The format for the upcoming Mythic Invitational is the brand-new Duo Standard. To quote the announcement, the rules are as follows:
In Duo Standard, players will submit two Standard-legal decks. Players will not sideboard between games, though they may submit a sideboard (for cards such as Mastermind’s Acquisition). Players can submit two different decks, two of the same archetype with different cards in both, or the exact same deck twice. Have at it!
Match play will take place using both of each players’ submitted decks. Matches will be conducted as follows:
Game 1
- Which deck each player plays and who goes first will be determined randomly.
Game 2
- In the second game, players will use the deck that they didn’t play in game 1.
- The player who went second in game 1 now goes first.
Game 3 (if necessary)
- Each player selects one of their two Standard decks to play for the third game.
- Then the player who goes first in this game is determined at random.
This format poses an interesting challenge in terms of which decks to submit. Instead of finding a single deck that has the best win rate against the expected metagame, you now have to find a combination of two decks that give you the best chance to win a match, taking into account that both you and your opponent have a simultaneous choice for game 3. Fortunately, the mathematical field of game theory, which was developed nearly a century ago by John von Neumann, can help model and analyze such interactive decision making.
The Game 3 Decision: Example 1
Let’s start with some examples to build our intuition. Consider a matchup between, say, Luis Scott-Vargas and Martin Jůza, who are going into game 3. So they have to choose a deck from the two they have submitted. Let’s suppose that the matchup matrix is as follows.
Jůza | |||
White Weenie | Mono-Red | ||
LSV | Anti-Aggro | 60% – 40% | 65% – 35% |
Anti-Control | 30% – 70% | 35% – 65% |
The first number in any cell is the probability that LSV’s deck from that row will beat Jůza’s deck from that column. So in case LSV chooses Anti-Aggro and Jůza chooses White Weenie, then LSV would be 60% to win the game and hence the match, and Jůza would be 40% to win. We assume that both players know and agree on these percentages. (The numbers and decks are all fictive—I don’t know what these players have actually submitted or how their decks might match up.)
Given the above matchup matrix, what should both players choose?
Well, this is an example where one player has a dominant strategy. Independent of what Jůza chooses, LSV is better off with Anti-Aggro. Indeed, it is better than Anti-Control against both of Jůza’s decks. Therefore, LSV should always choose Anti-Aggro.
Jůza | |||
White Weenie | Mono-Red | ||
LSV | Anti-Aggro | 60% – 40% | 65% – 35% |
Knowing this, Jůza should pick the deck that is best against Anti-Aggro, which is White Weenie. In the parlance of game theory, (Anti-Aggro, White Weenie) is a pure-strategy Nash equilibrium, which means that no player will find it profitable to unilaterally deviate from these deck choices. LSV will win the third game 60% of the time.
The general lesson here is that if you have two decks that share good and bad matchups (as is the case for Jůza, in this example) then your opponent will pick your bad matchup for game 3. So doubling up on the same type of deck (or even worse, submitting two copies of the exact same deck) puts you at a disadvantage when it comes to game 3. For this reason, it can be advantageous to submit two widely different strategies.
The Game 3 Decision: Example 2
In general, the game 3 decision need not be dominance solvable. Let’s see what would happen if Juza had submitted a more varied combination of decks.
Jůza | |||
Esper Control | Mono-Red | ||
LSV | Anti-Aggro | 40% – 60% | 65% – 35% |
Anti-Control | 70% – 30% | 35% – 65% |
Here, the choices are not as obvious. If Jůza picks Esper Control, then LSV will be better off with Anti-Control. But if Jůza picks Mono-Red, then LSV will be better off with Anti-Aggro. Likewise, the best choice for Jůza depends on LSV’s deck choice.
So it’s a guessing game. If one of the two players is a mind mage with the soul read on what the other might do, then they could select the best answer. But it’s nearly impossible to figure out with certainty what your opponent might do in such a situation. And any attempt to exploit a perceived flaw in your opponent’s deck selection would, in turn, make yourself exploitable.
A relevant game theory concept for these types of situations is the mixed equilibrium. Basically, each player chooses a deck with a probability that makes the other player indifferent between their two decks. For the above example, LSV could make Jůza indifferent by choosing Anti-Aggro with probability P and Anti-Control with probability 1 – P such that 0.60 * P + 0.30 * (1 – P) = 0.35 * P + 0.65 * (1 – P). Basic algebra yields P = 7/12 as the solution. Meanwhile, Jůza would choose Esper Control with probability Q and Mono Red with probability 1 – Q such that LSV becomes indifferent, that is, such that 0.40 * Q + 0.65 * (1 – Q) = 0.70 * Q + 0.35 * (1 – Q). This solves to Q = 1/2.
So the mixed equilibrium for the above example would be that LSV rolls a 12-sided die and chooses Golgari Midrange if the outcome is 7 or lower, and Jůza literally flips a coin to choose Esper Control exactly half of the time. This randomization may seem weird, but it ensures that their strategy is not exploitable and that there is no profitable counter-play for the opponent. Poker players often refer to this approach as “Game Theory Optimal.”
Under this mixed equilibrium, neither player will find it profitable to unilaterally deviate from their strategy, and LSV will win the third game of the match 52.5% of the time. In general, an equilibrium (pure-strategy or mixed) always exists for the game 3 decision in Duo Standard. I made a spreadsheet that determines equilibrium strategies and payoffs for any two decks here.
The Overall Match Win Probability
Consider decks A and B for player 1 and decks C and D for player 2. Assuming no play-draw dependencies—more on that later—the probability that deck A defeats deck C is denoted as W(A, C), and analogous notation is used for other deck combinations.
Given these probabilities, suppose that player 1 has an equilibrium probability G of winning game 3. The claim that this number is well-defined is only a conjecture at this point for me, but a formal proof would be beyond the scope of this article. Taking G for granted, some algebra reveals that the probability that player 1 will win the match is given by:
0.5W(A, C)W(B, D) + 0.5W(A, D)W(B, C) + G * [ 0.5W(A, C) + 0.5W(A, D) + 0.5W(B, C) + 0.5W(B, D) – W(A, C)W(B, D) – W(A, D)W(B, C) ].
This ugly formula is implemented in the aforementioned spreadsheet. In fact, the spreadsheet contains a slightly more involved formula that allows for the game win percentage to increase when you’re on the play and to decrease when you’re on the draw, but I found that the resulting impact on the match win probability is negligible.
If I apply the match win probability formula to the examples, then we get the following results.
Jůza | |||
White Weenie | Mono-Red | ||
LSV | Anti-Aggro | 60% – 40% | 65% – 35% |
Anti-Control | 30% – 70% | 35% – 65% |
The probability that LSV wins the match is 52.95%. So even though LSV is 60% to win game 3, his poor game with Anti-Control will still drag him down. It’s neat that we can calculate by exactly how much.
Jůza | |||
Esper Control | Mono-Red | ||
LSV | Anti-Aggro | 40% – 60% | 65% – 35% |
Anti-Control | 70% – 30% | 35% – 65% |
The probability that LSV wins the match is 53.64%. This is much closer to the equilibrium 52.5% probability of winning game 3, as there is no guaranteed bad matchup to drag him down.
The Deck Submission
Suppose that there are only four possible best-of-one decks in the entire metagame, with the following matchup matrix.
White Weenie | Mono-Red | Golgari Midrange | Esper Control | |
White Weenie | 50% – 50% | 41% – 59% | 40% – 60% | 70% – 30% |
Mono-Red | 59% – 41% | 50% – 50% | 35% – 65% | 65% – 35% |
Golgari Midrange | 60% – 40% | 65% – 35% | 50% – 50% | 35% – 65% |
Esper Control | 30% – 70% | 35% – 65% | 65% – 35% | 50% – 50% |
Again, I don’t know what the actual numbers would look like or what the typical card choices for these decks might be. The precise rules of the game aren’t even known because the opening hand algorithm is not public and may or may not have changed to three hands. But this matrix is based on the classic scenario where midrange beats aggro, control beats midrange, and aggro beats control. Basically, you beat a deck if you are slightly slower than they are, but you lose if you are much slower than they are. I tacitly assume that Esper is geared towards beating midrange and doesn’t overload on Cry of the Carnarium or Moment of Craving.
I included two aggro decks: Mono-Red and White Weenie. In this example, Mono-Red has a small advantage over White Weenie due to Goblin Chainwhirler, but due to its reliance on burn spells, you can see that it’s slightly weaker to Wildgrowth Walker and Absorb.
If this were a “single Standard” tournament where you just pick one deck, then the equilibrium metagame would be 3/9 White Weenie, 0/9 Mono-Red, 4/9 Golgari Midrange, 2/9 Esper Control. You may verify that in this metagame, each non-Mono-Red deck would have 50% to win an arbitrary game.
In Duo Standard, the deck combination choices and their corresponding match win rates (according to the formula from the previous section) explodes into this.
White + White | White + Red | White + Golgari | White + Esper | Red + Red | Red + Golgari | Red + Esper | Golgari + Golgari | Golgari + Esper | Esper + Esper | |
White + White | 50.00% | 41.00% | 40.00% | 60.00% | 36.65% | 35.68% | 50.68% | 35.20% | 49.60% | 78.40% |
White + Red | 59.00% | 50.00% | 40.91% | 65.34% | 45.50% | 36.64% | 56.50% | 32.80% | 47.05% | 76.30% |
White + Golgari | 60.00% | 59.09% | 50.00% | 55.88% | 60.91% | 49.00% | 54.03% | 45.00% | 47.13% | 63.70% |
White + Esper | 40.00% | 34.66% | 44.12% | 50.00% | 33.74% | 40.67% | 44.23% | 60.45% | 59.38% | 70.00% |
Red + Red | 63.35% | 54.50% | 39.10% | 66.26% | 50.00% | 35.00% | 57.50% | 28.18% | 41.83% | 71.83% |
Red + Golgari | 64.32% | 63.37% | 51.00% | 59.33% | 65.00% | 50.00% | 56.14% | 42.50% | 43.86% | 58.18% |
Red + Esper | 49.32% | 43.50% | 45.97% | 55.77% | 42.50% | 43.86% | 50.00% | 58.18% | 56.14% | 65.00% |
Golgari + Golgari | 64.80% | 67.20% | 55.00% | 39.55% | 71.83% | 57.50% | 41.83% | 50.00% | 35.00% | 28.18% |
Golgari + Esper | 50.40% | 52.95% | 52.87% | 40.62% | 58.18% | 56.14% | 43.86% | 65.00% | 50.00% | 42.50% |
Esper + Esper | 21.60% | 23.70% | 36.30% | 30.00% | 28.18% | 41.83% | 35.00% | 71.83% | 57.50% | 50.00% |
Right. Well, I’m not solving this by hand, I can tell you that much. Using an interleaved reverse search algorithm for polyhedral vertex enumeration, as implemented here, I found the following unique equilibrium (rounded to one decimal):
- 47.1% White Weenie + Golgari Midrange
- 30.9% Golgari Midrange + Esper Control
- 22.0% Mono-Red + Esper Control
You may verify that in this metagame, each of these three deck combinations would have 50% to win an arbitrary match. No edge can be gained by switching to another deck combination.
Four interesting insights arise from this example:
- All equilibrium deck configurations combine two different types of decks. So it’s aggro + midrange, or midrange + control, or control + aggro. We do not see a combination of the two different aggro decks, and we do not see anyone pick two copies of the exact same deck.
- Mono-Red, which was not a good choice for the “single Standard” equilibrium, becomes a valid option in Duo Standard. The ability to choose it against both White Weenie and Esper Control for game 3 makes up for its slightly subpar performance overall. This means that the winning Invitational decks need not be the best choices for the best-of-one ladder.
- As for the aggro + midrange option, the equilibrium metagame contains White Weenie + Golgari Midrange, but not Mono-Red + Golgari Midrange. The reason is because White Weenie is better at covering Golgari Midrange’s weakness: Esper Control. We see Mono-Red + Esper Control and not White Weenie + Esper Control for analogous reasons.
- For the possible non-mirror matches between the three equilibrium deck combinations, the game 3 decision is never dominance solvable. If things are similar at the Invitational, then we may see a lot of mind games, second-guessing, possibly exploitable strategies, and/or literal coin flipping to determine decks for game 3.
For those of you interested in the Mythic Invitational, which will begin this Thursday at 10 a.m. Eastern with live coverage on twitch.tv/magic, I hope that this article was insightful. I emphasize that my work was exploratory in nature, based on a single example that was heavily influenced by the decks and fictive matchup numbers I chose. Yet from a mathematical perspective, Duo Standard is complicated, deep, and deserving of further research.
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