The sideways view

Avalon has many complexities that become irrelevant given good enough play. If we remove those complexities the game can be played much more quickly (personally I also find the simpler version more aesthetically appealing):

• Only play the three largest (non-*) quests.
  (Skip the easier two quests.)
• The good players win immediately if a quest succeeds.
  (Rather than needing to succeed on three quests.)
• A quest automatically fails if it contains at least one evil player.
  (Rather than using pass/fail cards.)
• Any proposal is accepted  as soon as a majority of players announce support for it.
  (Rather than having a rotating leader who makes proposals.)

In this post I’ll show that these simplifications don’t change the game if the players are reasonably sophisticated, except for a small change with 6 players that I think should be ignored. (This also breaks if you play with Oberon, an evil player who doesn’t know the identity of other evil players, but that seems rare.)

Implementation notes

• Evil players can concede immediately (and move on to guessing Merlin) when a quest passes. When the evil players don’t concede within a few seconds, we assume the quest fails. So we don’t need pass/fail cards.
• It’s faster to use a machine to randomize roles and give people their non-public information. I wrote a simple program to do this here; it would be better to have a version that runs on a phone though.
• To announce support for proposals, I like the following procedure:
  • Any player can say “I propose {A, B, C}” or “I withdraw my proposal” at any time.
  • Any player can support a proposal by pointing at the player who made it. (They can also express support and reach agreement verbally, but pointing is what finalizes it.)
  • If at any time at least half of other players are pointing at someone, that person says something like “It looks like {A, B, C, D}, support my proposal {A, B, C}. Any objections?”
  • They then count down: “3, 2, 1, done.” If none of the supporting players object before “done,” then the team is finalized. (They should only object if there has been a misunderstanding, e.g. if they misheard the proposal.)
  • The evil players can concede immediately after “done.”
• I prefer playing with a clock, and strongly prefer sharing a single pool of time across all quests. Separate pools incentivize you to use all the available time for every quest rather than finalizing a team once you are reasonably confident; it’s more fun to allocate more time in rounds when something interesting is happening. You don’t need much time, I recommend [2 minutes] * [number of players], and a much shorter clock for the evil players’ deliberation.
• It seems useful to choose a random team as the “default” for the first quest. (The simple app above gives a random permutation, the canonical starting team is the first N people in that permutation.)
• The game should be interpreted as zero-sum: think of the winning team as getting 1 point which is divided equally across all players. So winning as an evil player is a slightly bigger deal.
• I prefer the version where whispering is either explicitly allowed (e.g. players can disperse to separate rooms to have conversations), or explicitly disallowed (e.g. strong norms against trying to send signals that will be observable only to some players). If whispering is allowed, the game should probably include Mordred (an evil player who is not known to Merlin).
• It’s probably better to choose an assassin to pick Merlin rather than using a majority vote or consensus. I also think it’s good to have an explicit procedure for finalizing that decision, I suggest reusing “3, 2, 1, done.”

Equivalence

Intuitive argument:

• Good players can pick the biggest three teams first, and then just send a subset of one of the big teams on each of the easier quests. The evil players reveal strictly more information by failing earlier so should only ever fail the hardest 3 quests, and in particular should always pass *- quests.
• Evil players can choose a fixed ranking of the players, e.g. clockwise starting from the initial king, and only ever have the highest-ranked player on a quest fail. So the good players should always see exactly 1 fail card if a team contains an evil player, and none otherwise. (This ceases to work if you play with an evil player who doesn’t know the other evil players’ identities, but I’ve never done that. It works fine if you play with Snape, since Snape gives themself up immediately if they fail to follow this procedure.)
• If a majority chooses a team for the next quest using any mechanism, then the good players can just commit to only proposing and voting for that team. So it doesn’t matter what mechanism the rules choose, the good players should adopt whatever voting mechanism they think works best.

There is a small subtlety in games with 6 players, which I describe in the appendix; the fast game is equivalent to the normal game only if we switch the order of the 3rd and 4th quests, which may slightly benefit the good players.

Formally, we prove the following theorem, which implies that the two games are exactly equivalent for any players who are able to implement the mappings F, F’:

There is a mapping F that turns strategies in the normal game into strategies in the fast game, and a mapping F’ that goes the other way, such that:

1. For every fast good strategy A and every normal evil strategy B’, the winning probability of A against F(B’) is at most the winning probability of F'(A) against B’. In particular, the fast game is at least as hard for the good players as the normal game.
2. For every normal good strategy A’ and every fast evil strategy B, the winning probability of A’ against F'(B) is at most the winning probability of F(A’) against B’. In particular, the normal game is at least as hard for the good players as the fast game.

Proof of 1. The normal strategy F'(A) is defined as follows: simulate the fast strategy A in order to select a team X for the first of the three quests. Good players then agree to only propose subsets of team X, and vote for a team iff it is a subset of X. This guarantees that a subset of team X is selected for the next quest.

(This only works if the hard quests are sorted in increasing order of size, which is what guarantees that you can always choose a team X for the next hard quest, and then use that team for the next quest. This isn’t the case for the 23434 ordering with 6 players, which is what presents a wrinkle.)

They keep doing this until a subset of X fails a quest. At that point they give up on X and resume simulating fast strategy A until it selects a new team Y. They repeat this process Y, and if Y fails they make their final pick Z.

• Once the good players commit to only proposing a subset of X, a team can only get a majority vote if it’s a subset of X. Because the number of allowed proposals is larger than the number of evil players, the good players are guaranteed to eventually propose and then accept a subset of X.
• Regardless of what the evil players do, the good players will win if any of the three proposals X, Y, Z has no evil players—since the good players will eventually commit to picking only subsets of that team, and they will never fail a quest after that point.
• Define F(B’) as the strategy of the evil players in the simulated fast game, when they play against good players who use this simulation strategy. Then this strategy in the normal game can only lose if A loses to F(B’) in the fast game, as desired.

Proof of 2. The fast strategy F(A’) is defined as follows: simulate the normal strategy A’ in order to pick teams. Whenever A’ selects a team for one of the smaller quests, assume that the players in A’ see zero fail cards (in order to continue the simulation). When A’ selects a team for one of the largest quests, all good players announce support for that proposal for the next quest in the fast game. Whenever a quest fails in the fast game, assume that the players in A’ see exactly one fail card (in order to continue the simulation).

(This assumes that players are allowed to reveal information simultaneously in order to simulate the voting procedure in A’. If players can’t make simultaneous reveals except as part of the voting procedure, then the voting procedure actually does have intrinsic significance and can’t be removed. Simultaneous reveals seem pretty easy to implement, and I’ve never played with a group that tried to ban them.)

For any fast strategy B by the evil players, let F'(B) be the strategy defined as follows:

• Evil players pick a fixed ranking.
• Evil players say exactly the same things in the normal game that B says in the simulated normal game, when playing against good players who use the simulation strategy.
• Evil players play a fail card iff they are on one of the three largest quests and they are the highest ranked player on that quest.

If the fast strategy F(A’) loses to B’, then each of the three largest teams chosen by A’ contains at least one evil player, and so A’ loses to F'(B).

Appendix: the wrinkle with 6 players

For 6 players, the fast game has a slightly different structure, which requires four quests and the use of pass/fail cards in the first quest. I recommend ignoring this and just playing with the rules above, using quests of size 3/4/4. This simplification is easier for the good players than the normal game, and is equivalent to swapping the order of the 3rd and 4th quests in the normal game. (Interestingly, this is a proof of the non-obvious fact that swapping the 3rd and 4th missions can only benefit the good players).

Alternatively, you could play with quests of size 3/4/4, but the good players need to choose the first two teams simultaneously. The good players should still choose the team of size 4 under the assumption that the team of size 3 failed (since otherwise they will win anyway), but the process of choosing the team of size 4 may leak information about Merlin that they could have avoided leaking if they chose the teams sequentially. This simplification is strictly easier for the evil players than the normal game.

The more complex “fast” version that is exactly equivalent to full Avalon is as follows:

• The good players propose a team of size 3.
• We hand out pass/fail cards.
• If there is at least one failure, then we play the fast game with teams of size 3, 4, 4, but the good players have already lost the first quest.
• If there are no failures, then we play the fast game with teams of size 4, 3, 4.
• If the evil players always fail the first quest (respectively never fail), then this reduces to the fast game with sizes 3/4/4 (respectively 4/3/4). I think that 3/4/4 is slightly worse for the good players, which is why I suggested that simplification.

To see the non-equivalence with the simpler fast game, consider trying to do the simulation argument in the proof of claim 1. The trouble comes from the fact that the three hardest quests come in the order 4, 3, 4, rather than the order 3, 4, 4.

If we set up the fast game with a first quest of size 4, then the good players choose a team X of size 4 and propose subsets of X for the first two quests in the normal game. If one of these quests fails, then the good players actually need to pick two more teams of size 4, because they still have two quests of size 4 left. So they are at a disadvantage. (They’ve now gained more information than they normally would when the first quest fails, because they can narrow the evil player down to one of three candidates rather than one of two candidates, but it’s not at all obvious whether that helps enough to offset the extra difficulty.)

To avoid this, we could have the first quest in the fast game have size 3. Then the good players use that team for the first two quests in the normal game. If those both pass, they now need to choose a team of size 4 for the third quest. If that team succeeds they win, but if that test fails they now need to choose a team of size 3 for the fourth quest. If they reuse the team of size 3 that they used for the first two quests, that team may now fail for the first time. So the good players need to actually try to win when picking their team of 4 for the third team, but they need to make this decision before they know whether the team of 3 fails.