Crazy Mathematics in Magic
General forum
Posted on June 2, 2014, 12:34 a.m. by DarkMagician
I was curious as to how many possible opening hands a deck can have (to get an idea how many times you should play test) so I decided to do some math and came back with some insane numbers.
In a 60 card constructed deck there are 32,441,381,340 possible opening hands. Assuming it would take a single minutes to properly shuffle your deck and draw a sample hand it would take more than 61,722 years just to draw all of the possible hands.
In an edh deck there would be 71,163,080,271,360 possible opening hands. Again assuming shuffling and drawing a hand requires one minute it would take over 135,393,988 hours to draw each possible hand.
Still think you're play testing enough?
On a side note this makes me feel really bad for Battle of Wits players.
DarkMagician says... #3
This math uses the idea that every card in the deck is a different card and only accounts for drawing a different hand each time.
June 2, 2014 12:46 a.m.
This kinda math has been done before. Also, what are you using as your baseline for an EDH deck and a 60-card deck? The amount of basic lands in each deck define the number of hands you can have, and the number of each card in the 60-card even more.
June 2, 2014 12:47 a.m.
DarkMagician says... #5
There's no way to account for redundant hands that I know of
June 2, 2014 12:47 a.m.
DarkMagician says... #6
erabel I'm sure it has and. Baselines are simply a 99 card edh and 60 card constructed.
June 2, 2014 12:49 a.m.
There is a way to account for it though. It's basic probability and statistics. You calculate the probability of drawing any given hand based on all known cards in the deck, and then using those calculations to statistically eliminate any redundancies created by multiple copies of any given card. If I run a deck with 4 Forest s, and I only draw one of them in my opening hand along with 6 other cards, that's 4 different ways that same hand appears. Statistically, that is still only one opening hand combination because it's the same hand regardless of which Forest I have.
June 2, 2014 1:05 a.m.
trentfaris242 says... #8
This is...not a complete thought. I got my undergraduate in Pure Mathematics. It's 2 AM here so I'm WAY to sleepy to think about this right now. Commenting for subscription. I'll do some awesome probability on MTG decks tomorrow for fun :)
June 2, 2014 1:58 a.m.
vampirelazarus says... #9
Hell yeah, LETS DO SOME MATH MOTHA FU......... screw it, Im too tired, like the guy above me....
Ill check back tomorrow.
June 2, 2014 2:49 a.m.
DarkMagician says... #10
ljs54321 There are no multiples in the imagined decks
June 2, 2014 3:18 a.m.
DarkMagician says... #12
trentfaris242 This is possibility not probability but I am looking forward to any mtg related probability math you post
June 2, 2014 3:21 a.m.
But, in your calculation, do you then count "Plains, Mountain and Lightning Bolt" and "Mountain, Lightning Bolt and Plains" as 2 different hands or the same? Because, it is the same hand....
June 2, 2014 5:03 a.m.
DarkMagician: No multiples in the imagined decks? So, you're gonna build a Standard legal, mono-colored deck without using multiple basic lands. Regardless of what color you choose, there are only 19 unique lands that can produce that specific color. Granted, you could include lands that produce only colorless or even colors you're not running to increase your total land count to a more optimal number, but in reality, are you gonna build said deck using no multiples of any one card, TO INCLUDE BASIC LANDS, just to make your math more accurate? Not trying to attack you or be a d!ck or take away from all the work you did to come up with these numbers, I know it took some time to come up with. Sorry if it may seem that way. However, when looking at how people actually build their decks, these numbers are off by a considerable margin. Even with using 60 unique cards, as peterlravn said, those 2 hands are the same and need to be counted as such. Drawing the same cards in a different order does not make them unique from each other.
June 2, 2014 10:20 a.m.
As I can't see how you computed your figures, I'll just show how I computed mine.
Every time you draw a card, you subtract a card from the deck. As order doesn't matter in your hand, the number of possible opening hands should be 60 * 59 *_58 all the way down to 53 for your seventh card.
Or 1.03 * 1014. In common language, a little over 100 TRILLION hands.
That said, you don't need to see all of those to get a good idea of what the deck looks like. You just need to know if it is a good idea to mulligan or not. That takes two or three hands, not 100 trillion.
June 2, 2014 10:47 a.m.
DarkMagician says... #18
Does nobody read the previous comments these days?
peterlravn By running 60 unique cards each possible hand is different. For example if my hand is Plains , Swamp , Island , Mountain , Forest , Black Lotus , Mox Opal and I do not run any cards as multiples there is no other way to draw that hand.
ljs54321 Who said anything about the deck being standard?
Egann That was the exact math I used except somewhere along the line you messed up your math as did I. The total hands for a 60 card deck is 1,946,482,876,800. For an EDH it's 75,030,638,981,760. How did you come up with 10 to the 14th though? The thing about play testing enough was a joke. If you account for possible hands under seven cards the numbers grow substantially. The math is simple:
60x59x58x57x56x55x54 = 1,946,482,876,800 (7 card opening hand)
60x59x58x57x56x55 = 36,045,979,200 (mulligan to 6)
60x59x58x57x56 = 655,381,440 (mulligan to 5)
60x59x58x57 = 11,703,240 (mulligan to 4)
60x59x58 = 205,320 (mulligan to 3)
60x59 = 3,540 (mulligan to 2)
And of course 60 hands at 1 card.
To figure out total hands add the figures together:
- 1,946,482,876,800 + 36,045,979,200 + 655,381,440 + 11,703 + 205,320 + 3,540 + 60 = 1,983,196,149,540 possible hands total
June 3, 2014 12:03 a.m.
vampirelazarus says... #20
But does your math account for a hand of Swamp , Plains , Black Lotus , Island , Mountain , Forest , and Mox Opal ?
To us, its the same hand, but to the math, its different, due to its order.
June 3, 2014 12:08 a.m.
DarkMagician says... #21
vampirelazarus I believe it does eliminate redundancies like that
June 3, 2014 12:38 a.m.
trentfaris242 says... #22
@DarkMagician Your computations don't account for the fact that we have duplicates. I've been working on a really long post about this today. I'll probably post it sometime tomorrow. It's actually a really complicated problem.
Generally we use nPr / N where N = m in N | n1! * n2! * ... * nm! but for combinations, but it's....wrong(?)....for large sets and small subsets? It's odd. I'll address it.
June 3, 2014 1:15 a.m.
I was only using standard as an example of one possible way to construct a 60 card deck.
On another note...if you're building a 60 card deck with 60 unique cards (to include basic lands), I think we'd be better off questioning your deck building skills instead of your math. lol/jk Even a typical EDH deck has multiples of basics to give some sort of consistency.
In all honesty, looking at the formula you used in your previous post, I'm not seeing how it would eliminate them. If you're physically drawing and redrawing 7 card hands (using pencil and paper to mark each individual hand), tedious, I know...that's why we use math instead, you would have to not mark any hand that comes up with the same combination of cards as one previously drawn except in a different order. When doing the math instead, that would require some subtraction to eliminate those redundancies since just multiplying some numbers is not biased toward those instances and can't eliminate them.
June 3, 2014 1:18 a.m.
DarkMagician: I have read every comment.... I'm sorry to be a d!ck but your calculation is wrong. Your calculation counts Plains and Mountain to be an other hand than Mountain and Plains. I understand that you don't have any copies af a single card. You wanna use this:
n! / ((n! - r!)!)
n is the amount of cards in deck. In this case 60.
r is the number of cards drawn. In this case 7, 6, 5, 4, 3, 2 or 1.
Im on my mobile, so I can't really calculate now... But this will give you the correct answer, when Plains and Mountain is the same as Mountain and Plains :)
June 3, 2014 1:44 a.m.
Oh, shit, sorry. I was wrong again. This must be the right formulae.... :
n! / ((n - r)! - r!)
Yea, this is right.....
June 3, 2014 2:27 a.m.
trentfaris242 says... #28
So close on the last one! It's addition in the denominator. I've almost finished my long post! I'll probably just end up making a new thread for it.
June 3, 2014 3:12 a.m.
Omg! Shit! F*ck this! I see the mistake! How can I type it wrong 3 times? I'll just stop posting here, lol...
June 3, 2014 3:17 a.m.
Oh, and by the way, it's multiplication not addition in the last denominator :)
June 3, 2014 3:19 a.m.
trentfaris242 says... #31
Oh, haha. Yeah you're right :P I guess that means it's bed time for me. Look for my post tomorrow!
June 3, 2014 3:21 a.m.
DarkMagician says... #32
peterlravn It's been a very long time since I've done this stuff so no offense taken when corrected
June 3, 2014 3:45 a.m.
@ DarkMagician: Oh, I included 53 (oops: you're completely right). You also included mulligans. I did not because technically all mulligan hands are included in the original number.
June 3, 2014 1 p.m.
By that I mean you remove a card from a hand and you get a mulligan hand. Technically the mulligan hand is mathematically unique, but from a playing perspective it's just lost card advantage.
ljs54321 says... #2
Does this math take into account how many hands would be drawn that are the same as a previously drawn hand due to there being multiples of a given card? If not, this number would significantly decrease. Even in EDH, the possibility of drawing the same opening hand due to having multiples of basic lands would decrease that massive number of possibilities. The number of EDH possibilities may not shrink as significantly, but there would still be a difference.
June 2, 2014 12:44 a.m.