Posted Tue Sep 12, 2006 6:06 pm GMT by redwizard
I know that this is a poker page but a friend told this would be a good place to post my question because it has to do with odds theory. that said I am a Magic the Gathering player and i am thrying to figure out what the formula of draw a paticular card when i have drawn X cards is. The deck is 60 cards. Idon't know how much more info you need so just ask. any help woiuld be Great and much apriciated. thanks
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Posted Tue Sep 12, 2006 6:41 pm GMT by xDiamond_CutteRx
(# of copies of your card) / (60 - (number of cards drawn))
Ie, say you draw your first seven cards and get no lands, but you know there are, say 15 lands in the deck. Your odds of drawing any land on the next card are = 15/53, or about 2.5 to 1 against.
Or, let's say you've drawn 20 cards so far, but you really need to draw the one Force of Nature in your deck. The odds of course would be 1/40, or 39-to-1 against.
Is that what you were asking?
Posted Tue Sep 12, 2006 7:35 pm GMT by redwizard
sort of but what i was looking for is what are may chances of drawing 1island,1 swampand 1 plain or 1 island 1 swamp and a land that taps for blue or white mana but thanks for the responds
Posted Tue Sep 12, 2006 8:13 pm GMT by xDiamond_CutteRx
I will try and come up with an appropriate formula for this later tonight when I come back.
Posted Wed Sep 13, 2006 10:19 am GMT by MasterShake
We played poker one time with a deck of tarot cards. Everybody died.
Posted Thu Sep 14, 2006 3:27 pm GMT by Adam Marshall
Diamond, you are an excellent poster. 
I think what you wanted here was "what are the chances of drawing 0-7 lands when I have a deck of x lands and y total cards?"
That would involve figuring out the chance to have a starting hand of 0, 1, 2, 3, 4, 5, 6, and 7 lands (assuming an initial draw of 7 cards), I think? You would have to do each of those seperately, I think. Is there any other way?
By the way, red, it's much easier to break up this calculation into just two clases, lands and non-lands. You start breaking lands up into five different things and then break up your non-lands into whatever then the math to calculate initial drawing chances are much, much tougher. In other words, nobody's going to figure that one out for ya.
Anyway, here's my solution with just algebra for the "chance of having __ lands in a starting hand, given x lands and y total cards.
| Quote: | x = number of lands in the deck
y = total number of cards in the deck
z =number of non-land cards in the deck = (y-x)
f(y) = combinations of seven cards = y! / ((y-7)! * 7!)
7 lands - ( x! / ((x-7)! * 7!) ) / f(y)
6 lands - ( x! / ((x-6)! * 6!) ) (z) / f(y)
5 lands - ( x! / ((x-5)! *5!) ) ( (z)(z-1)/2 ) / f(y)
4 lands - ( x! / ((x-4)! *4!) ) ( z! / ((z-3)! * 3!) ) / f(y)
3 lands - ( x! / ((x-3)! *3!) ) ( z! / ((z-4)! * 4!) ) / f(y)
2 lands - ( (x)(x-1)/2 ) ( z! / ((z-5)! * 5!) ) / f(y)
1 land - (x) ( z! / (z-6)! * 6! ) / f(y)
0 lands - ( z! / (z-7)! * 7! ) / f(y) |
It should spit out the probablity (multiply it by 100 to get a percent) of drawing 0 - 7 lands. Look right?
Did I use the f(y) there correctly? I think you get the idea, but I haven't used the actual function notation in a while..
I could've used the number of cards drawn as a variable (like if you took a mulligan and only drew 6 cards, you dork), but the equations would be extra messy and they're tough enough to look at as it is.
Posted Thu Sep 14, 2006 3:57 pm GMT by Sid Lambert
looks right to me....if you wanna stick in the chances for paris mulligans, you can make all those 7's a new variable...of course you'll have fewer equations since you can't have 7 lands when yer hand size is only 6...btw, mtgo has a little stats button that cranks this crap out for you....by turn number also ...however, their accuracy is to the nearest percentage
Posted Thu Sep 14, 2006 4:55 pm GMT by xDiamond_CutteRx
I'm about 99% sure those formulae are correct, Adam. I knew I was missing a factorial in the terms somewhere, and I couldn't find the right formula since I lent my Probability Theory book to a guy who needed it for summer school. Thanks for clearing up the question.
Posted Tue Sep 19, 2006 7:41 am GMT by redwizard
there are a few problems with the math. here are some that i have found
if goes into the realm of " In other words, nobody's going to figure that one out for ya. :D " or i have to learn calculus then dn't worry about it.
first one is what happens if there are exactly 7 lands in the deck because the equation makes you divide by 0. but u can use logic to find out the percantage since there is only one possible hand of getting exact same cards.
so proablity is 1/ f(y).
the other is when ther are less than 7 'lands' in the deck since the first equation( the funtction to find 7 lands) makes a negative number but in using logic it is impossible
plus i used 4 for the lands value and came up with 0 lands of about 60 % and the % for finding 1 to be about 50 % defies logic.????
i double check my number but will triple check them tommorrow
once again thanks very much for the help so far
Posted Tue Sep 19, 2006 1:22 pm GMT by groton
oh my They Still Play Magic
and Milk dude That was just the bomb post
Posted Tue Sep 19, 2006 2:28 pm GMT by Adam Marshall
| redwizard wrote: |
first one is what happens if there are exactly 7 lands in the deck because the equation makes you divide by 0. but u can use logic to find out the percantage since there is only one possible hand of getting exact same cards.
so proablity is 1/ f(y).
the other is when ther are less than 7 'lands' in the deck since the first equation( the funtction to find 7 lands) makes a negative number but in using logic it is impossible
|
True, you can break the equation. It's not quite a computer program, it's math. Which came before computers, I think.
Plugging in exact values or negative values (or imaginary numbers, or fractions, etc.) will make the result kooky. I guess I could describe a bunch of parameters for the values like x cannot be greater than y and x can't be = 0, etc. Some common sense will have to be in there sometimes when values are 0 or 1 and the equation doesn't work.
| redwizard wrote: | | plus i used 4 for the lands value and came up with 0 lands of about 60 % and the % for finding 1 to be about 50 % defies logic.???? |
I think the only way to respond to that is, "what?"
Posted Tue Sep 19, 2006 2:59 pm GMT by Sid Lambert
redwizard...i like that your profile says
| Quote: | | Interests: other games manly settl'ing |
i like settling too, but i dunno if its manly
Posted Thu Nov 16, 2006 7:38 am GMT by red_pen
| Quote: | | first one is what happens if there are exactly 7 lands in the deck because the equation makes you divide by 0. |
0! is by definition equal to 1 so the equation holds.
Posted Mon Mar 26, 2007 1:41 pm GMT by davepoker
The answer depends on the amount of multi color lands your deck runs. The more tundras,karpulsan forests, and stomping grounds you run, the more likely it is to draw on more colors. In addition, cards like flooded strand improve your chances as well of getting the right colors.
As an aside, cards like birds of paradise that require you to wait a turn for mana do not count towards your mana sources.
theres a book called advanced mtg players guide which has some great insight. specifically the section by henry stern, and his hymn hymn i win deck.
Posted Mon Mar 26, 2007 3:07 pm GMT by Geno
Brain. Hurts So. Much.
Posted Mon Mar 26, 2007 7:03 pm GMT by davepoker
I guess you dont play magic geno.
lol
comparable to poker logic/ game theory though
Posted Sun May 06, 2007 9:52 pm GMT by suitedaces84
The only thing you need to answer these types of questions are:
1) how combinatorials work
2) common sense
The formula Adam gave is simply the hypergeometric distribution, which is the appropiate distribution. (This is a distribution that's tougher to memorize than derive.)
Let's say you had 7 lands in a deck of 60 and you wanted to know the probability of drawing exactly 4 lands in 7 picks without replacement.
There are 60c7 different ways to pick 7 cards from 60 (this that 'how combinatorials work' thing) of cards you can pick.
There are 7c4 different ways to pick 4 lands (without reguard to the other cards).
If you pick 4 lands there are there will be 3 non-lands. There will be 53c3 ways to pick those non-lands.
There are a total of 7c4*53c3 ways to pick your 4 lands (with reguard to non-lands).
So the probability of picking exactly 4 lands is 7c4*53c3/60c7 (note that this is the same thing Adam got).
| red_pen0 wrote: | | 0! is by definition equal to 1 so the equation holds. |
n! =
n*(n-1)*(n-2)*...2*1 =
n*(n-1)!
so...
n! = n*(n-1)!
plugging 1 in for n shows:
1!=1*0!
1 = 0!
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