# Magic The Gathering: probability of casting a 3 drop on turn 3

So, I have a probability related question.

In MTG, which is the probability of landing a 3-drop on turn 3, but using the correct combination. For example, I want to cast a Geist of Saint Traft on turn 3 (assuming I started), what is the probability (given the assumption that I drop 1 land per turn) of having, by turn 3, one island, one plain and one other land and having the card in my hand ready for casting.

I know that as this is drawing without replacement, so some form of Multivariate Hypergeometric Distribution is needed, but I am a little lost.

Let say I have a 60 cards deck: 10 islands, 10 plains, 4 other lands (none of them giving either blue or white mana), 4 Geist of Saint Traft and 32 other cards.

Now, by turn 3 I would have drawn (assuming 1 draw per turn) 9 cards out of the 60 and for dropping the Geist of Saint Traft I would need at least 1 island, at least 1 plain, at least 1 other land (of the other 22 lands) and at least 1 Geist, so in that case would it be the same to calculate the probability of having by turn 3 none of those and subtract that from one? However, I am not sure how to deal with the case of the third land card I should pick it from 22 (the number of lands less the 2 already picked) or from 24 (the total)?

• This could maybe go on Mathematics, but I'll write up an answer anyway and if it's judged to be off topic here, a moderator can migrate it. Feb 11, 2017 at 20:43
• While deriving an answer is possible, a 10 million monte carlo simulation would take only a few minutes to code up and would probably be faster to execute even than plugging values into the solved formulas once the true, exact answer is found. Feb 11, 2017 at 22:32
• Deck = 10 I, 10 P, 4 L, 4 G, 32 O. Probability space: How many ways can the first 7+3 cards be drawn, in order? Possibly consider the entire deck to avoid dependencies. Now, how many of those are "success" events? Feb 12, 2017 at 2:12
• I would challenge the usefulness of the resulting figure, as being able to cast a single card in a single turn will hardly mean anything on it's own. You would have to incorporate playable starting hands if you want to get a useful number, which would be quite an ordeal (especially considering Mulligans, the Scry 1 if you're taking a Mulligan, and the two to three cards you draw until your T3 Main Phase). While I concur that math can help with deck building, this seems like a case where it's not worth the hassle, especially since you're likely not going to make decklist changes based on it. Feb 12, 2017 at 13:13

The thing is, having none of those cards is not the only way you'd be unable to cast the Geist on turn 3. You'd have to account for a whole bunch of possibilities:

• Zero islands, zero plains, and zero Geists
• Zero islands, one plains, and zero Geists
• Zero islands, two plains, and zero Geists
• ... (up to 9 plains)
• One island, zero plains, and zero Geists
• ... (up to 9 islands)
• Zero islands, zero plains, and one Geist
• ... (up to 4 Geists)
• Zero islands, one plains, and one Geist
• ... (up to 4 Geists)
• ... (up to 9 plains)
• ...

You get the picture. The number of combinations explodes.

I'd proceed like this. Group the possible arrangements of the deck which do let you cast the Geist by the number of lands you get in 9 draws, i.e. separately consider exactly 3 lands, exactly 4 lands, exactly 5 lands, and so on up to 8. You could also break these groups down further by the number of Geists you get in those same 9 draws, but it turns out not to matter; you only need the probability of getting at least one Geist. It's straightforward to get a computer to iterate over them and compute each probability using the hypergeometric distribution. This table shows the conditional probability that you have drawn a Geist in 9 cards, given the number of lands in those 9 cards:

Then for each number of lands, you have to calculate the conditional probability that, given that you have drawn that many lands, they include a combination of lands that will allow you to cast a Geist. There are two ways to go:

1. You can break these combinations down by (effectively) colorless lands vs colored lands, since that lets you immediately rule out any situation that doesn't have at least two colored lands. This table shows the conditional probabilities of getting specific counts of colored and total lands, given an otherwise arbitrary 9-card draw (that is, maybe it contains a Geist, maybe not, we're not worrying about that). E.g. the probability that you get exactly 5 lands, 3 of which are colored and 2 of which are colorless, is 0.160926.

After determining the number of colored lands, you just have to subtract off the probability that all your colored lands are of one type, because that's the only way you can fail to meet the colored mana requirements of the Geist with them. That can be calculated with a single-variable hypergeometric distribution. This table shows the conditional probability, given that you have a certain number of colored lands, of them being either all islands or all plains:

When I subtract these numbers from 1 and multiply by the corresponding figures in the previous table, I get the probabilities of drawing lands that actually let you cast a Geist. Then, I add up the probabilities for each number of total lands and multiply those by the corresponding figures from the first table, and finally sum over all possible numbers of Geists and all possible numbers of lands to get an overall probability of being able to cast the Geist of 0.2760477.

I haven't shown it here but I was also able to run the numbers using fractional arithmetic and it works out to 68014212/246385711

2. The alternative is to break down the possibilities by the total number of lands and calculate the probabilities for each total number individually using the multivariate hypergeometric distribution. For example, given that you get three lands exactly, here are the conditional probabilities of getting various numbers of plains and islands:

Or for five lands:

Then you can get the computer to select the cells corresponding to land configurations which allow you to cast the Geist, i.e. those which have at least one of each. This is a bit more tedious, but still can be automated with the help of a few tricks. By adding up all the selected numbers in each table, you get the conditional probability that you will be able to cast the Geist given that you draw that many lands. Then you can multiply those probabilities by the a priori probabilities of drawing that many lands and at least 1 Geist, and add them up to get the total probability of drawing a Geist and being able to cast it. I get the same answer of 0.2760477 or 68014212/246385711 by this procedure.

I have a messy Jupyter notebook with the code that computes this which I would be willing to fix up and publish, if people are interested, but it will take some time.

There is a legit mathematical answer to this question, but it's a bit outside my ability. What I can offer you, however, is an example of how Magic players of various levels of mathematical knowledge have attempted to tackle the question.

This article by Frank Karsten is a good start. Doing it mathematically, you get stuff like this:

Eventually, Karsten just gives up and decides to simulate it. A sample of his conclusions for 60-card decks:

• To have a good (90%) chance of casting a one-drop (e.g. Duress) on turn 1: 14 sources of its color.
• To have a good chance of casting a three-drop with one colored mana symbol (e.g. Thirst for Knowledge) on turn 3: 12 sources.
• To have a good chance of casting a three-drop with two of the same colored mana symbol (e.g. Liliana of the Veil) on turn 3: 19 sources.

Unfortunately, this particular analysis doesn't focus too much on multi-color non-land cards (more on multi-color lands, which are invaluable for building competitively powerful decks).

He also recommends looking at preexisting competitive decks as a guide, since players tend to build for consistency.

A 2012-season Delver deck is probably a good guidepost, since it's a deck that played Geist on-curve and as one of its most color-demanding cards.

How many white sources is it packing?

lands – 22:

9 Island

2 Plains

4 Glacial Fortress

3 Moorland Haunt

4 Seachrome Coast

card-filtering — 9 cards (plus the Snapcasters):

2 Thought Scour

3 Gitaxian Probe

4 Ponder

So that's 10 white mana sources, plus 4 card-filtering cards that dig deep (the Ponders). Pretty close to the 12 that Karsten said you want to get a 90% chance of casting a single-color three-drop on turn 3.

27.6% if you go first, or 34.5% if you go second. This does not account for mulligans.

Those numbers were obtained experimentally using the code posted below. There's an error in the math presented below. (David Z's theoretical result matches my experimental result, not my theoretical result which is slightly off.0 I'm leaving it up in the hopes that someone can correct me. Maybe I'm treating dependent events as independent? I intend to revisit this.

1. In the 9 cards drawn from the 60, there must be at least one of the 4 Geist of Saint Traft.
2. In the remaining 8 cards of 59, there must be at least one of the 10 Islands.
3. In the remaining 7 cards of 58, there must be at least one of the 10 Plains.
4. In the remaining 6 cards of 57, there must be at least one of the 22 remaining lands.

Or in other words,

1. At least one Geist of 4 in 9 of 60.
2. At least one Island of 10 in 8 of 59.
3. At least one Plains of 10 in 7 of 58.
4. At least one land of 22 in 6 of 57.

Useful "trick":

P(X) = 1 - P(not X)

so

P(at least one X) = 1 - P(no X)

1. P(at least one Geist of 4 in 9 of 60)
= 1 - P(no Geist of 4 in 9 of 60)
= 1 - (56/60 * 55/59 * 54/58 * 53/57 * 52/56 * 51/55 * 50/54 * 49/53 * 48/52)

Perl: my \$P1 = 1 - ( mul map { (60-4-\$_+1) / (60-\$_+1) } 1..9 );

2. P(at least one Island of 10 in 8 of 59)
= 1 - P(no Island of 10 in 8 of 59)
= 1 - (49/59 * 48/58 * 47/57 * 46/56 * 45/55 * 44/54 * 43/53 * 42/52)

Perl: my \$P2 = 1 - ( mul map { (59-10-\$_+1) / (59-\$_+1) } 1..8 );

3. P(at least one Plains of 10 in 7 of 58)
= 1 - P(no Plains of 10 in 7 of 58)
= 1 - (48/58 * 47/57 * 46/56 * 45/55 * 44/54 * 43/53 * 42/52)

Perl: my \$P3 = 1 - ( mul map { (58-10-\$_+1) / (58-\$_+1) } 1..7 );

4. P(at least one land of 22 in 6 of 57)
= 1 - P(no land of 22 in 6 of 57)
= 1 - (35/57 * 34/56 * 33/55 * 32/54 * 31/53 * 30/52)

Perl: my \$P4 = 1 - ( mul map { (57-(9+9+4)-\$_+1) / (57-\$_+1) } 1..6 );

Then you multiply those four probabilities.

#!/usr/bin/perl

use strict;
use warnings;
use feature qw( say );

use List::Util qw( shuffle );

sub mul { my \$acc = 1; \$acc *= \$_ for @_; \$acc }

{
my \$P1 = 1 - ( mul map { (60-4-\$_+1) / (60-\$_+1) } 1..9 );
my \$P2 = 1 - ( mul map { (59-10-\$_+1) / (59-\$_+1) } 1..8 );
my \$P3 = 1 - ( mul map { (58-10-\$_+1) / (58-\$_+1) } 1..7 );
my \$P4 = 1 - ( mul map { (57-(9+9+4)-\$_+1) / (57-\$_+1) } 1..6 );
my \$P = mul \$P1, \$P2, \$P3, \$P4;
say sprintf "Theoretical: %.2f%%", \$P*100;
}

{
my @reference_deck = ( ("G")x4, ("I")x10, ("P")x10, ("O")x4, ("x")x(60-4-10-10-4) );

my \$trials = 1_000_000;
my \$success = 0;
for (1..\$trials) {
my %hand;
++\$hand{\$_} for ( shuffle @reference_deck )[0..8];
next if !\$hand{G}--;
next if !\$hand{I}--;
next if !\$hand{P}--;
next if !\$hand{I} && !\$hand{P} && !\$hand{O};
++\$success;
}

my \$P = \$success/\$trials;
say sprintf "Experimental: %.2f%%", \$P*100;
}

Output:

Theoretical: 28.02%
Experimental: 27.65%

Theoretical: 28.02%
Experimental: 27.74%

Theoretical: 28.02%
Experimental: 27.58%

Theoretical: 28.02%
Experimental: 27.63%
• I think the 54% chance of just not drawing a Geist in the first 9 cards is kinda squishing the rest of the probabilities down a bit. Would it be more useful to treat that card as fixed and calculate "How likely am I to be able to cast it on turn 3 if it's in my opening hand?" Feb 11, 2017 at 23:24
• @Alex P, Experimentally: 56%. Feb 11, 2017 at 23:53
• I bet you're right that it's treating dependent events as independent in some case, though I can't immediately see where the mistake is. Feb 12, 2017 at 0:28
• P1 is really P(Geist in 9 of 60), and P2 is really P(Island in 9 of 60 | Geist in 9 of 60). We want P(Island in 9 of 60 ∩ Geist in 9 of 60), which is given by P(Island in 9 of 60 | Geist in 9 of 60) * P(Geist in 9 of 60), so as far as I can tell, the math is correct! But even in that simplified case, I get T:38.8, E:39.2. Feb 12, 2017 at 0:33
• Re "We want P(Island in 9 of 60 ∩ Geist in 9 of 60)", I think this is the part that's wrong. Feb 12, 2017 at 0:43