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Suppose we have a set $N$ of $n$ players and a set $M$ of $m$ items. We are given a matrix $P_{n \times m}$ whose element $p_{ij} \geq 0$ $(i \in N, m \in M)$ denotes the valuation of good $j$ by player $i$. Let $v = [v_1, v_2, \cdots, v_n]$ be a vector of valuation functions defined as $v_i(S) = \sum_{j \in S} p_{ij}$ for all $S \subseteq M$ and $i \in N$. We additionally have $v_i(\{\}) = 0$ for all $i \in N$.

Consider a distribution protocol or the order in which the objects will be picked up by a player:

Number players from 1 through n
While there are unclaimed objects:
    Let each player from 1 through n claim one unclaimed item one-by-one

Put simply, players take turns picking items in sequence from the first to the $n^{\text{th}}$ player, repeating the cycle until all items are chosen.

Game Description: Imagine a perfect-information sequential game that follows the distribution protocol described above. That is, the game starts with the first player choosing from a set of $m$ items. After that, the second player picks an item from the remaining $m - 1$ items, then the third from $m - 2$, and so on, until the $n$th player chooses one from what's left. Once all $n$ players have taken a turn, the game loops back to the first player, who now chooses one from $m - n$ items. This picking order continues in rounds until all items are allocated.

Along any path or in any sub-game, the utility of player $i$ is given by the valuation $v_i$ of the allocation he gets.

Example: Consider the following game with $2$ players and $4$ items.

  • The SPNE paths, obtained through backward induction, are marked in red.

  • Payoffs: The utility of player $i$, when employing strategy $s_i$, is given by $v_i(A_{(s_i, s_{-i})})$, where $A_{(s_i, s_{-i})}$ denotes the allocation resulting from player $i$ choosing strategy $s_i$ and the remaining players following strategies $s_{-i}$.

    Look at the first path: $a \to b \to c \to d$. Here, $1$ gets $\{a,c\}$ and $2$ gets $\{b,d\}$ and so the utility vector is $[v1(\{a,c\}, v_2(\{b,d\}))] = [6,7]$, written in ${\color{green}{\text{lemon green}}}$.

enter image description here

Objective: Show that in every SPNE outcome, if the equilibrium results in allocations $(A_1, A_2, \cdots, A_n)$, then it must be that $v_1(A_1^{*}) \geq v_1(A_j^{*})$ for all $j > 1$. In simpler terms, show that in every equilibrium outcome, the first player will weakly prefer his allocation to everyone else's.

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    $\begingroup$ @PeterTaylor We may not have $n \mid m$. I don't think it is trivial. Could it be that you are assuming that in each round, an agent will pick the item he values the most? If that is the case, it is indeed trivial. However, I am considering a completely different game. Say player 1 values a > b > c and 2 values b > c > a. If you pick your most valued item each time, the game proceeds as a -> b -> c which gives 1 {a,c}. In the above game, he will exploit his first-mover advantage and open with b so that 2 picks c and then 1 gets a. Eventually 1 gets {a,b} which is better than {a,c}. $\endgroup$ Commented May 2 at 16:19
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    $\begingroup$ @PeterTaylor As far as "in each round player 1 values their pick above the other picks" is concerned, note that in the game tree drawn in the post, there's a path c -> b -> a -> d where player 1 opens with c which is less preferred to 2's opening choice. (What you said is true when each player goes for the most valued item each time. Here, that may not always be the case as I explained in the comment above.) $\endgroup$ Commented May 2 at 16:25
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    $\begingroup$ With the non-negativity constraint we can assume wlog that $n | m$ because we can pad with items which everyone values at zero. $\endgroup$ Commented May 2 at 20:17
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    $\begingroup$ @PeterTaylor Yes, with the non-negative constraint, we can make that assumption. As for this being greedy, I was thinking from a backward induction perspective which is in some sense a backward-greedy. However, to avoid abuse of the term "greedy" or any unnecessary world-salad, I have removed "greedy" from the post entirely. $\endgroup$ Commented May 2 at 20:26
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    $\begingroup$ @PeterTaylor I have updated the entire question. Hope it is much simpler to read and understand now. $\endgroup$ Commented May 2 at 23:41

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