Clarification of this solution: Nash equilbirum of splitting 10 dollars between 2 people

Question

So the game is each player chooses an amount between 0 and 10, if the sum is greater than 10 they win nothing. If the sum is less than or equal to 10 each player gets what they selected.

The solution states that "the Nash equilibrium of the game is any two numbers that add to 10. This is obvious because any amount greater than 10 destroys the value of the game and any amount less than 10 induces regret".

However the solution goes on to further state that "the unique Nash equilibrium is for each player to choose $5".

Where I am getting confused is that the two statements contradict each other. The first states any pair that adds to 10 is a Nash equilibrium, but then the second statement seems to suggest only 5,5 is a Nash equilibrium.

Answer 1

Here is an explanation of why $(5,5)$ is not the only nash equilibria in this problem. The important thing to remember about game theory is that the two players have no idea of what the other shall choose.

Let us consider from each player's perspective.

Player 1's perspective

From player 1's perspective player 2 has picked some choice $c_2$ such that $ 0 \leq c_2 \leq 10$. So what is player 1's most profitable decision? He/she should pick $10-c_2$. Any higher and he gets 0, any lower and he isn't maximizing profit.

Player 2's perspective

This game is symmetric so the same hold from player 2's perspective. Player 1 has picked some choice $c_1$ such that $ 0 \leq c_1 \leq 10$ and so player 2's most profitable decision is to pick $10-c_1$. Any higher and he gets 0, any lower and he isn't maximizing profit.

So what is the optimal strategy? Well if we define the optimal strategy for player 1 as a best response function of player 2's choice ($c_2$), then $f_1(c_2) = 10-c_2$ and similarly for player 2: $f_2(c_1) = 10-c_1$, then the optimal strategy is the intersection of these two functions. These are the pure strategy nash equilibria.

This is demonstrated graphically below, where the filled circles are player 1's optimal decisions at each choice x of player 2 and the gray empty circles are player 2's optimal decisions for each choice y of player 1. Clearly the best response functions intersect at every point.

Edit: I managed to find an analogous problem to this one online. You should check it out too. It's called splitting a dollar (p.23).

Answer 2

I would interpret the first sentence as:

The Nash equilibrium must be given by a couple of numbers that sum to $10$.

Then the second sentence doesn't contradict the first one.

Answer 3

This doesn't make much sense. The $(5,5)$ Nash equilibrium might seem like the most "fair" equilibrium of all the possible Nash equilibria (and this can be formalized using notions that compare different Nash equilibria) but it is definitely not the unique Nash equilibrium.

Answer 4

I am Economist and do Game Theory. So I can assure you that while {5,5} is an equilibrium of the game, it is not unique.

Any pair {x,y}, such that $x\geq0$, $y\geq0$ and $x+y=10$ is an equilibrium.

To verify that {x,y} is an equilibrium we have to verify that no player has incentives to deviate given the actions of the other (in other words, he cannot do better by choosing different action then prescribed in equilibrium given the actions prescribed for the other player by our candidate equilibrium).

Here, if player 1 chooses $x \geq 0$, player 2 chooses $y\geq 0$ and $x+y=10$ then no player has incentives to deviate. To see this consider Player 1 (the case for player 2 is analogous). Player 1 can deviate by choosing z>x or zx then, given that the other player plays $y$, the sum will be z+y>10 and hence he will get nothing. This is worse then choosing $x$ since in that case he gets $x$. Clearly, choosing $z<x$ is also not an equilibrium. In this case $z+y<10$, and Player 1 will get $z$ which is less than $x$. So again, he has no incentives to deviate. Thus, we conclude that Player 1 has no incentives to deviate. Since the case of Player 2 is analogous, we conclude that {x,y}, such that x+y=10 is an equilibrium.

A little bit more counter-intuitive might be the case $x=0$ and $y=10$. But in this case Player that chooses $x$ is indifferent to playing the proposed strategy and deviating. In the Nash Equilibrium definition we require that player has no STRICT incentives to deviate. Hence this is still equilibrium.