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It is not clear, from the paper you linked, which version of PBE the author is using. PBE is somewhat of an elusive concept, see Fundenberg and Tirole (1991) for a discussion. A commonly used version is what Mas-Colell, Whinston, and Green (1995, aka MWG) call weak PBE, which is a strategy profile $\sigma$ and a system of beliefs $\mu$ such that : $\... 3 At the risk of misinterpretation, I'll reformulate your question as follows: Is there an equilibrium refinement criterion for any normal form game such that this criterion selects an equilibrium that would constitute a subgame perfect equilibrium in all possible extensive form games that can be reduced to the normal form game in question? Short answer: ... 3 Note that if$n$is odd then$d$is odd and$n+d$is always even. On the other hand, if$n$is even then$1$is a divisor of$n$and$n+1$is odd. Therefore it should follow that$n$is a winning position if and only if$n$is even. Starting from$1$, the second player has always a winning strategy: replace the current (even) number$n$with$n+1$. 3 Unfortunately, it turns out simply. Losing positions are where$(m,n,k)$are either all even or all odd. It is more complicated if the last stone loses, I don't have a formula for that. 3 First of all, it's clearly a Nash equilibrium for everyone to choose$100$. Nobody wins, and nobody can make themselves win. Let's call that the trivial equilibrium. If$n=2$, then there are also nontrivial equilibria where player 1 always chooses$100$and player 2 does... anything at all. Player 2 never wins, so it doesn't matter what they choose; and ... 2 Another standard (and very good) introductory book on Game Theory is Gibbons "Game Theory for Applied Economists," which is circulated in Europe under the title "A Primer in Game Theory." Gibbons on Amazon A more gentle introduction to game theory is "Games of Strategy" by Avniash Dixit and Susan Skeath. It puts more emphasis on intuition and less on math. ... 2 An Introduction to game theory by Martin J. Osborne is a very good introductory textbook. Here's the link: https://www.economics.utoronto.ca/osborne/igt/ Also, these lecture notes by MIT are very good but I'll suggest you go through with the book first, it has theory from basics and pretty good questions too. Lectures Link: https://ocw.mit.edu/courses/... 2 This is more of a discussion point than an answer. Let us assume$c_e < \sum_{i=1}^Nc_i$, otherwise this is a game nobody would play. It would also be illogical to terminate before the player has made the money$c_e$back during the game (otherwise the player would just not play). Let$R_j=\sum_{i=1}^jc_i$, and let$1 \leq r \leq N$be the minimum value ... 2 Thanks to Kuhn's theorem, there is no need to distinguish what you call Nature and Chance moves on a game tree. If randomization happens at multiple information sets throughout a game tree, it is without loss of generality to assume that it occurs only once at the beginning of the tree. We can think of randomization that occurs at multiple information sets ... 2 In a finite game with perfect information and no random elements, there are only three possibilities. The game, if perfectly played, is a win for the first player, a draw, or a win for the second player. This can be seen by considering the game tree. All the leaves can be labeled as win, loss or draw (from the perspective of the first player, say) then ... 1 This game is commonly known in combinatorial game theory as Kayles. And, yeah, it is well-explored. This seems to be a short paper on the process, although I haven't read it myself. 1 Stackelberg games are only a very small subset of sequential games with perfect information, where players (not necessarily just two) move one after another. The type of games you asked for -- one player acts first, followed by both players acting simultaneously -- falls under the set of sequential games with imperfect information. In the context of ... 1 In the order$(2,3,1)$we process$v_2$first. So Setting$i=2$, we first observe that$v_2$has neighbors$v_1$and$v_3$. As both$a_1$and$a_3$are at their initialized state ($=-1$), we set $$a_2=\operatorname{mex} \{-1,-1\}=0.$$ We then set$i=3$. The vertex$v_3$has a$v_2$as its sole neighbor. As$a_2=0$the algorithm tells us to set$$a_3=\... 1 Without more specification about how choices are made, there's not enough information to determine an optimal strategy. Here's one possible interpretation . . . Let$n,x$be fixed and known, where$n$is an integer with$n\ge 2$, and$x$is a real number with$0 < x < 1$. Suppose the players agree in advance on a probability$p\in [0,1]\$ (and are ...