I have been playing Magic for eight years now. Now that I have had internet to use where I live these past some months I decided to do research and learn different strategies and build more competitive decks.
One thing I've been having a difficult time figuring out is calculating the probability of a starting hand or a combination of cards in my starting hand.
I'm not fully sure why I'm so interested in figuring out how to do this
And the other thing, not long after I started playing Magic I was able to mostly set up the land base pretty well, but still a bit lose.
I would like to learn how to calculate mana curve to decide how much of what land would be most efficient in the decks that I build, as I realize having an efficient land base to match the mana curve is important in a competitive deck.
I've gone online and searched many tutorials that either didn't explain how to do it or wasn't making any sense to me, sometimes using abbreviation I'm not familiar with.
Any help would be greatly appreciated
And if anyone could explain to me why calculating hand / card combination in the opening hand could be helpful. I'm not so much sure it is but it would still be nice to know how to do
-also I'm not fully sure this is the right section but I'm assuming it is
And now that I think of it the title should have said help so you all know it's not a tutorial lol I don't see an edit
W may only be paid with white mana. U may only be paid with blue mana. B may only be paid with black mana. R may only be paid with red mana. G may only be paid with green mana. C may only be paid with colorless mana. 1 may be paid with white, blue, black, red, green, or clolorless mana.
Whenever i make a deck, i check my mana curve by laying out the cards that cost certain amounts, and seeing how they stack up. I know if i lean closer to an average of 4-5 mana, i'm going to need 24 land. Maybe more, or some acceleration. If i lean closer to 2-3 mana, i can chuck a few land for cards. Usually low cost cards, and if i am leaning towards 6 or more, i need to remake my deck, unless i really know what i'm doing. haha.
I've never personally calculated anything about what i will get in my hand, and random shuffles are really rarely random. It is hard to get true randomness, so i feel the best way to do that, is to play with it a lot, and pay attention to what you end up drawing in your opening hand.
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Whats the big deal about black lotus you ask? Well you see, there is no big deal about it. It IS the big deal.
Calculating probabilities is a high school/university level subject. You should try to get your hands on a textbook that explains this stuff.
Here's the 5-minute version. No guarantees that you'll get it.
Q1) How many ways are there of arranging the letters ABCDEF in a row without replacement?
A1) 6 choices for the first letter, 5 for the second (you can't choose the one that you picked for first again, because of "no replacement"), etc. 6*5*4*3*2*1 = 720. We write 6*5*4*3*2*1 as 6! ("6 factorial") for short.
Q2) How many ways are there of arranging 3 letters from ABCDEF in a row without replacement?
A2) 6 choices for first, 5 for second, 4 for third. 6*5*4 = 120. Note that this is equal to 6!/3!, the 3 comes from the number of letters you can't choose.
Q3) How many ways are there of choosing 3 letters from ABCDEF without replacement?
A3) Order doesn't matter when choosing, so for each arrangement, there are 5 other equivalent arrangements (e.g. ABC = ACB = BAC = BCA = CAB = CBA). Note that 6 (1 arrangement + 5 other equivalent ones) = 3!, the 3 comes from the number of letters you can choose. Using the result from the previous question, (6!/3!)/3! = 20.
In general, something of the form n!/(x!*(n-x)!) is written as nCx ("n choose x"). I use a TI-89, and the way to input this is nCr(n,x). I'll be using the TI notation for the remainder.
Q4) I have a 4-of in a 60 card deck, what is the chance of me drawing exactly one of the 4-of in a starting hand of 7 cards?
A4) You have 4 cards that you want and 56 others that you don't. You need 1 out of the 4 and 6 out of the other 56 for exactly one copy of the 4-of.
Number of successful hands = nCr(4,1)*nCr(56,6) <- 4 choose 1, then 56 choose 6
Total number of hands = nCr(60,7) <- 60 choose 7
Required probability = nCr(4,1)*nCr(56,6)/nCr(60,7)
This is the fabled "hypergeometric distribution": nCr(r,x)*nCr(b,n-x)/nCr(r+b,n)
Q5) I have 24 lands and 36 spells, what is the chance that I draw exactly 3 lands in an opening 7?
A5) nCr(24,3)*nCr(36,4)/nCr(60,7)
That should be enough to understand the maths behind calculating probabilities for drawing cards from a deck. Note that this business of calculating probabilities is very fiddly, because if you screw up it's kind of hard to tell that you got the wrong answer - it's not like other math problems, where if you get an answer of "John weighs 1000kg" you can be sure you went wrong somewhere. In probability you might get an answer of 0.617 when the actual answer if 0.601, and you won't know if you're on the right track.
If I have a 99 card deck and in that deck I have 43 lands and 56 spell. So what is the probability of me drawing 7 lands in my opening hand? Please explain why and show your work.
If I have a 99 card deck and in that deck I have 43 lands and 56 spell. So what is the probability of me drawing 7 lands in my opening hand? Please explain why and show your work.
Roughly .2%. Not even a quarter of 1%
I’m not going to show work, that’s way too much typing. That’s why I write software to calculate stuff like this.
I suggest you Google “hypergeometric probabilities” and it’s relation to Magic. There are people out there who spent quite a bit of time writing about it and it would be a waste not to utilize them.
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One thing I've been having a difficult time figuring out is calculating the probability of a starting hand or a combination of cards in my starting hand.
I'm not fully sure why I'm so interested in figuring out how to do this
And the other thing, not long after I started playing Magic I was able to mostly set up the land base pretty well, but still a bit lose.
I would like to learn how to calculate mana curve to decide how much of what land would be most efficient in the decks that I build, as I realize having an efficient land base to match the mana curve is important in a competitive deck.
I've gone online and searched many tutorials that either didn't explain how to do it or wasn't making any sense to me, sometimes using abbreviation I'm not familiar with.
Any help would be greatly appreciated
And if anyone could explain to me why calculating hand / card combination in the opening hand could be helpful. I'm not so much sure it is but it would still be nice to know how to do
-also I'm not fully sure this is the right section but I'm assuming it is
And now that I think of it the title should have said help so you all know it's not a tutorial lol I don't see an edit
Thanks for the reply
I've never personally calculated anything about what i will get in my hand, and random shuffles are really rarely random. It is hard to get true randomness, so i feel the best way to do that, is to play with it a lot, and pay attention to what you end up drawing in your opening hand.
Here's the 5-minute version. No guarantees that you'll get it.
Q1) How many ways are there of arranging the letters ABCDEF in a row without replacement?
A1) 6 choices for the first letter, 5 for the second (you can't choose the one that you picked for first again, because of "no replacement"), etc. 6*5*4*3*2*1 = 720. We write 6*5*4*3*2*1 as 6! ("6 factorial") for short.
Q2) How many ways are there of arranging 3 letters from ABCDEF in a row without replacement?
A2) 6 choices for first, 5 for second, 4 for third. 6*5*4 = 120. Note that this is equal to 6!/3!, the 3 comes from the number of letters you can't choose.
Q3) How many ways are there of choosing 3 letters from ABCDEF without replacement?
A3) Order doesn't matter when choosing, so for each arrangement, there are 5 other equivalent arrangements (e.g. ABC = ACB = BAC = BCA = CAB = CBA). Note that 6 (1 arrangement + 5 other equivalent ones) = 3!, the 3 comes from the number of letters you can choose. Using the result from the previous question, (6!/3!)/3! = 20.
In general, something of the form n!/(x!*(n-x)!) is written as nCx ("n choose x"). I use a TI-89, and the way to input this is nCr(n,x). I'll be using the TI notation for the remainder.
Q4) I have a 4-of in a 60 card deck, what is the chance of me drawing exactly one of the 4-of in a starting hand of 7 cards?
A4) You have 4 cards that you want and 56 others that you don't. You need 1 out of the 4 and 6 out of the other 56 for exactly one copy of the 4-of.
Number of successful hands = nCr(4,1)*nCr(56,6) <- 4 choose 1, then 56 choose 6
Total number of hands = nCr(60,7) <- 60 choose 7
Required probability = nCr(4,1)*nCr(56,6)/nCr(60,7)
This is the fabled "hypergeometric distribution": nCr(r,x)*nCr(b,n-x)/nCr(r+b,n)
Q5) I have 24 lands and 36 spells, what is the chance that I draw exactly 3 lands in an opening 7?
A5) nCr(24,3)*nCr(36,4)/nCr(60,7)
That should be enough to understand the maths behind calculating probabilities for drawing cards from a deck. Note that this business of calculating probabilities is very fiddly, because if you screw up it's kind of hard to tell that you got the wrong answer - it's not like other math problems, where if you get an answer of "John weighs 1000kg" you can be sure you went wrong somewhere. In probability you might get an answer of 0.617 when the actual answer if 0.601, and you won't know if you're on the right track.
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Big Johnny.
Roughly .2%. Not even a quarter of 1%
I’m not going to show work, that’s way too much typing. That’s why I write software to calculate stuff like this.
I suggest you Google “hypergeometric probabilities” and it’s relation to Magic. There are people out there who spent quite a bit of time writing about it and it would be a waste not to utilize them.