[xpsyuvz]

edit: I was trying to save StephenH some time, but it looks like he already posted while I was writing this.

Donkey69,

I came up with a counter-strategy to yours to show how it can be exploited ( -- and my opinion is that any strategy can be exploited if the opponent knows the strategy that is used).

First, just to be clear: let’s say the game requires a $1 ante. And we’re using a deck of 5 cards (1-5) where 5 wins.

Strategy for player-A will be:
When in the first position and player-A has:
1 -- then he folds
2, 3, or 4 -- then he checks and then calls (if he is raised).
5 -- then he raises indefinitely.
When in the second potion:
Basically he does the same as above (where for 2, 3, or 4 he checks or calls if raised.)

Strategy for player-B will be:
When in the first position, then with card
1-- Fold
2-- Check. If raised then fold.
3-- Check. If raised then fold.
4-- Raise. If re-raised then fold.
5-- Raise indefinably.
When in the 2nd position, then:
Basically do the same as above except for
4 -- If raised then fold. If player-A checks, then raise.


To look at the expected value we can play out the 40 different scenarios (20 where player-A is in the first position and 20 where player-A is in the second position). Here I’ll list the card combinations:

A in the 1st position vs. B in the 2nd
1 vs. 2 A loses -1 & B wins +1
1 vs. 3 A loses -1 & B wins +1
1 vs. 4 A loses -1 & B wins +1
1 vs. 5 A loses -1 & B wins +1
2 vs. 1 A wins +1 & B loses -1
2 vs. 3 A loses -1 & B wins +1
2 vs. 4 A loses -2 & B wins +2
2 vs. 5 A loses -2 & B wins +2
3 vs. 1 A wins +1 & B loses -1
3 vs. 2 A wins +1 & B lose -1
3 vs. 4 A loses -2 & B wins +2
3 vs. 5 A loses -2 & B wins +2
4 vs. 1 A wins +1 & B loses -1
4 vs. 2 A wins +1 & B loses -1
4 vs. 3 A wins +1 & B loses -1
4 vs. 5 A loses -2 & B wins +2
5 vs. 1 A wins +1 & B loses -1
5 vs. 2 A wins + 1 & B loses -1
5 vs. 3 A wins +1 & B loses -1
5 vs. 4 A wins +1 & B loses -1
(So far player-A wins +10 and loses -15 while player-B wins +15 and loses -10. And note how player-A’s losses against player-B’s 4 card outweigh his 4 card when up against B other cards.)

A in the 2nd position and B in the 1st position:
(I’m not going to bother typing this out but I think the results are that player-A wins +11 and loses - 15. And the only difference here is player-B’s 4 card raises into player-A’s 5 before folding.)

So the total expected value outcome is player-A loses $9.00 while player-B wins $9.00. (Here player-A doesn't breakeven like you suggested.)

And meanwhile I'm sure player-B's strategy is exploitable too... (If someone has an unexploitably strategy here, I'd be interested in hearing about it.)