Two prospecting partners, Curtin and Dobbs, find an area containing 40 ounces of gold. For each day that they mine the area, they extract 10 ounces of gold. At night, Curtin and Dobbs alternately serve as watchman to protect the gold from passing thieves: Curtin on the first night, Dobbs on the second night, Curtin on the third night, and Dobbs on the fourth night. (If they reach the fifth day, they leave with their gold.) (a) Suppose that whoever is watchman on any given night can choose whether or not to run away with all of the gold they have collected so far. If the watchman on night k runs away then the game ends, he gets a payoff of 10k, and his partner gets a payoff of 0. If neither runs away on any night then they amass a total of 40 that they split equally, giving each a payoff of 20. Find a subgame perfect equilibrium of this game. (Note that the only choices here involve whether to run away with the gold at night; as long as the game is not over, both always work on mining the gold during the day.) Solution: By backward induction, each of them always chooses to run away since if they do not the other will do so the following night. (b) Now suppose that each night, whoever is not serving as watchman chooses whether to monitor the watchman. Monitoring is costly: for each night that a player monitors, his payoff is reduced by 1. The watchman sees whether his partner is monitoring him before

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Chapter8: Game Theory
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. Two prospecting partners, Curtin and Dobbs, find an area containing 40 ounces of gold. For
each day that they mine the area, they extract 10 ounces of gold. At night, Curtin and Dobbs
alternately serve as watchman to protect the gold from passing thieves: Curtin on the first
night, Dobbs on the second night, Curtin on the third night, and Dobbs on the fourth night.
(If they reach the fifth day, they leave with their gold.)
(a) Suppose that whoever is watchman on any given night can choose whether or not to run
away with all of the gold they have collected so far. If the watchman on night k runs
away then the game ends, he gets a payoff of 10k, and his partner gets a payoff of 0.
If neither runs away on any night then they amass a total of 40 that they split equally,
giving each a payoff of 20. Find a subgame perfect equilibrium of this game. (Note that
the only choices here involve whether to run away with the gold at night; as long as the
game is not over, both always work on mining the gold during the day.)
Solution: By backward induction, each of them always chooses to run away since if
they do not the other will do so the following night.
(b) Now suppose that each night, whoever is not serving as watchman chooses whether to
monitor the watchman. Monitoring is costly: for each night that a player monitors, his
payoff is reduced by 1. The watchman sees whether his partner is monitoring him before
choosing if he will run away with the gold. If he runs away while being monitored, then
his partner catches him, they fight, and both receive payoffs of -1. But if, for example,
on the first night Dobbs monitors and Curtin does not run away, and then on the second
night Curtin does not monitor and Dobbs runs away, then Curtin gets a payoff of 0, and
Dobbs gets a payoff of 20-1=19. Find a subgame perfect equilibrium outcome of this
game.
Solution: If neither has run away by the fourth night, Dobbs will run away if and only
if Curtin is not monitoring. Hence Curtin will monitor him. Similar reasoning applies
on the third night, and also on the second night. On the first night, if Dobbs does not
monitor him, Curtin can get a payoff of 10 by running away, but will get a payoff of
20-2=18 if he stays. Hence he stays whether Dobbs monitors him or not, and Dobbs
can sleep soundly without monitoring.
Transcribed Image Text:. Two prospecting partners, Curtin and Dobbs, find an area containing 40 ounces of gold. For each day that they mine the area, they extract 10 ounces of gold. At night, Curtin and Dobbs alternately serve as watchman to protect the gold from passing thieves: Curtin on the first night, Dobbs on the second night, Curtin on the third night, and Dobbs on the fourth night. (If they reach the fifth day, they leave with their gold.) (a) Suppose that whoever is watchman on any given night can choose whether or not to run away with all of the gold they have collected so far. If the watchman on night k runs away then the game ends, he gets a payoff of 10k, and his partner gets a payoff of 0. If neither runs away on any night then they amass a total of 40 that they split equally, giving each a payoff of 20. Find a subgame perfect equilibrium of this game. (Note that the only choices here involve whether to run away with the gold at night; as long as the game is not over, both always work on mining the gold during the day.) Solution: By backward induction, each of them always chooses to run away since if they do not the other will do so the following night. (b) Now suppose that each night, whoever is not serving as watchman chooses whether to monitor the watchman. Monitoring is costly: for each night that a player monitors, his payoff is reduced by 1. The watchman sees whether his partner is monitoring him before choosing if he will run away with the gold. If he runs away while being monitored, then his partner catches him, they fight, and both receive payoffs of -1. But if, for example, on the first night Dobbs monitors and Curtin does not run away, and then on the second night Curtin does not monitor and Dobbs runs away, then Curtin gets a payoff of 0, and Dobbs gets a payoff of 20-1=19. Find a subgame perfect equilibrium outcome of this game. Solution: If neither has run away by the fourth night, Dobbs will run away if and only if Curtin is not monitoring. Hence Curtin will monitor him. Similar reasoning applies on the third night, and also on the second night. On the first night, if Dobbs does not monitor him, Curtin can get a payoff of 10 by running away, but will get a payoff of 20-2=18 if he stays. Hence he stays whether Dobbs monitors him or not, and Dobbs can sleep soundly without monitoring.
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