The Story So Far…
Two players are presented with an opportunity.  Each may remain loyal to the other player and betray her. Neither player will interact with the other in any way ever again. There be no out-of-game way to be rewarded and punished.

What could happen?

(1) If both remain loyal, each of them gets the “Cooperation” reward.

(2) If both betray the other, each of them gets the “Betrayal” reward.

(3) If one betrays the other while the other remains loyal, the betrayer gets the “Traitor” reward while the loyal one gets the “Sucker” reward.

This is the prisoner’s dilemma.

In a strict prisoner’s dilemma, the rewards are staggered with the “Traitor” reward best, followed by the “Cooperation” reward, then “Betrayal,” and finally “Sucker” the worst.  In mathematical terms, T > C > B > S.

Further, cooperation is much more likely if this decision point occurs repeatedly but obscures the exact number of times it will occur.

 

Correctly Staggering Rewards

In a strict prisoner’s dilemma, the rewards are staggered with the “Traitor” reward best, followed by the “Cooperation” reward, then “Betrayal,” and finally “Sucker” the worst.  In mathematical terms, T > C > B > S.

We discussed the Friend or Foe game show in detail in our last column.  Notice that this game deviated from this structure a bit–the “Betrayal” reward and the “Sucker” reward were identical; going home with no money.  Returning to mathematical terms, T > C > B = S.  If you opt to have two identical outcomes, this is the place to do it–at the bottom.  Notice also that in their scheme, T + S = C + C = total prize money.  Considering the demands of a of a game show–the need for clear rules that are easily parsed by the audience viewing at home–I can certainly see why they made this decision.

Personally, I would go a bit further than the standard T > C > B > S. I would also aim for 2C > T + S > 2B.  This was the reward structure of the peace war game and it suits my general design style.  Let’s go back to our Friend or Foe game show for an illustration.

Two players are going into the final showdown.  They have amassed $1000 in prize money.  Here are the possible outcomes under the standard game rules:

A) If both vote friend, they each get $500.

B) If both vote foe, they each get $0.

C) If one votes friend and the other votes foe, the foe gets $1000 and the friend gets $0.

Now imagine that we make a small tweak to the rules.  If both players choose “Friend,” we’ll throw in an extra 10%.  Now the decisions are:

A) If both choose friend, they each get $550.

B) If both choose foe, they each get $0.

C) If one chooses friend and the other chooses foe, the foe gets $1000 and the friend gets $0.

This is a small change but it has broad implications for the players.  If the players consistently choose friend, they end up collectively further ahead on repeated plays–$550 + $550 = $1100–than any other case–$1000 + $0 = $1000 in case (B) and $0 + $0 = $0 in case (C).  Of course, this game show doesn’t have repeated plays.  This decision is the last one of the game. But these players are conscientious, not amoral.  And that makes it all the more challenging for our players.

To Be Continued…

Our next column addresses ways to put Friend or Foe mechanisms into our own designs. See you Tuesday!

Have you played a game with a Friend and Foe mechanism? What did you think of it? Have you written one? How did your players respond to it?  Share with your fellow readers in the comments below.  And if you’re enjoying what you’re reading, create an account with WordPress and follow this blog.  You keep reading. I’ll keep writing.