How many initial hands would have a deck?

[Gocho]

I forgot a lot of my maths years ago, so I need that someone help me with this.

If I have a deck with 10 playsets (4 x 10) + 20 basic lands (forest for example). How many hands would I draw in my first 7?

[Sims]

I forgot a lot of my maths years ago, so I need that someone help me with this.

If I have a deck with 10 playsets (4 x 10) + 20 basic lands (forest for example). How many hands would I draw in my first 7?

1 hand, the hand of 7 you drew.

Alternately, no hands, because the deck has no hands.

Alternate alternately- did you mean lands?

[arebennian]

Not exactly sure what you are asking, but this might help:

http://www.mtgthesource.com/forums/showthread.php?t=15270&highlight=script

Sure it centres on lands, but it give you the chance for all amounts of land in a deck, and as such, a playset of cards as well.

[Gocho]

I didn't explain well.

If I have a deck of 60 cards, How many DIFFERENTS hands would I draw?

With 60 Forests, I only have a possible hand: 7 forests
With 59 Forests + 1 Llanowar Elf, I have five possible hands:
7 forest (a lot of times)
6 forest + 1 Elf
5 forests + 2 Elves
4 forests + 3 Elves
3 forests + 4 Elves

No matter how many times I draw the first seven, I can't get another hand.
So, If I have a deck with 10 playsets + 20 lands, How many differents hands would I draw?

[DrJones]

I know what you say, but it's hard to get 4 elves in a deck with only one. Also, I don't see what would be the usefulness of knowing that info. Try chaining statistics by suming the hands that have X forests and 7-X non-forest, and then start substituting non-forest by Y Llanowar Elf and 7-X-Y non-forest, non-llanowar elf and so. Then tell me if it was worth calculating it.

[DrJones]

Edit: Alternatively, just translate it to variables:
a: forests
b: llanowar elf
c: tarmogoyf
d, e, f, etc:

A hand will be a chain of letters, let's say: aaaaaaa, aabbccc, up to abcdefg.

Calculate the number of different patrons you can fabricate, taking into account that aaaaaab is the same than abaaaaa. Order not being significant makes it not as easy to calculate as simply doing (number of variables)^7. But the actual formula wasn't very difficult either, it's just that I can't remember it right now.

Edit: I just remember it. You have to divide (number of variables)^7 by the number of different permutations of 7 elements, that is, 7x6x5x4x3x2x1 = 7!

Edit2: I leave to others the work of pruning the permutations where a letter other than "a" appears more than four times.

[ParkerLewis]

Are you searching for the exact number, or just an estimate ?

The exact number would be complicated, but an estimate is easily possible. In your case of 20 identical lands + 10 quadruple of cards :

At first, let's ignore all hands were you have two identical non-land cards. The amount of other hands is as follow :

Let k be the number of non-land cards your initial hand has. If I'm not mistaken, the searched number of hands is Sum(from k=0 to k=7) of binomial(10,k), which is 968 (so you at least know the actual total will be more than that).

Now if you add hands which have multiple identical non-land spells... It would be difficult (it's possible to have multiple multpiles, etc), but a wild guesstimate of mine would be you'd end up with around 2000 hands total.

[Forbiddian]

The formula for a combination is n!/(k!*(n-k)!), or (n+k-1)!/(k!*(n-1)!) with replacement..

If you have 60 unique cards, the number of combinations for 7 draws is 3.86*10^8. Cool.

Let L = Land.
Let A-J=Non-land cards.

But you don't have 60 unique cards, you really have 11 unique cards. So let's work from there. With replacement, there are 19448 different ways to draw 11 things 7 times.

That's a slight overestimation, because it assumes that a hand like AAAAAAA exists (with replacement), when you obviously couldn't draw more than 4 non-lands of the same type.

So the question becomes: How many of those hands have more than 4 of a single type? And then subtract that number from 19448.

Let's just look at AAAAA + XX. At the end, we'll multiply this number by 10 for each of the other possibilities, BBBBB + XX, etc.

There are 66 ways to choose 2 from 11 elements, so that's how many ways you can have 5 or more As in a single hand. We multiply that by 10 because there are 10 different cards (A-J), giving us 660 non-existent hands.

19448 is an overestimation. It's an overestimation by 660.

The number of hands you can draw is 18788.

[fallenphoenix]

That's way to much since you still count several copies of the same card as individual, unique cards.

E.g. a hand of 3 Goyfs and 4 Forests should be one single possible hand.
With you calculation you get:
Hand 1: Goyf1, Goyf2, Goyf3, (4 Forest)
Hand 2: Goyf1, Goyf2, Goyf4, (4 Forest)
Hand 3: Goyf1, Goyf3, Goyf4, (4 Forest)
Hand 4: Goyf2, Goyf3, Goyf4, (4 Forest)
-> 4 Possible hands, not taking into account that the same is basically true for the basic lands.

I'd have to think a bit about the problem, there should be ~3-6k different possibilities, estimated.

[ParkerLewis]

That's way to much since you still count several copies of the same card as individual, unique cards.

Actually, Forbiddian is right (also, contrary to your belief, he's not counting arrangements, he is indeed correctly counting combinations).

I've mistakenly thought that seeing how multiple copies of a same 4some were a (relatively) rare sight, they probably weren't that much more hands with multiples than without, and contented myself with calculating number of hands without multiples (other than lands). Which is wrong since I didn't take into account the fact that each indivual one is simply far less probable.

The funny thing is that, among those 18788, we can see than only 968 of them are hands without any non-land multiples, which at first seems counterintuitive to our everyday experience. It just so happens that hands with multiples, even far more numerous in cardinal (~18 to 1 ratio), are so less probable that we still end up seeing more hands without multiples than hands with any.

[Forbiddian]

That's way to much since you still count several copies of the same card as individual, unique cards.

E.g. a hand of 3 Goyfs and 4 Forests should be one single possible hand.
With you calculation you get:
Hand 1: Goyf1, Goyf2, Goyf3, (4 Forest)
Hand 2: Goyf1, Goyf2, Goyf4, (4 Forest)
Hand 3: Goyf1, Goyf3, Goyf4, (4 Forest)
Hand 4: Goyf2, Goyf3, Goyf4, (4 Forest)
-> 4 Possible hands, not taking into account that the same is basically true for the basic lands.

I'd have to think a bit about the problem, there should be ~3-6k different possibilities, estimated.

How did you come up with ~3-6k?

I mean, obviously you're wrong (at the very least in analyzing my calculations -- I'm doing combinations and not permutations, so if I screwed up, it wasn't in the way you said it was), but I'm wondering how you got to that number, and why you're so convinced of its accuracy that you assume I was doing permutations.

I could have easily messed up, though. I think what I did was right, but whenever I invent a method to solve a problem, there's a chance that I fucked something up, so I was actually looking for someone to check my work, too.

[Dan Turner]

Actually if you count each card as a separate and unique item then it is a simple math problem.

60*59*58*57*56*55*54=1,946,482,876,800 possible combination's with 60 Unique cards.

[Phoenix Ignition]

Actually if you count each card as a separate and unique item then it is a simple math problem.

60*59*58*57*56*55*54=1,946,482,876,800 possible combination's with 60 Unique cards.

But that wasn't at all the question and is quite obviously not what he wanted to know. No one cares if they get Forest#1 or Forest #2...

[Dan Turner]

But that wasn't at all the question and is quite obviously not what he wanted to know. No one cares if they get Forest#1 or Forest #2...

ok well I will try and figure out a formula for it in the morning I got to get some sleep LONG week in front of me.

[Nihil Credo]

The funny thing is that, among those 18788, we can see than only 968 of them are hands without any non-land multiples, which at first seems counterintuitive to our everyday experience. It just so happens that hands with multiples, even far more numerous in cardinal (~18 to 1 ratio), are so less probable that we still end up seeing more hands without multiples than hands with any.

I'm not so sure that they're that unlikely. Remember the birthday problem? If you have 365 different "cards" (birth dates), you only need to "draw 23 cards" (pick 23 random people) to have better than even odds of two of them "being the same card" (sharing a birthday).

Keeping in mind that the birthday problem uses replacement, applying the formula for a deck with 11 different cards still gives a whopping 85% likelihood. For a more realistic number, say 21 different cards, you get around 65%. These numbers need to be adjusted down to account for the lack of replacement in Magic, of course.

[Forbiddian]

Actually if you count each card as a separate and unique item then it is a simple math problem.

60*59*58*57*56*55*54=1,946,482,876,800 possible combination's with 60 Unique cards.

No, that's a permutation. The hand: Forest, Tarmogoyf, Brainstorm is the same as the hand Brainstorm, Forest, Tarmogoyf. So you'd have to go through and remove duplicates. Or you could use the formula for combinations.

For instance, say instead of 60 cards in the deck, there are only 7.

There's obviously only one hand that you can draw, although with your formula, you'd get 7*6*5*4*3*2 = 5040.



ok well I will try and figure out a formula for it in the morning I got to get some sleep LONG week in front of me.

Incidentally, read my post where I solve the problem.

[Gocho]

I know what you say, but it's hard to get 4 elves in a deck with only one. Also, I don't see what would be the usefulness of knowing that info. Try chaining statistics by suming the hands that have X forests and 7-X non-forest, and then start substituting non-forest by Y Llanowar Elf and 7-X-Y non-forest, non-llanowar elf and so. Then tell me if it was worth calculating it.

Upps, I was thinkig in 56 Forest + 4 Elves and write only one :cry:

The formula for a combination is n!/(k!*(n-k)!), or (n+k-1)!/(k!*(n-1)!) with replacement..

If you have 60 unique cards, the number of combinations for 7 draws is 3.86*10^8. Cool.

Let L = Land.
Let A-J=Non-land cards.

But you don't have 60 unique cards, you really have 11 unique cards. So let's work from there. With replacement, there are 19448 different ways to draw 11 things 7 times.

That's a slight overestimation, because it assumes that a hand like AAAAAAA exists (with replacement), when you obviously couldn't draw more than 4 non-lands of the same type.

So the question becomes: How many of those hands have more than 4 of a single type? And then subtract that number from 19448.

Let's just look at AAAAA + XX. At the end, we'll multiply this number by 10 for each of the other possibilities, BBBBB + XX, etc.

There are 66 ways to choose 2 from 11 elements, so that's how many ways you can have 5 or more As in a single hand. We multiply that by 10 because there are 10 different cards (A-J), giving us 660 non-existent hands.

19448 is an overestimation. It's an overestimation by 660.

The number of hands you can draw is 18788.

Thanks!

[Maveric78f]

The formula for a combination is n!/(k!*(n-k)!), or (n+k-1)!/(k!*(n-1)!) with replacement..
I just don't get this.

n!/(k!*(n-k)!) is the formula without replacement. No doubt.

But the formula with replacement you used is just the same with n -> n+k-1. I don't get from where you get this. If it was an approximation, 60^k/k! seems better (but still an approximation).

[crow_mw]

This is not an approximation, this is an exact formula for combination with replacement.

I have double checked Forbiddian calculations and they seem correct to me. My only remark is that while this is the number of possible hands, you are not equally likely to draw each one of them. This is because you are more likely to draw one of the land cards, than the other 10 cards.

[Distortion]

Forbiddian's method looks good, so I'm probably not going to do this, but here's a correct approach for anyone with crazy amounts of time (I lost access to my math software when I came home, so I would have to do this by hand). This is actually a multinomial hypergeometric distribution with 11 variables (10 of 4 1 of 20). You want to sum over all the ways that you can make the choice variables add to 7. If anyone can use Fortran/Matlab/Mathematica it should be doable, if tedious to write out.

[Maveric78f]

This is not an approximation, this is an exact formula for combination with replacement.
I'm interested about the proof. I tried to figure out a way to compute this but I did not find it. For such a clean result, I expect an intuitive demonstration.

[Gui]

That seems a lot complicated!
You want to consider all the 20 lands as if they were the same, and also the same for playsets?

I mean, one example hand possibility would have 1 forest, but that would be 20 hands that are considered to be the same. Also, you want to ignore the alternate order results... Yeah, damn hard... call a math teacher!

But if noone solve this for you, I'll "monte carlo" that latter then...

[Distortion]

I think this is still on-topic since it provides the justification for someone else's argument, so I don't think this needs to be done in a PM. Sorry if I am wrong.

There are a number of proofs that n+k-1 C k is the fomula for choice with replacement. The easiest way I know uses generating functions. How many ways are there to choose n items from a 1 element set with repetition allowed? Clearly there is 1 way to choose 0, 1 way to choose 1, 1 way to choose 2 (you can choose numbers above 1 because of the replacement, but you always have to choose the same item, since there is only one in the set). It follows easily that for all n there is exactly 1 way to choose n items from a 1 element set with replacement. The generating function is thus <1, 1, 1, ...> = 1/(1-x). By convolution the generating function of choice from an n-element set with repetition is 1/(1-x)^n. MacLaurin expansion of this function yields that the coefficient of k is n + k -1 choose k.

Gui, if you have software that can run monte-carlo, then you can probably just use the method I described above to get an answer I assure you is correct. I hypothesize that Forbiddian's method is also correct, however, so that should not be necessary.

[Gui]

Gui, if you have software that can run monte-carlo, then you can probably just use the method I described above to get an answer I assure you is correct. I hypothesize that Forbiddian's method is also correct, however, so that should not be necessary.

Well, it's not actually monte-carlo, but it's a simulation... I'll have to come up with some crazy logic for it to work though... as soon as I get time. xD

I'll be impressed (with his skills) if my results matches forbiddian's ^^

EDIT: Actually I'll be impressed if nobody (mostly myself) messes up anything, I guess xD

[crow_mw]

I'm interested about the proof. I tried to figure out a way to compute this but I did not find it. For such a clean result, I expect an intuitive demonstration.

Assume we have a set {1,2..n}. (Say card1=1,card2=2,...,land=n). Each k-element combination (with replacement) can be sorted such that its elements (a1,a2,...,ak) fulfill

1 <= a1 <= a2 <= ... <= n.

this means that

0 < a1 < a2+1 < ... < a_{k-1} + k - 2 < ak +k -1 < n + k

Now, lets replace the elements of above equation with some other parameter b such that b1=a1, b2=a2+1 and so on.

0 < b1 < b2 < ... < b_{k-1} < bk < n+k

Note, that this means that no bi is equal to some bj (no replacements). Now, the number of solutions to the 'b' equation is equal to the number of combinations without replacement in the set with n+k-1 elements.

I understand the above explanation might be a bit complicated, but the thing is that as you have correctly observed this formula is the number of combinations without replacement in the set with n+k-1 elements. The increase in size of our set is caused by going from <= (replacements allowed) to < sign (replacements not allowed)

[ParkerLewis]

All right, if that can help convince that the 18788 that Forbiddian came to is true, here is another much more tedious, but different way to calculate it.

* Reminder : number of hands without any non-land duplicates : 968

* Let us count the number of hands with EXACTLY ONE double copy of a non-land card (and no other non-land multiples). Let k be the number of non-land cards in this hand. The result is, if I'm not mistaken

Sum(from k=2 to k=7) of binomial(10,1)*binomial(9,k-2)=10*Sum(from k=2 to k=7) of binomial(9,k-2)=3820.

* Let us count the number of hands with EXACTLY ONE triple copy of a non-land card (and no other non-land multiples) :

Sum(from k=3 to k=7) of binomial(10,1)*binomial(9,k-3)=10*Sum(from k=3 to k=7) of binomial(9,k-3)=2560.

* Let us count the number of hands with EXACTLY ONE quadruple copy of a non-land card (and no other non-land multiples) :

Sum(from k=4 to k=7) of binomial(10,1)*binomial(9,k-4)=10*Sum(from k=4 to k=7) of binomial(9,k-4)=1300.

* Let us count the number of hands with EXACTLY TWO double copies of different non-land cards (and no other non-land multiples) :

Sum(from k=4 to k=7) of binomial(10,2)*binomial(8,k-4)=45*Sum(from k=4 to k=7) of binomial(8,k-4)=4185.

* Let us count the number of hands with EXACTLY THREE double copies of different non-land cards (and no other non-land multiples, which isn't possible anyway) :

Sum(from k=6 to k=7) of binomial(10,3)*binomial(7,k-6)=120*Sum(from k=6 to k=7) of binomial(7,k-6)=960.

* Let us count the number of hands with EXACTLY ONE double copy and EXACTLY ONE triple copy of you know the following :

Sum(from k=5 to k=7) of binomial(10,1)*binomial(9,1)*binomial(8,k-5)=90*Sum(from k=5 to k=7) of binomial(8,k-5)=3330.

* Let us count the number of hands with EXACTLY ONE double copy and EXACTLY ONE quadruple copy of you know the following :

Sum(from k=6 to k=7) of binomial(10,1)*binomial(9,1)*binomial(8,k-6)=90*Sum(from k=5 to k=7) of binomial(8,k-6)=810.

* Let us count the number of hands with EXACTLY TWO triple copies of you know the following :

Sum(from k=6 to k=7) of binomial(10,2)*binomial(8,k-6)=45*Sum(from k=5 to k=7) of binomial(8,k-6)=405.

* Let us count the number of hands with EXACTLY ONE triple copy and EXACTLY ONE quadruple copy of you know the following :

binomial(10,1)*binomial(9,1)=90.

* Let us count the number of hands with EXACTLY TWO double copies and EXACTLY ONE triple copy of you know the following :

binomial(10,2)*binomial(8,1)=360.

Total of all of this : 18788, which is exactly equal to the previous result that was directly using the formula for combination with replacement.

[Kagehisa]

I'm not sure I understand the opening post...

I try to reply anyways..

You can only have 4 copies of a card except for basics land. Everybody knows that. Then, if I play 4 Counterspell for example, how many chances do i have to begin the game with (7 cards) ? I don't know...

My maths ( what i think maths are...) :

60/7=8.57142... say 8.6
This means I have to play with 8.6 or 9 copies of Counterspell to have it in the first hand. (most of times it would be 4 FoW+4Counterspell+1manaleak) It is all about redondancy.

Another example
I have to reach 4 Plains in 4 turns to cast WoG. How many Plains should i play ?

In 4 turns, I draw 3 cards ( draw phase ) if I've started the game. It means I've drawn 7+3=10 cards. In the 10 cards I've drawn, 4 amongs them have to be plains.

(60/10)*4=24
(60/(7+4))*4=21.818181... this result is for the situation you have not started the game, it means you have drawn 7+4=11 cards in turn 4.

If you have to cast turn 4 a WoG, play between 22 and 24 lands. That os why in landstill for example, it plays 23 lands plus Eternal Dragon (24 lands) plus 4 Brainstrom (played end of turn 3) to increase the chance to drop the 4th land turn 4.

And how many WoG do I need ?
Turn 4, I draw 7+3=10 cards One of these cards have to be a WoG.
with a deck of 60 cards i can make 60/10=6 groups of 10 cards and only one cards among these 10 cards need to be a WoG : 1*6=6. I have to play 6 WoG to be able (probable) to cast a WoG turn 4 So it means i have to play 4 WoG and 2 Damnation for this example.



I hope it can help... and with those simple maths, you can determine how many 1CMC you need, how many 2CMC etc...
it looks like 9 cards with 1CMC.
7.5 cards with 2CMC if you wanna a 2CMC turn 2 etc...

[Distortion]

Parker, I'm impressed with your patience. That's the method I was describing above, though written out without the 11 part summation.

As an aside, I assume by binomial you mean the binomial coefficient (generally written C for choice) and not the binomial distribution. Regardless, nice work on that.

[Forbiddian]

I hope it can help... and with those simple maths, you can determine how many 1CMC you need, how many 2CMC etc...
it looks like 9 cards with 1CMC.
7.5 cards with 2CMC if you wanna a 2CMC turn 2 etc...

Why would you do "simple maths" when Maveric78f exists?

Just ask him to do it.

[4eak]

I haven't had internet but for very brief moments over the weekend, and I only have it now for a brief moment. I'll need to be quick.

I had to write up a more generalized solution for generating these combinations of hands to solve for mulligans. Here's an excerpt (http://pastie.org/752632), which can give you the answer, which in this case is just the number of combinations. If you uncomment the last two lines, you'll be able to see all the hands generated.

You can quickly compute the answer (without much work) to this problem for any deck with this tool.




peace,
4eak

[AngryTroll]

@Forbidden and 4eak: Great work, thanks!





But if noone solve this for you, I'll "monte carlo" that latter then...

How would you Monte Carlo this problem? It doesn't seem to fall into the category of problems that are easy to solve with Monte Carlo.

[AngryTroll]

.

[Gui]

Well, it's not actually monte-carlo, but it's a simulation... I'll have to come up with some crazy logic for it to work though... as soon as I get time. xD

Just Quoting myself so that noone ask me why monte-carlo, since I did put it between " ".

[Gocho]

I haven't had internet but for very brief moments over the weekend, and I only have it now for a brief moment. I'll need to be quick.

I had to write up a more generalized solution for generating these combinations of hands to solve for mulligans. Here's an excerpt (http://pastie.org/752632), which can give you the answer, which in this case is just the number of combinations. If you uncomment the last two lines, you'll be able to see all the hands generated.

You can quickly compute the answer (without much work) to this problem for any deck with this tool.




peace,
4eak

I try it and my python 2.5 says that " 'module' object has no attribute 'combinations' "
Which version of python do you use? There aren't any combinations function in my itertools library.

EDIT:
I just see it in the code: python 2.6

[Maveric78f]

Assume we have a set {1,2..n}. (Say card1=1,card2=2,...,land=n). Each k-element combination (with replacement) can be sorted such that its elements (a1,a2,...,ak) fulfill

1 <= a1 <= a2 <= ... <= n.

this means that

0 < a1 < a2+1 < ... < a_{k-1} + k - 2 < ak +k -1 < n + k

Now, lets replace the elements of above equation with some other parameter b such that b1=a1, b2=a2+1 and so on.

0 < b1 < b2 < ... < b_{k-1} < bk < n+k

Note, that this means that no bi is equal to some bj (no replacements). Now, the number of solutions to the 'b' equation is equal to the number of combinations without replacement in the set with n+k-1 elements.

I understand the above explanation might be a bit complicated, but the thing is that as you have correctly observed this formula is the number of combinations without replacement in the set with n+k-1 elements. The increase in size of our set is caused by going from <= (replacements allowed) to < sign (replacements not allowed)
Brilliant. Thanks.

Forbiddian > What's your problem?

[Skeggi]

Thanks for the headache guys, that's quite some math. I'm sure it's all true.

Here's something I don't understand though: what's the point in knowing how many different opening hands there are? I'm much more interested in the chances of a Force of Will and another blue card in my opening hand for instance, but isn't the amount of different opening hands completely irrelevant?

[Forbiddian]

Here's something I don't understand though: what's the point in knowing how many different opening hands there are? I'm much more interested in the chances of a Force of Will and another blue card in my opening hand for instance, but isn't the amount of different opening hands completely irrelevant?

The OP explained, but I didn't really understand. It was just an interesting problem.

If you're interested in the Force of Will problem, I solved that in some thread. I think the Merfolk thread.



Forbiddian > What's your problem?

You've solved 2 or 3 hard problems I had no idea how to work in my recent memory. So I'm just pointing out that you're good at math and that people could ask you for help doing anything more complicated than counting up combinations. Sorry, I guess you're not into math charity, I can ship the shit I can't do to other people in the future.

[Gui]

The number of hands you can draw is 18788.

Vouch this result. My simulation got the same result. Good job you all ^^'

Even though I can't find a good use to this result. xD

[Maveric78f]

You've solved 2 or 3 hard problems I had no idea how to work in my recent memory. So I'm just pointing out that you're good at math and that people could ask you for help doing anything more complicated than counting up combinations. Sorry, I guess you're not into math charity, I can ship the shit I can't do to other people in the future.

Since the whole thing you quoted was complete mathematical nonsense and that I know you knew it. And since you called it "easy math", I got paranoid.

Actually, people need to know the whole Kagehisa post was complete nonsense. I have nothing against you, but before providing food to people it's better to know first if it's edible or everyone is gonna be sick.

[Lothian]

I'm interested about the proof. I tried to figure out a way to compute this but I did not find it. For such a clean result, I expect an intuitive demonstration.

Just ask Richard Garfield Mav ;-)



In 1985, he received a Bachelor of Science degree in computer mathematics. He joined Bell Laboratories, then decided to continue his education and attended the University of Pennsylvania, and studied combinatorial mathematics.

[Koby]

EDIT: reading is tech.

BTW, this link might help us derive the values:
http://en.wikipedia.org/wiki/Poker_probability#Derivation_of_frequencies_of_7-card_poker_hands

[Gui]

EDIT: reading is tech.

BTW, this link might help us derive the values:
http://en.wikipedia.org/wiki/Poker_probability#Derivation_of_frequencies_of_7-card_poker_hands

Not probable, since this isn't the kind of analysis we should be doing, even for poker... we should always consider the equal results as better chances of getting that result. That's why this hole math got no purpose at all...

[4eak]

@ "Purpose of this math" questions:


Here's something I don't understand though: what's the point in knowing how many different opening hands there are? I'm much more interested in the chances of a Force of Will and another blue card in my opening hand for instance, but isn't the amount of different opening hands completely irrelevant?


we should always consider the equal results as better chances of getting that result. That's why this hole math got no purpose at all...

I personally found the math useful simply as a way to double check that I hadn't screwed up in my programming.

I alluded to one general use for it (this is the only one I know) in my previous post: Providing justification for limiting the population of hands which we would evaluate when generating mulligan rules.

Through the use of this math, we are in a better position to realize how important it is for us to only generate the unique and non-duplicate hand combinations for the sake of creating mulligan rules. For anyone interested in creating a very, very thorough set of mulligan rules, using these combinations (which are much smaller quantity) rather than all possible 7 card hands (including duplicate hands with different ordering) saves us time.

For our burn simulations (http://www.mtgthesource.com/forums/showpost.php?p=410258&postcount=91), we saw a ~10% difference in the effectiveness of our decks when we moved from simple "no-lander" mulligan rules to the near perfect mulligan rules generated from testing the complete set of unique hands combinations. It would not have been nearly as easy to generate those mulligan rules if we hadn't used the unique and non-duplicate hand combinations.

If you aren't interested in simulations, there still could be use for those trying to define mulligan rules by hand and experience. For more stagnant metagames and/or crystallized decklists, improvement to archetypes often cannot come through innovation in the card choices of the decklists themselves, but rather through honed and fine-tuned mulligan and sideboarding decisions. Having a unique set of hand combinations to comb through would make it much easier to gain and show insight into mulliganing with crystallized decklists or within the context of a stagnant metagame.

It would be extremely easy to use rules to whittle down the unique sets to a much smaller subset of "questionable" mulligan hands. For example, I'd always keep a 2x Mountain, Vial, Lackey, Ringleader hand in game 1 against an unknown (and frankly, that goes for almost all game 2's and 3's). I could create several "auto-keep" hand rules to make a subset of the unique hand combinations which aren't so straightforward, and which might be considered "borderline" or "questionable".

Having an expert or a group of experts go through all the unique "questionable" hands could show newer players the general consensus among how to play the deck and even provide more targeted opportunities to point out the finer, delicate, and detailed aspects of the art of mulliganing with a particular deck.

That sounds like a lot of work, and I doubt it would concern most. It isn't useless though; the information is relevant.




peace,
4eak

[Gui]

4eak, all said is true, but on those results we got, we can't use them to consider muligan. We have to use other math. Because that one disconsider the diference between drawing forest1 and forest2 or forest3 and forest4... that means we don't have the right probability to estimate the amount of muligans, for example.
I think this math is great, and I did the program to check, but I wouldn't use those results to consider anything in my deck. I use probabilities to check things like "how many of this will I use" or "how many blues with fow do I need" but the number of possible hands isn't useful to that.