1. The problem statement, all variables and given/known data A and B put an amount of money in a bank for a period of two years. If both players take the money during the first year, each one gets 75$. If A (B) takes it y B (A) keeps it, A (B) gets 100$ and B (A) 50$. If none of them takes it, they move to the second year. Then, if both players take the money, each one gets 150$, which is the same amount of money that they get if they keep it. If A (B) takes it y B (A) keeps it, A (B) gets 200$ and B (A) 100$. Players act independently from one another. Questions: a) pure strategy equilibria, b) probability of both players taking the money in the first year. 2. Relevant equations Expected value for the second question. 3. The attempt at a solution Since I do not know how to draw the tree diagram in the present format, I can specify the pay-offs differently. First year A takes, so B can take (75,75) or can keep (100, 50). A keeps, so B can take (50, 100). Second year If B keeps then A can take, so B can take (150, 150) or keep (200, 100) then A can keep, so B can take (100, 200) or keep (150, 150). Regarding the questions: a) the game reminds me the prisoner's dilemma. If they collaborate, though inadvertently, and wait till the second year, they can earn 150$ instead of 75$ or at least 100$. But I am not sure whether their collaboration is a logical expectation. Does A take it in his first move because he knows that B will take it if he keeps it (50, 100)?. b) The pay-offs. T, T = 75, 75 T, K = 100, 50 K, T = 50, 100 K, K = 150, 150 --> is this correct? Be p and q the probabilities of taking the money for A and B: EV=75p+100(1-p) EV=50p+150(1-p) p=50/75=0.66 Given that they are symmetrical q=50/75 So p(T, T)=(50/75)*(50/75)= 0,4356, so they would keep the money. If this probability is right, which depends on the (K, K) pay-off, would elucidate a).