# Two-round game where one player commits to cooperate.

Quick question about game theory. Suppose you have a two round game with two players: A and B. There are two strategies, cooperate or cheat.

If both players cooperate, they each receive 10. If one cooperates but the other cheats, then the cooperative player gets 5 while the cheater gets 15. If they both cheat, then both get 7.

If player A commits to cooperating in the first round and adopts a tit-for-tat strategy on player B (if B chooses to cooperate in the first round, A will follow and cooperate in the second round and vice versa for cheating)

What is the likely outcome and payoff of the game?

• Any thoughts? First, I suggest clarifying the question. Usually "tit for tat" depends on the opponent's prior move, not "whatever $B$ chooses in the first round" as you say. If you mean something non-standard, you should explain. Second, how can we speak of likely outcomes when we have no information regarding $B's$ strategy? – lulu Apr 9 '18 at 11:59
• @lulu I think you're reading "whatever" differently than I am. I read it as "Player $A$ will in the second round do whatever $B$ did in the first round". Granted, the phrasing is a bit unclear. – Arthur Apr 9 '18 at 12:00
• @Arthur on re-reading, I hadn't focussed on the fact that the game is to only have two rounds. Given that, then yes...the only "prior move" is the first move. – lulu Apr 9 '18 at 12:12
• Edited for clarification. What I meant is that A has committed to cooperating in the first round and will play whatever B did in round 1 as A's move in the second round. – Wibble Apr 9 '18 at 12:31
• of course assuming player B prefers more to less, and has knowledge of player A's strategy, player B should play play (cooperate,cheat). Then player B gets 10+15=25. – palmpo Apr 9 '18 at 19:42

If we represent the decisions of both players in the 2 rounds as a $2x2$ matrix: $$M = \left[ {\begin{array}{cc} A_{0} & B_{0} \\ A_{1} & B_{1} \\ \end{array} } \right]$$ and say, $A_{i}$ or $B_{i}$ takes the value 1 if the player co-operates, else 0 if the player cheats. We have these conditions for our problem, that $A_{0} = 1$, and $A_{1} = B_{0}$. So the set of all possible such matrices as outcomes are: $$\left[ {\begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array} } \right], \left[ {\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} } \right], \left[ {\begin{array}{cc} 1 & 1 \\ 1 & 0 \\ \end{array} } \right], \left[ {\begin{array}{cc} 1 & 1 \\ 1 & 1 \\ \end{array} } \right]$$
We do not really know anything about player $B$'s strategy, so we consider every scenario equally possible. Calculating the payoffs, we get $(12,22)$, $(20,20)$, $(15,25)$ and $(20,20)$ for the above 4 matrices respectively.
So the average payoff should be $(A=14.25, B=21.75)$