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Consider a sequential game with two players who use the following procedure to share two desirable, identical and indivisible objects. Player 1 proposes an allocation in stage 1 which player 2 either accepts or rejects. In the event of rejection, neither player receives either of the objects and in the event of acceptance the objects are allocated according to the proposal of player 1.

The subgame perfect Nash equilibrium of the game is where player 1 allocates both the objects to himself and player 2 receives nothing. Is there any Nash equilibrium of this game other this?

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  • $\begingroup$ Hint: "player 2 receives nothing" only refers to the outcome, but what exactly is her strategy? This has implications on what player 1 would optimally propose in the first stage. $\endgroup$ – Herr K. May 6 '17 at 6:48
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    $\begingroup$ Ultimatum game $\endgroup$ – Henry May 6 '17 at 7:15
  • $\begingroup$ @HerrK. Player 2's strategy is whatever player 1 allocates, she will accept that. But still I couldn't figure out if there is any other NE $\endgroup$ – Abishanka Saha May 6 '17 at 7:47
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    $\begingroup$ @AbishankaSaha Accepting regardless of player 1's proposal is not the only strategy available to player 2. Consider scenarios in which player 2 rejects 1's proposal $\endgroup$ – Herr K. May 6 '17 at 18:00
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What is the difference for player 2 between accepting nothing or rejecting? None. Therefore, we may say that from the point of view of player 2, he is indifferent between accepting nothing or rejecting.

What is done, as in Rubinstein's (1982) Bargaining Game, is to assume that in indifference, player 2 accepts receiving nothing, which also has more implications when the game is extended to an extra round where players change positions in case of player 2 rejecting the proposal on the first round, or when is included an outside option.

EDIT:

Following the comment from denesp, I will complete my answer.

The previous reasoning is applied mainly to SPE. If we want to see if there is any other equilibrium, we can start by representing the game in normal form, such that, we can identify two equilibria.

  • Player A offers 2, and Player B accepts

  • Player A offers 0, and Player B accepts.

However, the last one is not SPE because precisely in Rubinstein's model (and in other bargaining models), it is assumed that if player B is faced with two options that have the same payoff (when involves accepting, and the other rejecting), then he accepts.

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  • $\begingroup$ It is worth noting that the OP is not asking for subgame perfect Nash equilibria but for any equilibria. $\endgroup$ – Giskard Jun 18 '17 at 5:27
  • $\begingroup$ Yes, you make a valid point. In that case, I will add some information to my previous answer $\endgroup$ – user13650 Jun 18 '17 at 12:26

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