Simple Cycle Games

Here Players A and B play alternately inside a loop. at each of A's nodes in this loop, he has the option to pass (continue the cycle) or quit in which case some designated payoff (a,b) is assigned and the game ends. Similarly for player B.

There is not always a pure Nash equilibrium in such games. Consider the two node game:

A (1,-2) <====> B (2,-1)
It is possible that in case of positive payoffs, there is always a pure NE.

Consider for a example an arbitary 4 node game A4-B3-A2-B1 with payoffs (ai,bi) for i=1,2,3 or 4 right to left.

A prospective algorithm to find mixed solutions to this game can naturally ignore deterministic NE, in particular any equilibrium where a player is quitting becomes determinstic. Using this property and the fact that a player at a node is willing to mix only if his expected payoff from adopting 'pass' is the same as when he adopts 'quit'.
Then assigning some variables (a,b) as the expectation of B1 when passes, we cam build a tree that at the first node asks if b1 The reason we never ask b1>b is because this would imply a deterministic NE. Going down the tree and checking for consistency allows us to determine the mixed equilibria, although this looks like it might be too big a process. Another process that is also probably too big involves the computation of Utility functions:

if ri,si are the probability of quits and passes and the corresponding node and say B1 is the first player to mix in a particular equilibrium.

This implies

b1= E(B's reward in one cycle where he passes at B1) * (1/(1-  s2*s3*s4)

where E denotes expectation. The term on the right is the end result of an infinite sum.
i.e. this is b1= (b4*r4 + b3*s4*r3 + b2*r2*s3*s4) (1/(1-  s2*s3*s4). Similarly for the other nodes.

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