3- way rock paper scissors (part 1)

 Rock paper scissors is a game for two people. But as we can see here, it does not easily scale up to three people.

 But is interesting that it has the following properties, and we would like to keep them for our 3 person game:

1.  Symmetry: No action is inherently different from any other action.
   In other words: The permutation group of isomorphic permutations over the actions is transitive.
   
2. Unambiguous: Unless all players select the same action, there is never a draw.
3. Single Winner: For all players, there should be a single winner. All losers are considered equivalent.
4. Order invariance: The winner is selected indifferently from the order the players go in. 

First, one can easily see that if two or more players (but not all of them) select the same action, they both lose, due to a combination of axiom 2, 3 and 4. (two or more winning break axiom 3, one of them winning breaks axiom 4, drawing breaks axiom 2). 

This means that for a number of players larger than 3, a system like this cannot be constructed. (because two players taking one action and all other player taking a different action can never be resolved).

Now let us introduce some notation. Note the different actions as 0,1,2,..., and note all player actions as {0,0,1} (this is two players play action 0, 1 player plays one). The winner is well-defined by its action, so if the player with action 1 wins, we can write this as such {0,0,1} = 1. Due to axiom 4 {0,0,1} = {1,0,0} = {0,1,0}. Here equal means "has the same winner".

For the isomorphisms, we use cyclic permutation group notation. So is the isomorphism switches 0 with 1 and 2 with 3, then we note this as (01)(23).

Another thing that can be noted is the following: no isomorphic permutation can contain a cycle of the same length as the number of players. This can easily be proven, as follows

Assume there are three players, and there is an isomorphic permutation with the cycle (abc), and a is the winner of the play {a,b,c}. Then if we apply the permutation, we get a play {b,c,a} where b is the winner, but this is the same play as {a,b,c}, with a different winner, so the permutation is not isomorphic. the proof works exactly the same for any other number of players.

2 players and 2 actions

For two players and two actions, there is only one non-identity permutation, being (01), and this permutation has length 2. To make the permutation group transitive, we have to make this isomorphic, but we cannot, as it has the same length as the number of players. So this is impossible. For the same reason 

2 players and 3 actions

This is the classic rock paper scissors. But we can build the game from the ground up. Over three elements, there are 6 permutations, being (1)(2)(3), (12)(3), (1)(23), (13)(2), (123) and (132). Only the identity and the rotations do not contain a cycle of length 2, so we will have to use them for our cyclic transmutation group. 

There are 3-non draw plays: {0,1}, {0,2} and {1,2} Non of them can have the same winner (because then a reflection would be isomorphic), so we can choose one arbitrarily. For example {0,1}=1 makes {0,2}=0 and {1,2}=2. This matches 0=rock, 1=paper 2=scissors. If we had made the opposite choice the results would have been equivalent.

3 players and 3 actions

Impossible because each third=order permutation group contains at least a 3-cycle.

In other words, there is never a good way to resolve the case where one person takes rock, one takes scissor, and one takes paper.

3 players and 4 actions

The permutation group [(0123),(02)(13),(0312)] is transitive and contains no 3-cycles.

Assume {0 1 2} = 1

Then {1 2 3} = 2, {2 3 0} = 3 and {3 0 1} = 0

In addition {0 0 1} = 1 is the only way to resolve that without breaking axiom 3.

So in short, the rules can be condensed as follows:

• If all three players take the same action, it is a draw
• If two of the three players take the same action, the third player wins
• If all three players take a different action, the one with the different parity wins.

What is notable on this game is that only the parity you play matters, not the actual action.

3 players 2 actions

There is only one permutation (01), and in the end, it doesn't matter.

Rules can be derived from previous statements:

• If all three players pick the same, it is a draw
• If two of them pick the same colour, the third player wins.

This is fully consistent with all rules, but in practice, this is not a very good game.

Imagine you have finished a certain game a certain way. What would happen if you made a different choice. The results are the following:

• If you won, switching would have resulted in a draw.
• If you draw, switching would have resulted in a win.
• If you lose, switching would have just resulted in a loss, but the winner changes.

In addition to this, two players can work together, agreeing to pick a different colour than each other, ensuring that the third player can never win or draw, everything the third player can do is choose which of the two original players wins.

Therefore, to make the game better, we could add an extra axiom,

• Each player must have a winning move, given he knows the moves of all other players.

Sidestepping this issue. Adding this axiom makes constructing a game with 3 players and two actions impossible.

How to construct this game with the axiom can be read in a future blog