I've been a huge fan of using Hypergeometric calculators for deckbuilding, mostly as a tool for establishing likelihood of drawing lands and whatnot. I recently stumbled upon a Binomial Probability Calculator, which calculates the odds of success over a number of trials. I combined the two to calculate some odds of winning games in the long run. I have always found that opening hands with at least 3 lands gives me the most confidence in winning a game. If we assume a 3 land opening hand is a Winning hand then we can calculate the odds of consistently Winning in the long run. Using the Binomial Probablity calculator I ran the numbers for the likelihood of having Winning hands in 200 out of 300 games, which would just be having a record of winning 100 matches over 300 games. Now, this doesn't actually mean anything because the premise is so vague, but I thought it was an interesting exercise in looking at deckbuilding. What I found was pretty surprising. While adding 1+ land to the list had little effect on the likelihood of drawing 3+ lands each opening hand, the cumulative effect on consistently seeing those 3 lands over hundreds of games increased significantly with each land added. What I found , if we're assuming having 3+ lands in our opening hand is the benchmark for winning a game, then it takes running 27 lands in a 60 card deck to consistently see those results in the long run. What I'm wondering now is if anyone else has things they look for in opening hands that give them the most confidence in winning a game, and how we can build to maximize that likelihood.
Number of Lands in a 60 card deck Probability of 3+ lands in 7 (A Winning Hand) Probability of 200+ Winning Hands in 300 games
22 0.509838301 < 0.000001
23 0.549341464 0.0000231
24 0.587929496 0.003075374
25 0.625348194 0.077150495
26 0.661372406 0.449759624
27 0.695807315 0.876423503
I disagree with your conclusion about the results, unless I am misunderstanding things.
I used two different binomial calculators (https://stattrek.com/online-calculator/binomial.aspx and http://statisticshelper.com/binomial...ity-calculator). Both calculators gave me slightly different numbers from yours. I have rounded the percentages in this post, but the numbers should be close enough for our purposes.
For example, with a 23 land deck, the probability is 23/60 =~ .3833333. The trials = 7. The successes = 3. The probability is 0.28510161444. Doing this for each number (1-7) yields:
0: 3.4%
1: 14.8%
2: 27.5%
3: 28.5%
4: 17.7%
5: 6.6%
6: 1.4%
7: 0.1%
For a 27 land deck (27/60 = .45 probability):
0: 1.5%
1: 8.7%
2: 21.4%
3: 29.2%
4: 23.9%
5: 11.7%
6: 3.2%
7: 0.4%
23 land deck probability of 2-4 land (73.7%) and 3-4 land (46.2%) in the opening hand.
27 land deck probability of 2-4 land (74.5%) and 3-4 land (53.1%) in the opening hand.
So far, this agrees with your analysis. The numbers make it seem like the 27 land deck is better. However, extending to look at the first 10 cards (going first + draws for the first 3 turns):
23 Land Deck
0: 0.8%
1: 4.9%
2: 13.8%
3: 22.9%
4: 24.9%
5: 18.6%
6: 9.6%
7: 3.4%
8: 0.8%
9: 0.1%
10: 0%
27 Land Deck
0: 0.3%
1: 2.1%
2: 7.6%
3: 16.6%
4: 23.8%
5: 23.4%
6: 16%
7: 7.5%
8: 2.3%
9: 0.4%
10: 0%
23 land deck probability of 3-5 land in the first 10 cards (66.4%).
27 land deck probability of 3-5 land in the first 10 cards (63.8%).
So a 23 land deck will have a “better” land situation in 2.6% more games. This depends a lot on what kinds of spells are in the deck, both in terms of things like cantrips and the average cmc.
When looking at the 3-4 land situation, it gets even better for the 23 land deck:
23 land deck probability of 3-4 land in the first 10 cards (47.8%).
27 land deck probability of 3-4 land in the first 10 cards (40.4%).
EDIT: None of this takes fetchlands into account. With fetchlands, it is likely that your numbers are better than mine. I will ask some others about the math regarding fetchlands.
Binomial calculators will give less accurate results than hypergeometric calculators, since binomial calculators assume sampling with replacement.
tl;dr - It's possible to optimize this with math, but it's also very easy to get the math wrong depending on assumptions and calculations used. In cases where the math contradicts common sense from real games, consider the possibility that the math is off.
With the two calculators, just have to be sure you're using the right calculator for the right thing. Playing 300 games is sampling with replacement. Each new game you replace the deck, shuffle, and draw from a fresh deck. Binomial makes sense to compare different games. For drawing a certain hand within one game, the cards are not replaced so it's sampling without replacement and a Hypergeometric calculator.
Hypergeometric Calculator - for chance of seeing a certain opening hand (e.g 3 lands in 7 cards)
https://en.wikipedia.org/wiki/Hyperg...c_distribution
Binomial Calculator - for chance of seeing a certain outcome after many games (e.g. 200 of the target hand in 300 games)
https://en.wikipedia.org/wiki/Binomial_distribution
@Whoshim: You're using a Binomial calculator where a Hypergeometric should be used.
@jrw: There's a major flaw in your analysis, not related to that.
Re-run using 59 lands in a 60-card deck. You'll get 100% probability of winning >290 of 300 games. What a winning deck!
Why does it do that? The problem is you've defined anything with 3+ lands as a winning hand. Including 7 lands, 6 lands, or 5 lands with 2 slow cards. Those are obvious mulligans, but the calculation counts everything with at least 3 lands and ignores the possibility of flooding. With a low land count those events are rare, maybe negligible, but as you increase the number of lands they become more common. As you add more lands, of course the chance of having more lands in your hand will go up, so your analysis will always point you towards the highest number of lands in your table. If you keep going past 27, it will tell you to add even more lands. The assumption that "3+lands = win" is too strong and makes the math do things like recommend a 60-land deck.
In a real game there's a tradeoff between not having enough mana and being flooded. 27 seems like too many.
Last edited by FTW; 03-03-2021 at 03:56 PM.
I think the Hypergeometric probability you want is not for 3+ lands in 7 cards but for 3-4 lands in first 7 cards and no more than 5 lands in the first 11 cards (3 or 4 lands in the opening hand but you don't hit more than 5 lands after 4 draws - turn 5 on the play).
That represents seeing the 3-4 lands in your opening hand, but also not flooding over the critical first turns of the game when you need business and interaction. Unlike the condition of "3+ lands = win", this would have diminishing returns as you increase the land count too high. It would not give the unreasonably high win %s for land-heavy decks.
*Hypergeometric probabilities only
Lands 3-4 lands in hand 3-4 lands in hand AND no more than 5 total after 4 draws 18 32.7% 28.7% 19 36.1% 30.8% 20 39.4% 32.6% 21 42.5% 34.0% 22 45.4% 35.0% 23 48.1% 35.6% 24 50.5% 35.8% 25 52.6% 35.6% 26 54.4% 34.9% 27 55.8% 33.9% 28 56.8% 32.5% 29 57.4% 30.8% 30 57.6% 28.8% 31 57.4% 26.6% 32 56.8% 24.3% 33 55.8% 21.9%
If you just try to get 3-4 lands in your opening hand and don't care about flooding, you optimize it with 30 lands in the deck (57.6% of hands will have 3-4 lands). But that will lead to flooding during the game.
If you want to see 3-4 lands in the opening hand but then don't want to flood in the early turns, that's optimized around 24 lands (35.8%), with very close odds for 23 and 25 lands. That's the default land count for many Standard and Casual decks.
Goblins runs 22-23 lands. Should Goblins be running 24? You could test using a card like Mutavault, Ghost Quarter or Teetering Peaks as the 24th land and see how that performs. Goblins has other constraints though too. It needs a high enough Goblin count for Ringleaders, Muxus and Lackey to do their job, which means there is a penalty for having too many lands. Meanwhile the land odds don't increase that much between 22-24 lands.
Accelerants also help cheat on mana costs. Would anyone ship a 2-land hand with Lackey or Vial?? Most 2-land hands are keepable. So what if we expand it to include 2 land hands where you hit your 3rd land in time but don't flood past 5 lands?
Lands 2-4 lands in hand AND 3 lands by 2nd draw AND no more than 5 lands after 4 draws 18 42.0% 19 45.0% 20 47.5% 21 49.4% 22 50.6% 23 51.3% 24 51.3% 25 50.7% 26 49.5% 27 47.8% 28 45.5% 29 42.9% 30 39.9%
This one optimizes around 23-24 lands (51.3% of games). Because of the need for high Goblin count, 23 seems better than 24. And that's what Vial Goblins has already been running for ages.
I would argue that extensive playtesting and tournament results provide better data than a simulation with simplifying assumptions, especially for an interactive aggro-control deck. I think these calculators are most useful when optimizing fast combo decks. Those games are closer to goldfishing and more dependent on opening hand, while an interactive deck like Goblins depends on a lot of factors.
Last edited by FTW; 03-03-2021 at 04:50 PM.
Long time lurker, first time poster since I haven't played competitively in over a decade. Finally can contribute with statistics. I give jrw1985 credit for bringing up deck odds and FTW provided needed corrections. Can take this further with fetch lands and the draw after London mulligan(s) but the math gets complex enough that simulating makes more sense. As in, writing a computer program that shuffles and draws from deck. Is beginner-level coding. I still like the pure math approach to start and validate any code results with upper and lower bounds.
Hypergeometric limitations are naturally only 2 classes of cards, the success (land) and failure (non-land) and simple 1 draw = 1 card removal. You can actually work out the math without such a distribution. Let me show where the 24 land values come from with spreadsheet formulas. The 0 or 1 at the end denotes being cumulative that your spreadsheet may need as false or true, respectively. For instance, cumulative of (x=3) means sum of the probabilities of 0, 1, 2 and 3 successes:
24 lands: 3-4 in hand (50.5%)
=HYPGEOM.DIST(3, 7, 24, 60, 0) + HYPGEOM.DIST(4, 7, 24, 60, 0) = 0.505152419330032 (exact)
24 lands: 3-4 lands in hand AND no more than 5 total after 4 draws (35.8%)
=(HYPGEOM.DIST(3, 7, 24, 60, 0) * HYPGEOM.DIST(2, 4, 21, 53, 1)) +
(HYPGEOM.DIST(4, 7, 24, 60, 0) * HYPGEOM.DIST(1, 4, 20, 53, 1)) = 0.358184438166005 (exact)
Really good work by FTW here. Took me a few tries to setup correctly.
-----------------------------------------------
Warning: Math Intensive below
The "long way" for HYPGEOM.DIST(3, 7, 24, 60, 0) for instance is:
=COMBIN(7, 3)*(24/60)*(23/59)*(22/58)*(36/57)*(35/56)*(34/55)*(33/54) = 0.308704256 (exact)
Here you take COBIN(7, 3), the number of possibilities of arranging 3 successes in a group of 7 and multiply by the probability of it occurring once. I think easier to understand since odds of first draw being a land are obviously (24/60) and you see next draw is from pool of 23 lands and 59 cards. Can work in fetch lands and mulligans with this approach and enough chaining. You could find poker hand odds worked out online strictly with COMBINs and smart counting.
This is the most difficult for me to setup correctly even though it's the simplest in theory. The "poker math way" is:
=(COMBIN(24, 3)*COMBIN(36, 4))/COMBIN(60, 7) = 119223720 / 386206920 = 0.308704256 (exact)
Where (60, 7) denominator is the total count of starting hands that can exist and the numerator is the total count of 3 successes in a hand of 7 with 24 lands and 36 non-lands. In other words, (total hands we want) / (total hands possible).
Thanks for joining The Source, adding to the discussion, and showing the calculations.
I probably should have explained how I got those numbers. I worked through cases and conditional probabilities. I started to type out the equations, but they got long and ugly so I left it out. I figured those who knew enough math would know what I meant and others wouldn't want to see the formulas anyway. It's much easier to do in Excel or in code than on the Hypergeometric calculator. I just programmed a spreadsheet to generate those tables for me.
The easiest way to conceptualize the math is like a tree splitting into multiple branches. You multiply the probabilities on each branch and then add the branches together. Each branch represents a different case for how you could draw your lands.
#1> 2 lands in 7-card hand ---- 1 land in next 2 cards ---- 0-2 lands in next 2 cards
#2> 2 lands in 7-card hand ---- 2 lands in next 2 cards ---- 0-1 land in next 2 cards
#3> 3 lands in 7-card hand ---- 0-2 lands in next 4 cards
#4> 4 lands in 7-card hand ---- 0-1 land in next 4 cards
The first 2 branches are the extra cases I added to include 2-land hands that hit at least 3 lands.
The 3rd and 4th branch are what you calculated above.
#3: HYPGEOM.DIST(3, 7, 24, 60, 0) * HYPGEOM.DIST(2, 4, 21, 53, 1)
#4: HYPGEOM.DIST(4, 7, 24, 60, 0) * HYPGEOM.DIST(1, 4, 20, 53, 1)
The big thing to remember is to adjust the 2nd (and 3rd) Hypergeometric function to account for the fact that you've already drawn some cards and lands, so the numbers left in the deck are smaller. 53 cards left in the deck. 20 or 21 lands in this case.
If you don't care about flooding, only the opening hand, you can leave out the 2nd part. I think the flooding is an important part to include because it's the main drawback of having too many lands in the deck.
You can use the same logic to work out the probabilities for fetchlands.
It's just messier. It leads to a lot more branches and cases. In the end, it's easier to solve it with code, either to iterate the different cases for Hypergeometric calculations or to simulate random draws from a deck.
This is a "simpler" example.
3 lands in opening 7-card hand, on the play.
No more than 4 lands by turn 4
Deck has 24 lands: 6 fetches, 18 other lands
Assume you use fetchlands as early as possible
A> 3 lands in 7-card hand ---- 0/3 lands are fetches ---- 0 lands in next 3 draws
B> 3 lands in 7-card hand ---- 0/3 lands are fetches ---- 1st draw is a fetch ---- 0 lands in next 2 cards
C> 3 lands in 7-card hand ---- 0/3 lands are fetches ---- 1st draw nonland ---- 2nd land fetch ---- 3rd draw nonland
D> 3 lands in 7-card hand ---- 0/3 lands are fetches ---- 1st draw nonland ---- 2nd draw nonland ---- 3rd draw fetch
E> 3 lands in 7-card hand ---- 0/3 lands are fetches ---- 1 land in next 3 draws, not a fetchland
F> 3 lands in 7-card hand ---- 1/3 lands are fetches ---- 0 lands in next 3 draws
G> 3 lands in 7-card hand ---- 1/3 lands are fetches ---- 1st draw is a fetch ---- 0 lands in next 2 cards
H> 3 lands in 7-card hand ---- 1/3 lands are fetches ---- 1st draw nonland ---- 2nd land fetch ---- 3rd draw nonland
...
and it keeps going
Each case of drawing fetchlands is handled separately because the thinning affects odds for other draws. They're conditional on whether a fetch was drawn and used.
A> HYPGEOM.DIST(3, 7, 24, 60, 0) * HYPGEOM.DIST(0, 3, 6, 24, 0) * HYPGEOM.DIST(0, 3, 21, 53, 0)
What does all that mean?
The first one is drawing exactly 3 lands in a 7-card hand, with 24 lands in a 60-card deck.
The second is 0 of those 3 lands being fetchlands, with 6 fetches among 24 lands.
The next is drawing 0 lands in 3 cards, with 21 total lands remaining out of 53 cards left
B>HYPGEOM.DIST(3, 7, 24, 60, 0) * HYPGEOM.DIST(0, 3, 6, 24, 0) * 6/53 * HYPGEOM.DIST(0, 2, 19, 51, 0)
The first 2 parts are the same.
Then your first draw is 6/53 to draw one of 6 fetchlands in the remaining 53 cards.
Assuming you crack that fetch, you take a land out of the deck, so there are only 19 lands and 51 cards left (instead of 20 lands andf 52 cards).
The last part is drawing 0 lands in 2 draws, with 19 lands in the remaining 51 cards.
C> HYPGEOM.DIST(3, 7, 24, 60, 0) * HYPGEOM.DIST(0, 3, 6, 24, 0) * 32/53 * 6/52 * 31/50
The first 2 parts are the same.
32/53 chance to draw one of the remaining 32 nonland cards in the remaining 53 cards (36 nonlands total - 4 in opening hand)
6/52 chance to draw a fetch next, with 6 in the remaining 52 cards
If you crack that fetch, you take a land out of the deck, so there will be 19 lands and 50 cards left
31/50 chance to draw one of 31 nonland cards in the remaining 50 cards
D> HYPGEOM.DIST(3, 7, 24, 60, 0) * HYPGEOM.DIST(0, 3, 6, 24, 0) * 32/53 * 31/52 * 6/51
The first 2 parts are the same.
32/53 chance to draw one of the remaining 32 nonland cards in the remaining 53 cards
31/52 chance to draw a nonland again, with 31 nonlands left in 52 cards
6/51 chance to draw one of 6 fetches in the remaining 51 cards
E> HYPGEOM.DIST(3, 7, 24, 60, 0) * HYPGEOM.DIST(0, 3, 6, 24, 0) * HYPGEOM.DIST(1, 3, 21, 53, 0) * 15/21
The first 2 parts are the same.
The next is drawing 1 land in 3 draws, with 21 lands left among 53 cards
15/21 chance that land is not a fetch, with 15 regular lands left among 21 lands
F> HYPGEOM.DIST(3, 7, 24, 60, 0) * HYPGEOM.DIST(1, 3, 6, 24, 0) * HYPGEOM.DIST(0, 3, 20, 52, 0)
The first part is the same.
The second part is now 1 of those 3 lands being a fetch, with 6 fetches in the 24 lands.
On turn 1 you crack that fetch, taking a land out of the deck.
The last part is drawing 0 lands in the next 3 draws, with 20 lands remaining in the thinned 52 card deck
G> HYPGEOM.DIST(3, 7, 24, 60, 0) * HYPGEOM.DIST(1, 3, 6, 24, 0) * 5/52 * HYPGEOM.DIST(0, 2, 19, 50, 0)
Again on turn 1 you use a fetch and thin a land.
On turn 2, 5/52 chance of drawing one of the other 5 fetches in the thinned deck of 52 cards
Then you crack that and thin another land out of the deck.
The last part is drawing 0 lands in 2 draws, with 18 lands left in a 50-card deck after the thinning.
H> HYPGEOM.DIST(3, 7, 24, 60, 0) * HYPGEOM.DIST(1, 3, 6, 24, 0) * 32/52 * 5/51 * 31/49
32/52 chance of drawing one of 32 nonland cards in the thinned deck of 52 cards
5/51 chance to draw one of the other 5 fetches in the remaining 51 cards
Then you crack that and thin the deck of another land
31/49 chance to draw one of 31 nonland cards left in the remaining 49 cards
It's possible to work through all the math for fetchlands thinning the deck, but messy. You'd have to calculate each branch, and all the branches I didn't even include, and add them all up. There are so many separate cases because it matters what turn the fetchland is drawn and cracked, because it affects the remaining cards in the deck. Mulligans or being on the draw add even more to consider.
The calculation gets crazy. It's easier to do with code.
Thank you! Happy to be here. Only one person needs to work through the math for everyone to benefit. I guess I wanted to show the details for someone who may adjust for running 22 or 24 lands and validate your work at the same time.
For real, working out the calculation is so much effort, I wouldn't be confident that the result is right. Better to show easily understood calculations (for people good at stats) for the lowerbound odds since fetches should slightly reduce odds of mana flooding.
So you do still crunch through fetchlands with hypergeometric. Impressive work again. We need not be concerned for the case of less fetchable lands than fetches that pull them in all the decklists I see but there is the edge case of all colorless lands that forces a mulligan. Tracking that I think makes using straight combinatorics ("poker way") easier than further modifying hypergeometric. I started a probability matrix this way with 1-4 count Lackey rows and 1-5 count land columns that gives odds of Turn 1 Lackey Go. Sum boxes of probabilities for keepable hands. Also for turn 1 Winstigator with Chrome Mox. Manageable but just barely due to cases such as 3x Winstigator 2x Mox not needing a land.
Also started coding where I'd need to calculate hypergeometric odds so I finally looked up the formula. Really is a (COMBIN*COMBIN)/COMBIN chain and can mathematically convert between hypergeometric and its COMBIN chain without thinking how to logically setup as I did.
Bringing back odds of exactly 3 lands in starting hand with 24 lands in 60 card deck:
=HYPGEOM.DIST(3, 7, 24, 60, 0) = 0.308704256 (exact)
In actual mathematical notation, the first four numbers are (usually) named from left to right k, n, K, N:
k is how many lands we want to draw (observed successes)
n is the total cards drawn (total observations)
K is the number of lands in the deck (successes in population)
N is the deck size (population size)
Actually switching the 2nd and 3rd numbers n and K in the distribution gives the same answer for all hypergeometric inputs. With COMBIN work:
=COMBIN(24, 3)*COMBIN(36, 4)/COMBIN(60, 7) is equivalent to =COMBIN(7, 3)*COMBIN(53, 21)/COMBIN(60, 24)
More useful to us is the hypergeometric mean being n * K / N and variance is ( n * K ) * ( N - K ) * (N - n ) / [ N^2 * ( N - 1 ) ]
Variance at sufficiently high N approaches 1/N. If you thought that hypergeometric at N approaching infinity (no replacement) becomes the binomial distribution, you'd be correct.
I saw in the Legacy Goblins Discord pinned messages a chart for expected number (EV) of goblins pulled with Goblin Ringleader based number of goblins in 61 card deck. Let's upgrade this. Is easy to do with hypergeometric. For instance, odds of exactly 3 goblins from 25 goblins remaining in a 50 card deck is =HYPGEOM.DIST(3, 4, 25, 50, 0) = ~0.25 or 25%. Multiply that by 3 goblins for the weighted average. Sum the weighted averages for k=1, 2, 3 and 4 goblins for the expected goblin count.
But really can just go n * K / N = 4 * (goblin cards left in deck) / (total cards left in deck) and solve even more easily, unless you must see odds for 0, 1, 2, 3 and 4 goblins separately. Ringleader EV for 30% goblins is 1.2, down from 1.6 at 40% and 2.0 at 50%. Use 6 for Muxus but reduce count from Tarfire and Warren Weirding. You then have the n, K and N you need to solve variance and square root of variance is standard deviation.
Basically, smaller deck size reduces the variance (randomness or luck factor) which is one reason why more than 60 cards is bad. A deck 50/50 in goblins maximizes variance of drawing them.
Our [Matron / Boggart] searching reduces that goblins to deck ratio, as does [Ringleader or Muxus] + shuffle effect. Fetching a land increases it. I don't want to use all that to say what goblin count is optimal. I agree with FTW in not mathmagically saying what to do without factoring the gameplay itself, but perhaps you could justify as few as 30.
Going down to 30 seems bad, especially now with Muxus, Goblin Grandee.
Although the EV doesn't seem to change much, the probabilities of having very good or very bad flips changes a lot.
This is a table of Hypergeometric probabilities for Turn 2 Muxus off Lackey or Instigator.
60 card deck.
Fetchlands ignored.
Muxus + 2 non-goblin cards + 2 Goblin cards are known to be out of the deck.
That could be Lackey + 2 lands + removal or WInstigator + land + Mox/SSG + removal/imprinted card. Other unused cards in hand don't affect the odds, as they could be interchangeable with cards in the deck.
# Goblins in Deck EV P(4-6 goblins) P(0-2 goblins) 20 1.9 7% 74% 22 2.1 10% 66% 24 2.3 14% 58% 26 2.5 19% 50% 28 2.7 25% 43% 29 2.8 28% 39% 30 2.9 32% 35% 31 3.1 35% 32% 32 3.2 39% 28% 33 3.3 43% 25% 34 3.4 47% 22% 35 3.5 50% 19% 36 3.6 54% 17%
For math-checkers, example with 30 goblins: P(0-2 goblins flipped) = HYPGEOM.DIST(2, 6, 30-3, 60-5, 1) = 35%
Between 30 and 35 goblins in the deck, the EV only changes from 2.9 to 3.5. Just 1/2 a card. What's the big deal?
Well with 30 goblins there's a higher chance of a dud flip of 0-2 goblins than a flip of 4-6 goblins (35% vs 32%)! Muxus will disappoint more often than it will be busted. With 35 goblins there's less than a 1/5 chance of a dud flip, while 1/2 will be stronk (19% vs 50%).
The best Turbo Muxus lists even cut Vial for more acceleration and a higher Goblin count.
Last edited by FTW; 03-15-2021 at 07:35 AM.
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