Game theory: $100 in a box

Question

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?

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?

3. What if there are 10 boxes with $z in each one of them, where p(z=40)=0.5 and p(z=50)=0.5? How much would you pay if you are given the opportunity to examine the number in each box?

4. How much would you pay if you are given the opportunity to examine the number in each box, when p(z=40)=0.5 and p(z=100)=0.5?

5. If there are 1000 boxes, each with \$100 inside. Before the game, your opponent claims that he will report \$80 every time. What would be your optimal strategy

I assume the Nash equilibrium would be (50,50) or (x, 100-x). So a good strategy would be to start higher and eventually reach the 50,50. Though I am not sure if there are better strategies or how to prove that choosing 50 is the most optimal. For the second part, does it mean that I should write 20? or should I stick to the original $50? The probability of whether he will commit to 80 isn't given, so I don't know if this information should influence the choice at all.

Answer 1

All pairs of the form $(x,100-x)$ are Nash equilibria because there is clearly no way for either player to improve by unilaterally changing their strategy. There isn't really a way to objectively say that one equilibrium is "more optimal" than another without defining what exactly "more optimal" means. Which strategy to choose isn't too clear without making some guess about the distribution of strategies the other player will choose. Any number you choose is part of a Nash equilibrium, and it seems reasonable to conjecture that the $(50,50)$ equilibrium would be commonly chosen by the other player.

In the second, your opponent is telling you he will play in the $(20,80)$ equilibrium. Since there is no way to improve your payoff by deviating from $20$ (unless your opponent is lying), it is optimal to respond with 20. The opponent also has no incentive to lie because it is still a Nash equilibrium.