Rock. Paper. Scissors. A classic game, because it is so simple. Many Magic writers have long theorized that the Magic metagame is like rock, paper, and scissors. Deck A beats Deck B which has a winning percentage against all of the decks in Class C. Since the decks of Class C beat Deck A, we have a balanced metagame. In theory, at least.
That’s not always true in reality. In Odyssey Block, for example, you were either with Blue/Green Madness or with Mono-Black Control or with something from left field. A few extra deck types appeared at the tail end of the season, but for the lion’s share, OBC was lacking in diversity.
We’ve seen, therefore, that the theory of rock-paper-scissors does not hold up in every metagame. What about the perfect metagame? In casual Magic, unless a specific group has established rules for itself, virtually every card in Magic is game to be played. With so many strategies available (and so many silver bullets to those strategies), do multiplayer groups follow this pattern?
In the perfect world of casual Magic, are there enough cards, enough options to break out of the rock-paper-scissors trichotomy? Or is casual Magic still bound to a basic system of rule?
Control beats Aggro beats Combo beats Control…..
Or does it? And what about other archetypes that defy description? Tempo? Aggro-control? No matter how focused a Five Color deck is, there are going to be elements of all three”overtypes” in most decks. Obviously, Five Color has a bit more room.
Shouldn’t Casual Magic have more room as well?
Which led me to an interesting dilemma. This happened in a multiplayer group recently, and it’s an interesting question to me. What is the right play?
There is a famous situation in philosophy called the Prisoner’s Dilemma:
There are two thieves who are partners. The police discover the crime, and capture both thieves. There is not enough evidence to convict the thieves of a major crime, just a few misdemeanors. The police detective, being a smart man, separates the two thieves. He approaches both Thief A and Thief B with the same offer. If the thief confesses to the crime, and helps the police convict his partner, then the police will look kindly upon him. In fact, a thief who confesses and convicts their partner will only be subject to one year in jail and another year of probation. However, the thief who is convicted will go away for ten years plus five probationary years. If both thieves happen to confess, each will receive three years of jail time. Of course, if neither confesses, then both receive minimal jail time and a year of probation.
What does a thief do? The interesting thing to consider is that the thief does not know which choice his companion selected. So, all a thief can do is select a pair of options, not the exact result. The thief can choose from A). 10 years jailtime or 1 year probation by not confessing, or B). 3 years jailtime or one year jailtime for confessing. Although the second option seems more attractive, there is a possibility that choosing the first option will result in the least jail time. What choice to make?
The Prisoner’s Dilemma is an interesting enough dilemma on its own merits. Philosophy has a lot of ways of analyzing it. Rationally, the second choice, to confess, seems the better option, yet two actors taking non-rational decisions will yield better results. Interesting. It’s also a fairly common situation. Take another example which is similar although not exactly the same:
A state university continues to increase tuition. The student government organizes a huge protest on the Governor’s Mansion lawn. There is reason to believe that the Governor might overturn the latest tuition increase if enough college students show their support of such a measure. You are contacted to go to the capital and show your support. You can either go, or not go. And either the Governor will choose to overturn the tuition increase or he will not. Going all the way to the capital will be expensive, take time, cut into your partyi…er…studying, and so forth. Do you go?
Again, one decision, but since you have no way of knowing how many students will go and if the Governor will overrule the tuition increase, you have to choose from two sets of results. A). You go, and the Governor doesn’t decide to change the rule (the worst possible outcome) or you go and the Governor does elect to repeal the tuition increase, or B). You stay home, and the Governor doesn’t decide to the change the rule (the status quo) or do not go and the Governor does change the rule.
It does not seem like a difficult choice. Obviously, since you cannot control or know how many will go or what the Governor’s decision will be, you should choose to not go to the demonstration. However, note that if every student made the same perfectly logical choice, then the Governor certainly won’t halt the spike in tuition and nothing was accomplished. Only by having a cornucopia of students who are willing to set aside the rational choice and storm the capital will the tuition be overturned.
These sorts of dilemmas are arguably witnessed with such social phenomenon like low voter turnout or dozens of spectators watching a crime occur while nobody calls the police or steps in to intervene.
Philosophical musings of this sort are terribly interesting, and have nothing to do with Magic. Or so I thought, until this happened during one multiplayer session recently.
This situation is a little different because it is neither exactly like the Prisoner’s Dilemma, with two actors, nor is it like the classic Magic rock-paper-scissors view of the metagame because, again, we have multiple actors.
This particular situation occurred at the end of a multiplayer game, when the table had been reduced to three players. It just as easily could have occurred in a game that started with three players as well. Anyway, without further ado, here is the Multiplayer Dilemma:
Player A is playing a Red/Black deck chock full of evasive creatures (many with Shadow), a bit of removal, and a smattering of burn. Player A is able to race past the defenses of both of the other players at the table, and deal a bit of damage to the head. He cannot remove too many permanents and simply is incapable of destroying enchantments, although he does have a few emergency cards for the other permanent types.
Player B is playing a Blue/Black deck with a large portion of countermagic. Player B has a few walls plus a few more defensive creatures, and just a couple of creatures that serve as winning conditions. Player B has practically no permanent removal except for a couple of creature removal spells, but he has recurring bounce effects.
Player C is playing a White/Green deck with a small number of very efficient creatures, a variety of ways of stopping creatures, and several permanent effects through enchantments that gain board position. This deck also sports Wrath of God effects that can sweep the board, although the way Player C wins is usually through just a couple of spells.
In the above case, the decks were made up, but they’ll suffice. Now, assume that each player is of approximately the same skill and that each is set up accordingly when our dilemma begins. Obviously, the dynamics change if somebody gets mana screwed, so we’ll assume an established board position for each player.
Player A is able to player enough creatures to get past Player B’s countermagic and feeble defenses. Since most of Player A’s creatures will evade Player B’s defenses, Player A should be able to take out Player B with very little difficulty.
Player B has enough countermagic to stop and prevent Player C’s key spells. Likewise, Player C’s smaller number of efficient creatures will run into Player B’s defenses, countermagic and bounce. Once Player B establishes control on the board, bounce effects can be used to clear a path for the few winning conditions in the deck. Player B should be able to defeat Player C with ease.
Player C has a variety of non-creature ways of handling a creature rush, from sweeping board clearers to enchantments that slow or stop attacks, even from evasive creatures. The creatures that Player C does have are bigger and just as easy to play, so Player C should get an advantageous position over Player A quickly. There should be no way that Player A should overcome these disadvantages, and Player C should be able to take out Player A.
What we have here is the Multiplayer Dilemma. Player A defeats Player B defeats Player C defeats Player A.
This is where the dilemma enters. How do you win the game? No matter which deck you are playing, you can easily take out one player, but if you do, then you are left with a duel that you should lose, by all rights.
What happens then? How different players react when under the Multiplayer Dilemma is interesting. In my real life situation, only one player (Player A) realized the situation, and actively helped his target (Player B) defeat the third player who could defeat Player A (Player C). In other words, a player helped the only player at the table he could beat defeat his enemy.
Suppose that each player recognized the Multiplayer Dilemma for what it was. We would have a pretty cycle: Player A helps Player B defeat Player C who helps Player A defeat Player B who helps Player C defeat Player A.
Helping the only opponent that you can defeat has a hint of irony.
The beauty of the Multiplayer Dilemma is that is doesn’t have to apply to group of three players. In multiplayer it often takes five or ten turns to figure out what everyone is playing. No matter how many players are around the table, I then can identify who I can take out and who I have little chance of defeating. Those who present the greatest threats to me are those I help defeat.
Helping another player is not just a way to build good vibes at a multiplayer table. It’s not just a way make something cool happen (we once had Shahrazad cast only to have it Forked. Twice). It’s also a way of ensuring your own victory.
Now that I start looking for it, the Multiplayer Dilemma happens in just about every multiplayer match, to one degree or another. If one knows to look for it. And now I do.