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In the game "War of attrition NE", three players are fighting for one good worth $v_i$ for each player $v_1>v_2>v_3>0$. Each player decides the time it wants to spend on fighting: $t_1,t_2,t_3$. Until they both fight they loose 1 per time unit. Whoever fights longer than an enemy gets the good they are fighting for (if $t_1=t_2=t_3$ they split equally). Find best response correspondence and find Nash equilibrium.

I know how this works for two players (simply find $u_1(t_1,t_2)$ and $u_2(t_1,t_2)$, combining those two i can easily find best response correspondence ). But how to compute those functions when we have to deal with three players?

Any hints how to construct those functions and/or find Nash equilibrium? Any ideas?

EDIT: question unanswered....

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  • $\begingroup$ I have taken the liberty to modify your title in order for it to proceed "from the general to the particular" $\endgroup$
    – Jean Marie
    Oct 17, 2017 at 9:22
  • $\begingroup$ Please add information about $v_3$ in your question. Is it $v_1 > v_2 > v_3 > 0$? $\endgroup$
    – mlc
    Oct 17, 2017 at 11:06
  • $\begingroup$ Yes i guess we can assume that. $\endgroup$ Oct 17, 2017 at 11:13

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I'm guessing time is discrete so the best reply correspondence is well defined. From what you wrote it seems that the payoff function of player $i$ looks something like \begin{equation*}u_i(t_i,t_{-i})=\begin{cases}v_i-t_i & \text{if } t_i>\hat t_{j}\\ -t_i & \text{if } t_i<\hat t_{j} \end{cases} \end{equation*} (let me not write ties), where I let $\hat t_{j}=\max_{j\neq i} t_{j}$. Now the best reply correspondence would be \begin{equation*} r_i(t_{-i})= \begin{cases} \hat t_j+1 & \text{if } v_i-\hat t_{j}-1>0,\\ 0 & \text{else}. \end{cases} \end{equation*} (I also guess you are playing the strategic form of the game.) Intuitively, you are going to stay at war for just long enough to win the good, given you're valuation. If valuations are integer numbers, then the unique Nash Equilibrium is that a) 1 plays $v_2$, b) 2 and 3 play 0. If valuations are not integer numbers things can get subtle if $v_1-v_2<1$, in which case I think the only equilibrium could be-and could be not- in mixed strategies.

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