Probabilities and Combinations

[Savage]

Well, we wouldn't have to separate the power of 3, but rather what is being cubed. 1/60+1/59+1/58+1/57+1/56+1/55+1/54 is the chances of 1 event on the very first turn, no outside factors. Now, if we begin to include SoB, assuming all cards are upgraded, we would need 1 fire card SoB would only profit 1 card since your opponents hands would be filled and so only 1 can be used, so even if you had 6 in your deck, only 1 becomes useful (and so SoB is a card draw event on it's own). Now, before I can calculate the new equation, I need to say:

1: this deck is using a 3:1 pillar count ratio.
2: All pillars are fire pillars
3: Based on a 60 card deck, there are 20 pillars.

So, there is a (20/60+20/59....20/54) chance of drawing just one pillar on the first draw and a

assuming 6 SoB

(6/60+6/59+...+6/54) chance of drawing a SoB on first turn.

Issue above^those sums add up to pass 100% (the pillar one). So, where did I go wrong with the calculations?

[Keeps]

Because you didn't answer the challenge above...  the answer becomes the same as I have been saying.
You have to break it down to the odds of drawing cards at each individual draw. 

Card 1
Odds of having one of the two is
26/60 The odds of drawing one of the two cards

Card 2
Odds of having one of the two is
(26/59) The odds of drawing one of the two cards if * (34/60)  The odds one of the cards wasn't already drawn + (25/59) The odds of drawing one of the two cards if * (26/60) The odds you did draw one of the two cards.

Repeat.

Also note, this is still an over simplification, as you actually have two sets of variables...  the odds the Card 1 was SoBr, versus odds Card 1 was a Pillar.

The only reason why your 1 card scenario worked is because of the math behind multiplying numbers by 1, it was an over simplification avoiding the actual complications.

[Savage]

wait...no, the sum of the first 7 cards is incorrect

the chances 1 card on the first hand is 1/54

Now, if I have 20 pillars, that is 20/54

SoB is 6/54, and so the chances of drawing 1 pillar and 1 SoB is really (6*20)/54^2=4.1% chance (rounded).

[Keeps]

No the probability should around 50%

Here this might help you:
http://www.unseelie.org/cgi-bin/cardco.cgi

[Savage]

Your link suggests a 22% chance of being drawn. Also, does this link take into consideration the element that some cards have a less chance of being drawn. My issue with your last post was that we cannot add 6+20 because we are not looking to see if we draw 1 of those cards. Both have very different draw outcomes and my 4.1% is almost based on the fact that we need a permutation of those cards.

[CrockettRocket]

...Im out. You guys are beyond me.        8) 

[Keeps]

Savage,
You are still off mathematically, but I do thank you for providing this exercise as I actually re-learned something, and it does result in a simplification of what I had way above...

So to make sure everyone is falling and to confirm you and I are on the same page...

From the beginning of that site:

One can calculate the probability of drawing at least ONE of a SET of TARGET cards from a deck by using the following formula.
in which D=deck size, T=number of target cards and H=number of cards to be drawn into the hand:

probability=1-((((D-T)!)-((D-T-H))!))/((D!)-((D-H)!)-1))

Notice how this equation looks very different from Savages.  This is the thing I was trying to get too...  what Savage was doing was actually WRONG once you expanded it beyond 1 Count of 1 Card.

I was hoping that once he realized the complexity of TWO targets that his method was off base. 
Now if we step back and use this equation for ONE target (20 Pillars)
This is what that equation says:

The probability of a Pillar showing up in a 1 card draw = 33.3%
The probability of a Pillar showing up in a 2 card draw = 55.9%
The probability of a Pillar showing up in a 3 card draw = 71.1%
...
The probability of a Pillar showing up in a 7 card draw = 95.1%

Now if we use this equation for our second target (6 SoBr)

The probability of a SoBr showing up in a 7 card draw = 54.1%

Now this is fine for 1 TARGET but only accounts for 1 TARGET. 

Explaining the equation:

Spoiler for Hidden:

First is the nomenclature
Notice the ! for those who don't know this means repeat for each EVENT, in this case an EVENT is a single DRAW of a single CARD.
probability=1-((((D-T)!)-((D-T-H))!))/((D!)-((D-H)!)-1))

Next notice the  Subtraction of the deck size in the denominator by 1, this is the Deck going down by one card for each EVENT (Draw).
probability=1-((((D-T)!)-((D-T-H))!))/((D!)-((D-H)!)-1))

Finally, notice the core of the equation is the subtraction of the probability getting the target card for EACH EVENT.
probability=1-((((D-T)!)-((D-T-H))!))/((D!)-((D-H)!)-1))

Now many posts above I had a really broken down version of what's coming next.  To add 2 or more targets, this is the thing I've learned, apparently one just needs to apply the JOIN PROPERTY to EACH EVENT.  Fortunately, this equation is already handling each event.

The Join Property says:
P(A n B) = P(A) P(B)

So now we just need to join the probability of each event occurring
Or
probability(2)= (1-((((D-T)!)-((D-T-H))!))/((D!)-((D-H)!)-1))) * (1-((((D-T(2))!)-((D-T(2)-H))!))/((D!)-((D-H)!)-1)))

Or simply in this instance

95.1% * 54.1% = 51.4% That both a SoBR and a Pillar Show up in the first 7 cards.

[CuCN]

probability=1-((((D-T)!)-((D-T-H))!))/((D!)-((D-H)!)-1))

This is wrong, and your "explanation" doesn't make mathematical sense at all. I get that the probability is 1-(((D-H) choose T)/(D choose T)), which is equal to 1-(((D-H)!(D-T)!)/(D!(D-H-T)!))

Spoiler for Explanation:

The number of ways that all T cards can be in the bottom D-H cards is (D-H) choose T, and the number of ways that all T cards can be in the deck is D choose T, so the probability that all T cards are not in the top H is ((D-H) choose T)/(D choose T). We're looking for the opposite of this, so we subtract from 1.

X choose Y is equal to X!/(Y!(X-Y)!). Using this formula to replace the chooses, the probability is 1-(((D-H)!/(T!(D-H-T)!))/(D!/(T!(D-T)!))), which simplifies to 1-(((D-H)!(D-T)!)/(D!(D-H-T)!)).

Now many posts above I had a really broken down version of what's coming next.  To add 2 or more targets, this is the thing I've learned, apparently one just needs to apply the JOIN PROPERTY to EACH EVENT.  Fortunately, this equation is already handling each event.

The Join Property says:
P(A n B) = P(A) P(B)

This is only correct for independent events, which A and B aren't.

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.

For non-independent events, we should use the inclusion-exclusion principle:
P(A∧B)=P(A)+P(B)-P(A∨B)

All of the quantities on the right side can be calculated using the formula for drawing one card.

[Keeps]

CuCN you are usually right on this stuff (along with one of the few who can simplify my tedious writtings.)  But I challenge this.

probability=1-((((D-T)!)-((D-T-H))!))/((D!)-((D-H)!)-1))[/quote]
This is also from several sites, are you saying they are all wrong?

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.
For non-independent events, we should use the inclusion-exclusion principle:
P(A∧B)=P(A)+P(B)-P(A∨B)

All of the quantities on the right side can be calculated using the formula for drawing one card.

1.) The spot doesn't matter in the hand, only if you are concerned with order. We are not here.
2.) IF Card A is drawn, the next draw actually improves the odds that Card B will be drawn next, not decreases.

Draw 1 Assuming we Draw A, Means Draw 2, the odds of drawing B actually go up Count(B)/59.

Since the two cards have no relationship to each other, thus a JOINT.

[CuCN]

probability=1-((((D-T)!)-((D-T-H))!))/((D!)-((D-H)!)-1))

This is also from several sites, are you saying they are all wrong?

Considering that the parentheses aren't even balanced (and it's written exactly this way on the sites that have this formula), yes I'm saying they are all wrong.

Since the two cards have no relationship to each other, thus a JOINT.

If your hand is only one card, do you really think that the probability of having both A and B is the product of P(A) and P(B)? Of course not, the probability of having both A and B in one card is zero. The probabilities are clearly not independent.

[Keeps]

Fuck, your right.

So in Conclusion:
1-(((D-H)!(D-T)!)/(D!(D-H-T)!))
and for each T
P(TA∧TB)=P(TA)+P(TB)-P(TA∨TB)

(or did I typo something)

Did you post this earlier, if so I am truly sorry.
If you did not post this earlier, where have you been?

Oh, and thanks for schooling me too...
I think you're my hero on this site.

[Savage]

Wait, the game draws 7 cards at the same time, so it is possible to draw SoB and a Pillar on the same event, thus the events are not independent which brings me back to our 51% (pillar % * SoB %).