By solution I mean is there a certain percentage of 0/1 A should pick to minimize his loss? Although this has nothing to do with poker, it's an interesting game... do you have a pointer to study more (or did you make it up on the fly
) ?
Quote:
Here we have two equilibria when all players pick the same number.
I suppose it's an equilibrium... but there's no pure equilibrium _strategy_ that I see. A can't take the strategy of "pick the same number as the others", he can only take the strategy of "pick 0 x% of the time, pick 1 the rest". Same goes for B and C.
(Or do you allow for a strategy of "pick the number the majority picked the last time", i.e. is there memory? How does that work on the first round?)
BTW, since B and C both prefer picking 0, how does the game reach a state where everyone picks 1?
EDIT: hmm.... since B and C prefer 0, does that mean that there is exactly one pure equilibrium strategy for everyone, i.e. pick 0 100%?
In shallow 1 street 1CP (or any other poker I guess), BTN can say "I'll be going all in with X or better" and that'll set the equilibrium strategies for the other players. At least as far as I can see... is that correct?