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Part 1:

You play a game where there is a box with 100$ and there are two players, each of you should write a number 0-100 on paper, then you show your numbers, if the sum is higher than 100 then each of you get 0 dollars, else you get what you wrote. What is your strategy?

Part 2:

You play the same game but your opponent told you that he is putting 80 (he might change his mind) - what is your plan?


For part 1, I think the Nash equilibria are (my number, opponent's number) = (x, 100 - x) for x between 0 and 100? Is this right, and if so, how to choose between them to pick your move? I was thinking about saying 50 as my number for symmetry/aesthetic reasons but would be interested in how one is supposed to tackle this problem. I drew out a reward matrix and tried iterated removal of weakly dominated strategies to get (50, 50) as the only cell remaining - would be grateful for any thoughts on the validity of this.

For part 2, I thought, if you're allowed to say something to the opponent, you could say that you were going to put 50, to try to force your opponent down to 50? Again I have no idea how to approach this. Any help would be much appreciated.

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  • $\begingroup$ Strategy for what, though? Is it implicit that we want to win money? Ruin of the other player could also be a goal. But then we write 100 in any case.. $\endgroup$
    – AlvinL
    Commented Dec 9, 2020 at 16:50
  • $\begingroup$ Sorry this was not clear. I think the goal is to win as much money as possible regardless of what the other player receives. $\endgroup$
    – asfjbkjabf
    Commented Dec 9, 2020 at 16:52
  • $\begingroup$ Just to clarify - do you think writing down 100 is the best thing to do even if your goal is to win as much money as possible? Could you explain your reasoning for this please? I would have thought that it’s not a good play because it only wins any money if the opponent puts 0, which is weakly dominated by all of their other options? $\endgroup$
    – asfjbkjabf
    Commented Dec 9, 2020 at 16:56
  • $\begingroup$ No, I merely remarked if ruin is the goal, then writing 100 is optimal. I don't have a solution for the other goal. $\endgroup$
    – AlvinL
    Commented Dec 9, 2020 at 16:58

2 Answers 2

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The question of choosing a NE among several options is an open question - there is no general mechanism to choose among them (think about Battle of the Sexes, for a very simple example). In many papers, the "punch line" is adding some restriction or addition to the model that ruins other equilibria or somehow helps players choose one. Here, the only symmetric equilibrium is indeed $(50,50)$, but all the $(x,100-x)$ equilibria are also good.

The same goes for the second part. Without ability to credibly commit to $80$, his $80$ is more or less worthless.

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  • $\begingroup$ Thanks for this great answer. Do you think there's any validity in justifying (50,50) because it is the unique solution obtained by iterated removal of weakly dominated strategies (I believe)? Does symmetry of a NE have any significance (like if there exists a symmetric NE, is there an argument to be made that it is the best one to choose?)? $\endgroup$
    – asfjbkjabf
    Commented Dec 9, 2020 at 20:47
  • $\begingroup$ Try a simple $2\$$ game. I don't see any option to remove weakly dominated strategies. The same goes for the $100\$$ game (if you say 50 and he said 0, you should have said more, so the strategy of 50 is not dominating any higher one). What is the purpose of "justifying" any of the equilibria? $\endgroup$
    – YJT
    Commented Dec 9, 2020 at 20:54
  • $\begingroup$ For the $2 case I think we can remove 0 for each player because 0 is weakly dominated. Then once 0 is removed, we can remove 2 for each player, leaving (1,1)? Is that wrong? The only reason I wanted to justify my choice of equilibrium is that when playing the game you have to actually choose one - and if my opponent and I both went through the iterated removal of weakly dominated strategies process then we might both choose 50. $\endgroup$
    – asfjbkjabf
    Commented Dec 9, 2020 at 21:02
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    $\begingroup$ Yes, you are correct. So the $(50,50)$ can be a plausible option if it the only one that remains, but nothing says that other equilibria cannot be played (because it's not strong dominance). $\endgroup$
    – YJT
    Commented Dec 9, 2020 at 21:10
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For the second part, I think I'd say to her "I'll choose the same number as you do". This hints to the other player that she should choose $50$, because both larger and smaller numbers result in lower payments ($0$ if higher, $x<50$ if lower). Of course, this depends on the credibility of your "threat".

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