On Probabilistic Quantified Satisfability Games

We study the complexity of some new probabilistic variant of the problem Quantified Satisfiability(QSAT). Let a sentence ∃v1∀v2 . . . ∃vn−1∀vnφ be given. In classical game associated with the QSAT problem, the players ∃ and ∀ alternately chose Boolean values of the variables v1, . . . , vn. In our game one (or both) players can instead determine the probability that vi is true. We call such player a probabilistic player as opposite to classical player. The payoff (of ∃) is the probability that the formula φ is true. We study the complexity of the problem if ∃ (probabilistic or classical) has a strategy to achieve the payoff at least c playing against ∀ (probabilistic or classical). We completely answer the question for the case of threshold c = 1, exhibiting that the case when ∀ is probabilistic is easier to decide (Σ2 –complete) than the remaining cases (PSPACE-complete). For thresholds c < 1 we have a number of partial results. We establish PSPACE-hardness of the question whether ∃ can win in the case when only one of the players is probabilistic, and Σ2 -hardness when both players are probabilistic. We also show that the set of thresholds c for which a related problem is PSPACE is dense in [0, 1]. We study the set of reals c ∈ [0, 1] that can be game values of our games. The set turns out to include the set of binary rationals, but also some irrational numbers.