turn we'll have it on, on average.

Your chance of having it out turn 1 is 0, you don't have quanta! But your chance of drawing 1 is 1-(26/30*25/29*24/28...*20/24).
That works out to around 67.7%. by turn 2, you have a 73.3% chance.

So basically, it goes:
P=1-((total-amount of desired card)/total*((total-1)-amount of desired card)/total-1*...((total-n)-amount of desired card/total))

at least, that's the way it works when you only need that one card, like this case, because you only need 1 rustler for the deck to function perfectly.

Case 2: Rainbows

Alright so, for rainbows, we're going to analyze a very common deck. It's gonna be:

Code: [Select]

4sa 4sa 4sa 4sa 4sa 4sa 4sa 4sa 4sa 4sa 4sa 4sa 4sa 4sa 4sa 4vm 52n 52r 592 5c6 5c6 5f8 5oj 5ro 5rp 5rp 5rp 5rp 5rp 5rp 5up 61t 621 621 74b 74b 7q5 7q5 7q5 7q5
So we should analyze the quanta balance vs. creature cost first.

First, we average out the fact that quantum pillars effectively produce 1/4 of a quanta for each element. There's 15 pillars. 15/40*7=105/40=2.625. So we produce 2.625*1/4 quanta for each element which equals 0.65 quanta per element+1 time, with a gain of 15-2.625/40*1/4 quanta per turn, which rounds to 0.07 quanta per element more, every turn.

The cost is 3 Entropy, 10 gravity, 7 air, 9 death, 10 life, 7 fire, 28 time, 4 darkness, 14 aether, 2 earth.
The average card takes 0.12 entropy, 0.36 death, 0.40 gravity, 0.28 air, 0.28 fire, 1.12 time, 0.16 darkness, 0.56 aether, and 0.08 earth quanta. In short, the first turn is generally enough to cover playing costs (not skill costs though).

However, hourglasses add a new mechanic to the mix. How do you calculate it, you might ask? Well, we draw 1/10*7= 7/10 hourglasses per turn, so we draw 1.7 cards per turn, and 3.3/40=0.0825 cards more every turn.

nCr: the better way

Alright so, first thing you should know about probability. Probability is the number of favourable outcomes divided by the total amount of outcomes.

So everyone above grade 6 should know this classic question:
If you flip a coin twice, what's the chance of getting 2 heads?
There is 1 favourable outcome: HH, and 4 total outcomes. HH, HT, TH, TT.

Now we ask this question:
How many ways can you pick 3 cards out of 5, if the cards are all different.
The answer uses nCr.

n! is n*(n-1)*(n-2)...*2*1
nCr is defined as: n!/(n-r)!
It counts the number of ways of choosing r things out of n objects.

Here the answer is 5!/2!=60.

In elements, we do things like this.

remember this deck?

The Example

Code: [Select]

5f0 5f0 5f0 5f0 5f0 5f0 5f0 5f0 5f0 5f0 5f1 5f1 5f3 5f3 5f4 5f4 5f4 5f6 5f6 5f6 5f7 5f8 5fa 5fa 5fa 5fa 5fa 5fa 5fb 5fb
I told you that 1/3*7 wasn't the most accurate way. nCr is more accurate, but harder to use. So, we use it this way:

Step 1: Strategy

Same as before.

Step 2: Using quanta

Same as before

Step 3: Pillars

Alright, we now find the number of ways we can pick 3 pillars out on turn 1, the magic parity number in step 2.

That'd be ₁₀C₃*_(30-10)C_(7-3). I should explain why I'm doing this. So basically, I find the number of way I choose 3 pillars first. So that means there's 4 cards that are non-pillars in my hand. So I'm calculating the set of pillars and the set of non-pillars and multiplying them to see the total amount of ways to draw 3 pillars in 7 cards.

Next, I find the total number of ways to draw 7 cards, which is ₃₀C₇. ₃₀C₇ is the number of ways of choosing 7 cards from 30 cards.

Finding the first term divided by the second term is the chance of drawing parity on the first turn. As of yet, I'm not too sure on how to average using nCr. Maybe I'll get a brainwave tomorrow, but although it's more accurate, I don't see how to average.

TTW

I had this great idea. So, let's average out all the creature's attack to determine TTW of rushes!

(finishing tomorrow)


Adding more later
Checklist that will be addded:

• Key card probability
• Formulas
• Potential App that calculates this

[/list]

[Stickmasterluke]

Well said words.

[jumpoffduck]

We draw 7 cards in the starting hand. you draw an average of:
(1/3)*7 pillars per hand which equals ~2.3 pillars per hand

I don't think that accounts for automulligan. I'm calculating 2.422 pillars per hand, can anyone confirm?

[kitty45]

How are you possitive you are going on team 

[agentflare]

How are you possitive you are going on team 

I'm probably not being on any team, and probably sitting this one out. I do however want to see  win over those overglorified elemental teams

@Jumpofftheduck
It doesn't account for Automulligan, but automulligan is simply drawing twice. I'm calculating the average draw, or average over infinite games, so automulligan doesn't apply because the second draw is, in theory the same as the first draw.

[kitty45]

How are you possitive you are going on team 

I'm probably not being on any team, and probably sitting this one out. I do however want to see  win over those overglorified elemental teams

@Jumpofftheduck
It doesn't account for Automulligan, but automulligan is simply drawing twice. I'm calculating the average draw, or average over infinite games, so automulligan doesn't apply because the second draw is, in theory the same as the first draw.

I might be with you  look at my appellation anyways this is a good thread

[OldTrees]

Using average quanta cost is interesting. I am making an Excel file to calculate the speed of playing a card based on its casting cost (turns 1-10 going first or second considered). My results might be useful. I will post a summary here when the file is complete.

[Stickmasterluke]

Usually I just build a deck, and make a guesstimate on how many pillars/pendulums it needs. Then test it out. I tent to be right on the money, but if I'm not I just tweak it a bit. It's like natural selection. xD Also, when building a deck, keep in mind that it is better to have too little quanta than too much.

[kitty45]

Fahrenheit and Firebolts runs on quanta

[jumpoffduck]

How are you possitive you are going on team 

I'm probably not being on any team, and probably sitting this one out. I do however want to see  win over those overglorified elemental teams

@Jumpofftheduck
It doesn't account for Automulligan, but automulligan is simply drawing twice. I'm calculating the average draw, or average over infinite games, so automulligan doesn't apply because the second draw is, in theory the same as the first draw.

Automulligan is drawing twice only if you didn't draw any pillars the first time. I'll admit the difference is small, but it does give you an increased chance of drawing a pillar.

[Raptor6789]

the draws are 2 independent events, so the theoretical probability is the same.
first draw - 30 card deck, 7 cards drawn.
second draw - 30 card deck, 7 cards drawn.
no difference at all.

[OldTrees]

the draws are 2 independent events, so the theoretical probability is the same.
first draw - 30 card deck, 7 cards drawn.
second draw - 30 card deck, 7 cards drawn.
no difference at all.

No the second hand is only drawn if no pillars were in the first hand

Probability of no pillars is = Probability no pillars draw 1 * Probability no pillars draw 2
Probability of pillars is = Probability of Pillars draw 1 + Probability of no pillars draw 1 * Probability of pillars draw 2

N = n1 * n2
Y = y1 + n1 * y2
Y+N = y1 + n1 * y2 + n1 * n2 = y1 + n1 * (y2 + n2) = y1 + n1 = 1