So mana is a bitch... and you can blame the RNGesus gods for just so long... so here are some resources to help build mana bases.

RESOURCES



GENERAL GUIDELINES

Coloured sources needed




Lands needed


YORION UPDATE

That looks like a lot of work. Thanks for putting it all together

I don’t think you’re supposed to call them coloured sources anymore

So, using the Hyper Geometric Calculator to figure out frequency of lands in an agro deck:

I set population size to 60 (cards in deck)
Number of successes in population- 22 (lands in deck, RDW for simplicity, so all mountains)
Sample Size - 7 (cards in initial draw, barring no mulligan)
Number of Success in Sample - 2 (amount of lands I want in opening hand)

And that gives me the likely hood of getting a 2 mountain hand, right? I dont understand how to extrapolate the likeliness of a 2 land hands from the results it gives in the next 5 fields.


Going further, I now want to know the probability of getting the third land by turn 3 if I'm on the play. I assume I would change Sample Size to 9 (7 cards plus 2 draw steps), and number of success to 3. Is that how I would do it? But doesnt that take into consideration the 7 card hands that have 3 lands anyways. So I guess I would have to change all the numbers:

Population size -53 (60 minus starting 7)
Successes in population- 20 (starting lands minus 2 in starting hand)
Sample size - 2 (2 draw steps)
Successes in sample - 1 (looking for that 3rd land) [can I do "1 or 2" to account for drawing 2 lands in a row, or is 2 faceted in as a success because I'm looking for at least 1, or does it ONLY count the chance of 1 success, meaning the other is always a failure]

Based on your questions:

1) you'd want pop: 60, pop successes: 22, sample size: 7, Samp successes: 2. answer is 30% P(x=2)

The next four probabilities are less than 2, less than or equal to 2, greater than 2, and greater than or equal to 2.

You can use them for a number of reasons: for example what is the probability I draw AT LEAST 2 lands in my opener? 81% (this includes instances drawing 2, 3, 4, 5, 6, and 7 lands)

2) you'd want pop: 53, pop successes: 20, sample size: 2, Samp successes: 1. answer is 48% P(x=1)

Okay. So I would take the 3rd field, the greater than or equal field.

But then I need to run it again, with Successes in Sample of 5, and figure out that percentage. That would be hands I wouldn't keep due to to many lands. Then I take the greater than or equal field and subtract that from the results of my first calculation to determine successes of 2, 3 or 4, but not more or less than.

Correct?

Yes: holding Pop, Pop successes, and sample size equal, P(x>=2) - P(X>=5) = exactly the Probability that you draw 2, 3, or 4 lands, AND it would be totally appropriate to use that number to decide how likely your deck would be (in terms of lands) to have a hand worth keeping. In this case (using the numbers we used before), it's ~75% 'good' hands.

I should mention that the first two links are using a somewhat more complicated formula, but doing something very similar. (the main calculation is still a hypergeometric calculation though)

Edit: I should also mention that a 25% failure rate is still highly noticeable, which is why mana screw comes up sooooooo often in discussions.

Nice job Cucho, but is any of these resources updated taking into account london mulligan?

Good point. Frank Karsten has used the Vancouver Mulligan in his latest iteration from 2018.

I wouldn't know how the hyper geometric can model that. I mean, whatever you're calculating, you have an X % chance for your first Y cards to contain Z amount of whatever.

To account for a London mulligan you'd get the exact same values for the second try, for which the base is the initial failure rate. Could be worded badly, so here's an example:

70% chance to hit what you want in the starting hand.
If you allow 1 Mulligan, you get 70% + (70% x 30%) = 91% total success (failure would be 30% twice in a row, ergo 9%)


Another aspect is the hidden hand selection in bo1 games, where the game draws 2 hands for you and only shows you the one it deemed "better".

This is what Karsten says about it...

Can You Cut a Land From Your Deck?

As I’ll show, under the new mulligan rule, you may want to run more lands to optimize your probability of getting a keepable opening hand.

Of course, this is all based on the specific definition of “keepable hand”. In an insightful Reddit comment by aldeayeah, a “keepable hand” was defined as one with at least two lands and two spells. For my analysis, I slightly tweaked this definition to exclude hands with four or more lands. For the context of an aggro deck whose mana curve tops out at 3-drops, which is typical for Modern, such hands may flood out and therefore may deserve a mulligan. More importantly, by excluding 4+ land hands, the optimal deck configuration won’t be the obvious symmetric 30 lands and 30 spells, which allows us to get some actual insight on deck construction.

So suppose that we play an aggro deck with 21 land and 39 spells, and that we mulligan until we get a keepable hand containing two or three lands and at least two spells. My script yields the following:
Vancouver probability of keepable opening hand London probability of keepable opening hand



Suppose we define a non-game as a game where a player has to mull down to four in search of a keepable hand. Then given this deck and mulligan strategy, the number of non-games was roughly 1 in 19 under the Vancouver mulligan but will be roughly 1 in 190 under the London mulligan rule. That’s a major improvement.

Next, suppose that players are building their decks with the sole goal of minimizing the amount of non-games. That is, they want to optimize the probability of a keepable hand when willing to mull to 5. By enumerating over all possible land counts, my script determined that under the London mulligan rule, the probability of a keepable hand is indeed optimized with 27 lands. Under the Vancouver mulligan rule, however, this probability was optimized with 25 lands.

Now, real aggro decks in Modern don’t run 25 lands. This is because deck construction is not only about opening hand optimization but also about flood protection in the late game. Also, after a mulligan under the Vancouver rule, you could keep one-landers in the hope of scrying towards another land. These factors were not considered in this preliminary analysis.

Nevertheless, there is a difference, and it indicates that under the London mulligan rule, decks should run more lands than before. This is counter-intuitive, and requires further research.

UPDATED THE OP WITH THE FULL ARTICLE.

Isn't that the "Pudel's Kern"? London Mulligan shows you 7 new cards of which you may keep 6. Chance to hit 2 lands is the same as before. Vancouver Mulligan shows you 7 cards (6 at once, 1 on the scry) with the option of trading the 7th for an unknown 8th. If you ignore the possibility of going down to 5 (which means not seeing #7), the chance to hit 2 lands (by turn 2) should be higher.

edit: That's probably not it. After rereading I think he doesn't factor in the scry at all.

A land in your hand is worth two in your deck, or some other such nonsense.

Lol, now I’m going to end up thinking about this today.

That’s exactly how you’d do it (although the second calculation .3*.3 is more correct, even though they give you the same answer*). You see seven cards both times, right?

Anyway, you’re touching on why I prefer the mana base failure rate method. It’s not just a simple Hypergeometric calculation, it’s directly tackling the good game question: ‘what are my odds of having cards I can’t play.’ This goes beyond opening hand, and arrives at how will the deck perform.

*tldr; the reason is that failure is usually the simpler calculation of the two, so working from failure in probabilities tends to be easier. Then you just take 1-failure to get your probability of success. With Hypergeometric models, it’s no big deal, but some models require summations (or multiplication) across many possibilities.

Your tldr part is longer than the rest

tldr, for me: 'too long DON'T read' lol

Yes, we see seven card with every London Mulligan, ergo the odds are the same for every iteration. You're just chaining them.

Your method is more accurate for it has another layer. It's not (simply) asking "Will I have 1BB by t3 for Murder?" but "What are the odds that I do have Murder but not 1BB?" So you rule out the fail cases that don't hurt.

Sooooo
If I'm splashing a color on a 24 land deck and I have seven 3 cmc gold spells that I want to be able to cast on turn 4 how many sources do I need?
I'm thinking 12...

It's probably 11, but 12 won't hurt you if you have the room. Unless it has a double mana cost of the splashed color, in which case you'd need 18(!).

Interestingly, those 7 cards are likely to completely define your mana base on their own. Between them they will probably drive ALL decisions about mana colors for the entire deck. Crazy how that works when you think about it. Even more so if they were 2 CMC instead.

DJ, help me with the odds:

In a regular 2-color draft deck, what are the odds of getting at least one basic of both colors?

For simplification purposes, I assume 8 of each color. Not getting one of 8 Swamps for example is ~18%, that's how far I get. I don't know how to combine this with the odds for the other color. Multiplying is certainly wrong for the odds of failing on either color have to be higher.

As an aside, why is this 18% x 18% = ~3% not the same as the odds for getting no land at all, which I have at ~2%?

But back to the original question: Adding these up would give me 36% chance to fail, can this high be correct? I certainly feel like I get these hands all the time in draft, so that would explain it.

The reason the math doesn’t work is that the two events are not independent. Hyper geometric distribution calculators (online one) only work when you ask a simpler question, like the odds of at least one land in a 7 card hand. As soon as you ask a more complicated question, you need to think in terms of combinations instead. What you want here, depends on what you are asking.

For example, do you mean exactly one of each? At least one of each? Percent chance of getting at least one but not the other?

Unfortunately, each one has slightly different calculations.

The basic form can be found here: http://birdsofparadise-mtg.blogspot.com ... etric.html

For your question, I think you want (8 choose 1)*(8 choose 0)*(24 choose 6)/(40 choose 7)

Let me englishify this for you: the first term is how many ways we can pick one mountain from 8, next is how many ways to get 0 swamps from 8 (slightly tricky concept, but the answer is just 1 by definition*), then how many ways can the rest of the deck be ordered (huge number here), divided by how many ways can the whole deck be ordered (even bigger number).

Why the number choices? 8 land one mountain, 8 lands no swamps, 24 remaining cards (40-16 lands) with 6 chosen (doesn’t matter which), 40 and 7 just are deck size and hand size.

The problem is that your question is more complicated. Depending on what you specifically want to ask, you may need to sum across multiple calculated probabilities (for the at least one, I think we need to sum 8 calculations, but I need to think about it). That said, I think I figured out an easier way to do this last time I looked at it, let me think on it, and I’ll share it when I remember.

*complex mathematical proof is not needed, but it exists, in case you ever feel like going full nerd.