[Literature] Game Balance ch9 - Intransitive mechanics

My notes from course 9 of the Game Balance class of Summer 2010, by Ian Schreiber.

Reminder: “intransitive” is just a geeky way of saying “games like Rock-Paper-Scissors” – that is, games where there is no single dominant strategy, because everything can be beaten by something else. Ex:

• fighting games: blocks > throws > normal > blocks
• RTS: fliers > infantry > archers > fliers
• Advance Wars, and several FPS: Tank > Recon > Anti-tank weapon/infantry > Tank
• visible long-range atk unit > visible med range atk unit with radar > invisible short-range atk unit > visible long-range atk unit without radar
• Magic cards: (X/Y = X atk and Y HP). 3/2 > 1/3 > 2/1 with first strike > 3/2.

Intransitive mechanics are more interesting than transitive ones, esp in PvP games. Players can change strategies in mid-game, or bluff. Balancing is achieved by putting a probability ratio on how often each element of the chain will be used: Rock: 50%, Paper: 20%, Scissors: 30% (= 5:2:3).

Solving RPS

1. Normal-form representation: The payoff matrix consists of: my choices in the rows, the opponent's choices in the columns, and my payoffs in the cells. r, p, and s are the probabilities that my opponent throws rock, paper, and scissors. R, P, and S are the expected payoffs for each of my choice when I choose them during the game.
2. The matrix leads to 3 equations of 3 variables (R = s-p, etc ...)
3. If I know what my opponent plays on average, I know r, p, and s. Therefore I can get my payoffs: R, P and S.
4. The best strategy to use is the one with highest payoff.

Theorems from game theory

• Symmetric games have the same optimal strategy for all players. (Opponent's proba of choosing Rock is same as ours)
• All actions worth taking at all have equal payoffs. (R=P=S)
• In symmetric zero-sum games (eg RPS), all actions worth taking at all have a payoff of zero.

Games are solved considering each player plays optimally (unlike the previous section where I knew r, p, and s). As a side note, a symmetric game is represented using an anti-symmetric matrix. Also, note that the opponent's probabilities r, p, and s must always satisfy r + p + s = 1.

A choice is dominated if one of the other choices is always better. In that case, optimally, the dominated choice will never be picked. Hence, the dominated choice can be removed from the list of choices.

Solving "Rock wins count double"

1. Wins by R count as double. The game is still symmetric: payoffs are the same for the 2 players, the matrix remains anti-symmetric.
2. Obtain 3 equations from the payoff matrix. Each of those equations equals to 0 (cf 3rd theorem above).
3. Solve
4. Conclude: P will be played 50% of the time if wins by R count double.

RPS with costs and partial wins

Solving K(night) - A(rcher) - F(lier), the three being units of a RTS game with costs and partially beating each other. For example, a Knight costing 50g will kill an Archer but lose 20% of its life in the process. What differs is only the payoffs matrix. Solving the equations is the same. Result: k:a:f = 14:10:13.

-	k	a	f
K	50-50 = 0	(-50*0.2)+75 = +65	-50
A	-75+(0.2*50) = -65	75-75 = 0	(-75*0.4)+100 = +70
F	+50	-100+(75*0.4) = -70	100-100 = 0

Asymmetric payoffs

1. If player A wins double with R, but not player B, then there are 2 matrices to look at (one per player).
2. The game is not symmetric anymore, therefore R, P, and S are not 0, and they are different for each player (strictly positive for the player with a bonus, strictly negative for players with handicap). Solve A's matrix: r:p:s = 3:5:4 (interestingly, R has highest payoff but is played least often).
3. Solve B's matrix: r:p:s = 4:5:3.
4. With each of the solution, it's possible to get X and Y.

3-player RPS

In the 3-player case, let's assume I score two points when I beat both of the 2 other players (I play S, they play P and P), and I score one point when I am tie with one player and beat the other (I play R, they play R and S). In that normal case, the intuition is 1:1:1 because it's not different from the 2-player RPS. Hence let's say that wins with R count double (but not the losses). Hence, R vs SS gives +4 (instead of +2), and S vs RS gives -2 (instead of -1). The solution is: r:p:s = 3:4:3.