Probabilities and Combinations

[Keeps]

OP, there isn't just one equation for this, there is at least two, possibly many more.
This is a dependent probability equation based on draw factors.
Each draw is a different event and so a different equation.

And the first question is a certain combination of cards repeatable. For example, is the combo two novas, and a shard of patience?  or are each card distinct. 
The equations change because the odds of certain draws change
Assuming you want distinct cards.  Here are the equations:

X = # card 1
Y = # card 2
Z = # card 3
T = Total Cards in Deck

Draw One
P1a = X/T
P1b = Y/T
P1c = Z/T
P1 = P1a + P1b + P1c
Draw Two
//* Odds that X is picked but was not picked before *//
P2aaa = X * - (1*(1 - P1a))/ T-1
//* Odds that X was picked and was picked before*//
P2aab = X *- 1(P1a) / T - 1
//* Odds that X is picked*//
P2aa = P2aa + P2ab
//* Odds that X is picked and you were missing it before *//
P2a = P2aa * P2aaa

Repeat for Y and Z to get:
P2b = P2ba * P2baa
and
P2c = P2ca * P2caa
//* The odds you now got two of the three cards of the combination *//
P2 = P1a + P2a + P2b + P2c

Draw Three
//* Odds that X is picked but was not picked before *//
P3aaa = X * - (1*((1 - P1a) + 1*(1 - P2aa)))/ T-2
//* Odds that X is picked but was picked before *//
P3aab = X - (1*P1a) - (1*P2aa) / T- 2
//* Odds that X is picked*//
P3aa = P3aa + P3ab
//* Odds that X is picked and you were missing it before *//
P3a = P1aa * P2aaa * P3aaa

Repeat for Y and Z to get:
P3b = P3ba * P3baa
and
P3c = P3ca * P3caa

//* The odds you now got all three cards for the combo of the three cards of the combination *//
P3 = P1a + P2a + P2b + P2c + P3a + P3b + P3c

Continue to repeat the process for the number of cards drawn.

[CrockettRocket]

Does this account for bravery, precognition, and I would also include silence.

[Keeps]

Does what account for bravery, precognition, and silence?

[CrockettRocket]

the multiple equations

[Keeps]

I think you might be missing the point of the comment.

When just asking what are the odds of drawing a card, it's the same equation up to the event of getting the card.
But when you start asking what are the odds of drawing a card combination you have to understand that each draw is a separate event.  Because of this, the conditions change at each event and so the equation for the event has to change a bit. 

The above showed a method to get the odds of a 3 card combination all in your opening hand.  Before play.  If you were asking for a 4 card or a 2 card combination the question changes and so the equations change, but stay in the same vein.  Also at play the equation changes, but the gist is the same.  In other words, what are the conditions of your next draw draw after the opponent plays will mean a new equation for that turn.

If Silence is played against you, no event
If Nightmare was played against you and hand is full, no event
If Bravery is played against you (calculate next 1 to 3 draw events for bravery) if hand is full, no event at normal draw time.
Calculate odds of normal draw if event is possible
If you can play precog, calculate next 1 draw event odds for that draw.
If you play SoBr, proceed with calculating odds of the next 1 to 3 draw events.
Etc.

Also once you get playing, you would need to know the opponents deck to be able to calculate the odds of every turn.  Otherwise it's a null condition, leaving things a dual state like a famous cat or an entangled string.

[Savage]

To be honest, you guys sound confusing. Let's assume we are not using hourglass, precognition, SoB, silence, or SoS. Also, if we include these factors, did we take into consideration that when using SoB that sanctuary cancels it out? (Maybe I missed this).

Card 1=A
Card 2=B
Card 3=C
As stated before, I will use 'T' as total number of cards in deck
Capital X will represent the amount of cards of A B and C you have in the deck

Probability of A is not A/(T-1) as someone stated in another post.
P(A)= {XA/T + XA/(t-1) + XA/(t-2)..... + XA/(t-7)} [You draw 7 cards on the first turn and so the sum of the first 7 draws counts as your 'first draw.'] then begin adding XA/(t-8) + XA/(t-9)...where each of these elements are a single event draw by themselves. I do understand that if we were to actually calculate the chances for drawing a certain card given how many of it we have and how many cards are in the deck that we would only add up the sums to the point of how likely is it to draw that card on this turn.

Before I continue, does this look correct? I am only doing P(A) at the moment. My thinking of that first seven card draw is throwing me though.

[Keeps]

Savage,
Sorry to sound confusing to you.

You are correct for a single card:
As I showed P1a = X/T which is your P(A)= {XA/T} For the first draw.
And if looking for the odds of drawing a specific card on the first turn, (P(A)= {XA/T + XA/(t-1) + XA/(t-2)..... + XA/(t-7)}) is correct.

That is standard probability, and is easy.

To go into probability of combinations of cards though you will find it becomes a lot more complex, hence my: "This is a dependent probability equation based on draw factors. Each draw is a different event and so a different equation." post about 4 or 5 posts up from this one.

What Brandon started getting into gets into the realm of AI.  Probabilities based on a large set of variable actions.

TL;DR  Odd of drawing a single card are easy to calculate, but dependent probability for combinations, you have to look at the probability of each event separately, there is no master equation.

[Savage]

Thank you for the confirmation:

Couldn't we calculate the probability of each independent card up to a certain number of draws and then multiply all 3 together? Again I am assuming my initial conditions at the moment?

Like if I had 1 of A, 1 of B, 1 of C, A=/=B, A=/=C, B=/=C, then our chances would be (1/60+1/59+1/58+1/57+1/56+1/55+1/54)^3? (I think the equation should change on the first event to t-6 since the 7nth card likely hood becomes 1/54) Also, I am doing the beginning of the game.

Again I am just establishing some preliminary stuff...but need confirmation on my calculations.

[CuCN]

If you draw A, your likelihood of drawing B goes down slightly because that A is taking up a spot in your hand that B now can't be in. Therefore the chance of drawing A and B is slightly less than the product of the chances of drawing them individually.

[Keeps]

I do think my thoughts above could be simplified a lot, the equations can too, part of what I put down was an extended version to 1.) Show my point.  2.) Show the various questions that could be asked.

My point was simply it depends on each question and each event, so since you seem to understand this we could continue on expanding this, a simplified version or subset of equations could be helpful to those here.

So,

Your question for your second thought is also correct for the conditions of 1 of each card 3 cards for the combination 60 total cards.

But you will need to separate out the power of the ^3 for each one once you go beyond that, as A,B,C can have different counts, also the required cards for a combination so 2 of nova to play a SoPa, could have different accounts, if you want to simplify that to the Count of (A,B,C) is the Same and the required Count(Combo(A,B,C) is always 1, we could produce a much more simplified model along the lines you are going for.

So it leaves only the challenge of simplifying if you draw A, B, or C...  and you don't need it anymore, what are the odds of drawing the cards you are missing,

[Savage]

Again thank you for confirming.

Ok, so to extend, we can actually model an equation based off this model ==> (1/60+1/59+1/58+1/57+1/56+1/55+1/54)^3

Now, lets only talk about the person who goes first in a game. They will only have 7 cards. There is, how ever, the possibility to get even more cards on this turn. Again, lets still use a 60 card deck.

Nightmare: has no effect (if we were talking about the second person, this actually would add to the probability)
SoB: will only draw 1 card because our opponent has 1 card slot
SoS: You could technically use up to all 6 of them assuming they draw you pillars and pillars (but to calculate that out would be a waste of time IMO)
Precognition: Can add up to 1-6 cards
Hourglass: has no effect if we stay on the 'first turn' assumption
Fractal: if you are looking for 6 of the same card, this could affect the probability

I believe we can model a good equation based off the 'first turn' and talking about the player going first.
If we stick to this, hourglass and nightmare become useless and since fractal adds such a minuscule adjustment I think we can throw out that option as having any real difference.

So, lets talk about SoB (and SoB without any of the others)
How would you model the changes in the equation? Again, 60 card deck and also, what parameters should we set? (type of deck and depending on what type of deck, a good pillar to non pillar card ratio).

[Keeps]

Again thank you for confirming.

Ok, so to extend, we can actually model an equation based off this model ==> (1/60+1/59+1/58+1/57+1/56+1/55+1/54)^3

Just to clarify, Only if it's 1 card of each distinct card in a 3 card combo is that correct.  If that's the question then this can be used, otherwise, it can't.

Now, lets only talk about the person who goes first in a game. They will only have 7 cards. There is, how ever, the possibility to get even more cards on this turn. Again, lets still use a 60 card deck.

That's fine so our model now has these assumptions:
1.) The player goes first
2.) The player has a 3 card combo, The 3 cards are in the deck of only 1 instance.
3.) The deck has 60 cards.

Card that can immediate cause addition draw events
Time Pillar, Pendulum or Nova + Precognition can generate 1 draw event
Fire Pillar, Pendulum or Nova + SoBr can generate up to 1 draw event

On this turn

Cards that can generate the required card for the combination
If SoS has said card in it's list
2 Novas or 2 Entropy Pillars, Pendulums + SoS will have a 3 card generation.

Since per the model the desired card combo has 3 distinct cards, Fractal does not apply



Now regardless you are getting ahead of yourself, because you haven't answered the last challenge I stated.   Because of that, you won't realize what is needed to be calculated for this event.