Implementing equal division with an ultimatum threat

“Practically speaking, why threaten such punishment if the threat is likely to be empty and there are, moreover, ‘softer’ ways of inducing reasonable behavior?” (Brams and Taylor 1996, p. 173)

Abstract

We modify the payment rule of the standard divide the dollar (DD) game by introducing a second stage and thereby resolve the multiplicity problem and implement equal division of the dollar in equilibrium. In the standard DD game, if the sum of players’ demands is less than or equal to a dollar, each player receives what he demanded; if the sum of demands is greater than a dollar, all players receive zero. We modify this second part, which involves a harsh punishment. In the modified game $(D\!D^{\prime })$, if the demands are incompatible, then players have one more chance. In particular, they play an ultimatum game to avoid the excess. In the two-player version of this game, there is a unique subgame perfect Nash equilibrium in which players demand (and receive) an equal share of the dollar. We also provide an $n$-player extension of our mechanism. Finally, the mechanism we propose eliminates not only all pure strategy equilibria involving unequal divisions of the dollar, but also all equilibria where players mix over different demands in the first stage.

Appendix

In this appendix, we provide an extension of our mechanism to $n>2$ players. First, we introduce some new notation. $i=1,...,n$ still denotes players. If $\mathop {\textstyle \sum }\nolimits _{i=1}^{n}d_{i}\le 1$, then the game ends in the first stage and for all $i, \pi _{i}=d_{i}$. If $\mathop {\textstyle \sum }\nolimits _{i=1}^{n}d_{i}>1$, then the game reaches the second stage and the player with the minimal first-stage demand is chosen to be the proposer in the ultimatum game. If $\mathop {\textstyle \sum }\nolimits _{i=1}^{n}d_{i}>1$ and there are multiple players with the common minimal first-stage demand, then each such player is chosen to be the proposer with equal probability. We assume that players are risk-neutral. We denote the excess by $x\equiv \mathop {\textstyle \sum }\nolimits _{i=1}^{n}d_{i}-1$. If player $j$ is the proposer (i.e., either $d_{j}$ is the unique minimal demand or player $j$ is chosen among players with the minimal demand as a result of a random draw), to avoid $x$ he makes a proposal $(x_{1},...,x_{n})$ to other players such that $\mathop {\textstyle \sum }\nolimits _{i=1}^{n}x_{i}=x$. Each other player can either accept or reject this proposal. The proposer’s action set in the second stage is for every $i, x_{i}\in [0,d_{i}]$ and the responders’ action sets in the second stage are $\{accept,reject\}$. ¹² If each responder accepts the proposal, then for all $i, \pi _{i}=d_{i}-x_{i}$. If at least one responder rejects the proposal, then $\pi _{1}=...=\pi _{n}=0$. Therefore, the mechanism requires a unanimous agreement. For clarification, below we provide some examples on how our mechanism works.

Example 1

Let $n=5$ and $(d_{1},d_{2},d_{3},d_{4},d_{5})=(0.2,0.3,0.4,0.8,0.9)$. In this case, $\mathop {\textstyle \sum }\nolimits _{i=1}^{n}d_{i}>1, x =1.6$ and the minimal demand is unique. Player 1 proposes a split of $x$. A possible allocation would be $(x_{1},x_{2},x_{3},x_{4},x_{5})=(0,0,0,0.7,0.9)$. This proposal would be accepted by all responding players and thus, for all $i=1,2,3, \pi _{i}=d_{i}; \pi _{4}=0.1$, and $\pi _{5}=0$.

Example 2

Let $n\!=\!4$ and $(d_{1},d_{2},d_{3},d_{4})\!=\!(0.3,0.3,0.6,0.7)$. In this case, $\mathop {\textstyle \sum }\nolimits _{i=1}^{n}d_{i}>1, x =0.9$ and the minimal demand is not unique. A random draw determines the proposer. Without loss of generality, assume that player 2 is chosen to be the proposer. A possible allocation would be $(x_{1},x_{2},x_{3},x_{4})=(0,0,0.4,0.5)$. This proposal would be accepted by all responding players and thus, for all $i=1,2, \pi _{i}=d_{i}$; $\pi _{3}=0.2$, and $\pi _{4}=0.2$.

Note that the proposals in these examples are just given for clarification purposes and hence may not be optimal in the subgame perfect sense. Characteristics of sequentially rational ultimatum proposals will be provided in the following proof.

Theorem 3

There exists payoff-equivalent multiple subgame perfect Nash equilibria of $DD^{\prime }$ and $(d_{1},...,d_{n})=(1/n,...,1/n)$ in all equilibria.

Proof

We first show that the subgame perfect Nash equilibrium play induces $x=0$ and there are multiple ways to avoid the excess in the second stage if it is reached (Claim 1). Then, we show that in subgame perfect Nash equilibrium, the minimal demand is equal to the maximal demand in the first stage (Claim 2). The result directly follows from these arguments.

Claim 1 Subgame perfect Nash equilibrium play induces $x=0$.

The following exhaustive set of cases needs to be analyzed. In each case, we show that at least one profitable unilateral deviation exists.

Case 1 $x<0$. Without loss of generality take player $j$. Clearly, a unilateral deviation $d_{j}^{\prime }=1-\mathop {\textstyle \sum }\nolimits _{i\ne j}d_{i}$ brings $\pi ^{\prime }=1-\mathop {\textstyle \sum }\nolimits _{i\ne j}d_{i}>d_{j}=\pi _{j}$.

Case 2 $x>0$. There are two subcases here. Before analyzing these subcases let us first take a look at the second-stage behavior. If and when the second stage of the game is reached, without loss of generality assume that player $j$ is the proposer. As in Brams and Taylor (1994), we employ iterated elimination of weakly dominated strategies here.

For any responding player $k$, playing accept brings $\pi _{k}\in [0,d_{k}]$ and hence it weakly dominates (also there exist cases in which strict dominance occurs) playing reject. Thus, responding players choose to accept any proposal. This establishes the second-stage behavior of responding players.

Now, let us denote the set of players with the common maximal first-stage demand as $H$ and the common value of the maximal demand as $h\equiv \max \{d_{1},...,d_{n}\}$. Given that any responding player chooses to accept any proposal,there are two cases to be studied: if $x\geqslant \mathop {\textstyle \sum }\nolimits _{i\in H}d_{i}$, then “offering $x_{i}=d_{i}$ to all players in $H$ and distributing the remaining amount $x-\mathop {\textstyle \sum }\nolimits _{i\epsilon H}d_{i}$ to $n-1-\left| H\right|$ players in such a way that for all remaining players, $\pi _{i}\ge 0$” weakly dominates all other strategies. Note that there are (infinitely) many ways of distributing $x-\mathop {\textstyle \sum }\nolimits _{i\epsilon H}d_{i}$ to $n-1-\left| H\right|$ players, which is why there exists multiple equilibria. On the other hand, if $x< \mathop {\textstyle \sum }\nolimits _{i \epsilon H}d_{i}$ then player $j$ distributes $x$ among the players in $H$. In these two cases, weak domination follows from the fact that playing such, player $j$ can always gurantee $\pi _{j}=d_{j}$. This establishes the proposers’ second-stage behavior.

Case 2.1 Minimal demand is unique. Without loss of generality, assume that $d_{j}=\min \{d_{1},...,d_{n}\}$. Denote the set of the remaining demands as $A$, where $A\equiv \{d_{i}\!\!:d_{i}\ne d_{j}\}$. Clearly, for player $j$, a unilateral deviation $d_{j}^{\prime }=\min (A)-\epsilon$ brings $\pi _{j}^{\prime }=\min (A)-\epsilon >d_{j}=\pi _{j}$.

Case 2.2 Minimal demand is not unique. This means that there exists a set of players with the common minimal demand. Denote this set as $M$ and the common value of the minimal demand as $m\equiv \min \{d_{1},...,d_{n}\}$. There are two subcases.

Case 2.2.1 $\mathop {\textstyle \sum }\nolimits _{i\epsilon M}d_{i}<1$. Then there are two subcases.

Case 2.2.1.1 $\mathop {\textstyle \sum }\nolimits _{i\epsilon H}d_{i}\leqslant x$. Without loss of generality, take player $j\in H$. Then for player $j$, a unilateral deviation, $d_{j}^{\prime }=m-\epsilon$ brings $\pi _{j}^{\prime }=d_{j}^{\prime }=m-\epsilon >\pi _{j}=\epsilon$.

Case 2.2.1.2 $\mathop {\textstyle \sum }\nolimits _{i\epsilon H}d_{i}>x$. Without loss of generality, take player $j\in M$. Let $d_{j}^{\prime \prime }$ be the demand that satisfies $\mathop {\textstyle \sum }\nolimits _{i\epsilon H}d_{i}=x.$ Then, for player $j$ a unilateral deviation, $d_{j}^{\prime }=\min (h-\epsilon , d_{j}^{\prime \prime })$ brings $\pi _{j}^{\prime }=d_{j}^{\prime }>\pi _{j}=d_{j}$.

Case 2.2.2 $\mathop {\textstyle \sum }\nolimits _{i\epsilon M}d_{i}\geqslant 1$. This implies $x\geqslant \mathop {\textstyle \sum }\nolimits _{i\epsilon H}d_{i}$. Without loss of generality, assume that player $j$ is the one with the maximal demand. Thus, a unilateral deviation, $d_{j}^{\prime }=m-\epsilon$ brings $\pi _{j}^{\prime }=d_{j}^{\prime }=m-\epsilon >\pi _{j}=\epsilon .$

Arguments above imply that in the subgame perfect Nash Equilibrium $x=0$.

Claim 2 In the subgame perfect Nash equilbrium, $x=0$ induces $h=m$.

Case 1 $h<m$. This case is impossible by definition.

Case 2 $h>m$. Without loss generality player $j$ is the one with the minimal first-stage demand. Therefore, a unilateral deviation to $d_{j}^{\prime }=h-\epsilon$ brings $\pi _{j}^{\prime }=h-\epsilon >\pi _{j}=d_{j}$. $\square$

The following theorem is a generalization of Theorem 2.

Theorem 4

There is no subgame perfect Nash equilibrium of $DD^{\prime }$ (with $n>2$), which involves mixing in the first stage.

Proof

We will prove this result in steps.

Step 1 Claiming $d_{i}<1/n$ is strictly dominated by claiming $d_{i}=1/n$ , for every player $i=1,2,...n$. Without loss of generality, let us focus on player $1$. It is sufficient to investigate the following two cases $x\le 0$ and $x>0$.

Case 1 $x\le 0$. Let us analyze a deviation from $d_{1}<1/n$ to $d_{1}^{\prime }=1/n$. Note here, we assume that (as in Theorem 3) in case the second stage is reached, the proposer will distribute $x$ as follows: (i) he starts from $h$ and if possible deducts the whole $x$ from $h$, if not possibe deducts as much as possible from $h$; (ii) then he moves to the agent with the second highest demand follows the procedure in (i); and (iii) continues in the same fashion until $x$ vanishes. Under this assumption, a deviation from $d_{1}<1/n$ to $d_{1}^{\prime }=1/n$ is always profitable for player $1$. Note that, even if this deviation creates an excess, $x>0$, it brings $\pi _{1}^{\prime }=d_{1}^{\prime }>\pi _{1}$. To see why, observe that $h>d_{1}^{\prime }$ and $x<h$.

Case 2 $x>0$. In this case, since $x>0$ there must exist a player $j$ with $d_{j}>1/n$. Now, consider a deviation from $d_{1}<1/n$ to $d_{1}^{\prime }=1/n$. Clearly, $\mathop {\textstyle \sum }\nolimits _{i\in b}d_{i}<1$ where $b \equiv \{i:d_{i}\le 1/n\}$. Therefore, under the assumption made in Case 1, $x$ will vanish before reaching player $1$. This implies that $\pi _{1}^{\prime }=d_{1}^{\prime }>\pi _{1}$. Since player $1$ was chosen without loss of generality, the result is valid for any player.

Step 2 In any subgame perfect equilibrium $E(\pi _{i})\ge 1/n$, for every player $i=1,2,...,n$.

From Step 1, we know that for any $i$ setting $d_{i}=1/n$ guarantees $\pi _{i}=1/n$ independent of what other players do. Therefore, in any subgame perfect Nash Equilibrium of $DD^{\prime }$, the expected payoff of any player should be greater or equal to $1/n$.

Step 3 Subgame perfection implies $\mathop {\textstyle \sum }\nolimits _{i=1}^{n}\pi _{i}=1$.

Suppose that, $\mathop {\textstyle \sum }\nolimits _{i=1}^{n}d_{i}>1$ and without loss of generality player $1$ is the proposer in the second stage. Observe that there is always a way to divide $x$ such that player $1$ is not punished: by definition $x=\mathop {\textstyle \sum }\nolimits _{i=1}^{n}d_{i}-1$ and$\ d_{1}\le 1$; therefore, $x\le \mathop {\textstyle \sum }\nolimits _{j\ne 1}d_{j}$. This implies, player $1$ have uncountably many ways to divide the $x$ to other players. Then, in any subgame perfect equilibrium $\mathop {\textstyle \sum }\nolimits _{i=1}^{n} \pi _{i}=1$.

Therefore, combining Steps 1, 2, and 3 implies that in any subgame perfect equilibrium, $E\left[ \pi _{1}\right] =...=E\left[ \pi _{n}\right] =1/n$.

Note that this equal division outcome, in principle, can be reached in different ways. For instance, all players may mix uniformly over $[1/n,1]$ ¹³, or one player may (purely) demand $1/n$ and the others mix between $(1/n,1]$, or yet some of the players (purely) demand $1/n$ and others mix over different demands. Nevertheless, Step 4 shows that this cannot be the case in equilibrium.

Step 4 Suppose without loss generality that players $2,...,N$ mix over $[1/n,1]$ using a probability distribution $F$. First, we show that $F$ cannot have atoms above $1/n$. Suppose for a contradiction that it has an atom above $1/n$. Hence, we can pick an arbitrary point $a$ such that $a>1/n$ and $P(a)>0$. Now, suppose player $1$ mixes using a distribution $G$, which has a support $[1/n,a)$, which is first-order stochastically dominated by $F$ and $G(1/n)\ne 1$. As a result, we have $E(\pi _{1})>E(\pi _{2})=...=E(\pi _{n})$ and hence $E(\pi _{1})>1/n$ (a contradiction to $E(\pi _{1})=E(\pi _{2})=...=E(\pi _{n})=1/n$), where the second inequality follows from subgame perfection. This shows that $F$ cannot have atoms above $1/2$. In other words, we should not worry about ties in demands.

Suppose $k<n$ players mix over demands within the support, $[1/n,1]$. For simplicity, assume that they use the same distribution, $F$, which has a density above $1/n$. $\mathcal F$ denotes the joint distribution of demands for $k$ players. Therefore, for player $i$ playing $d_{i}>1/n$ will bring an expected payoff , $\pi _{i}(d_{i})$,

$$\begin{aligned} \pi _{i}(d_{i})&= (1-F(d_{i}))^{k}d_{i}+\mathop {\displaystyle \sum }_{h=1}^{k}\left( {\begin{array}{c}k\\ h\end{array}}\right) (1-F(d_{i}))^{h}\mathop {\displaystyle \idotsint }_{1/n}^{d_{i}}[\mathbf I _{\{\varphi >x\}}d_{i} \\&+((1-\mathbf I _{\{\varphi >x\}})\max (0,d_{i}-x+\varphi ))]f(d_{1})...f(d_{h})\mathbf d (d_{1})...\mathbf d (d_{h}) \\&+\max \left( 0,\mathop {\displaystyle \idotsint }_{1/n}^{d_{i}}(d_{i}-x)f(d_{1})...f(d_{n})\mathbf d (d_{1})....\mathbf d (d_{n})\right) , \end{aligned}$$

where $\varphi \equiv \sum _{j\in L}d_{j}, L \equiv \{j:d_{j}>d_{i}\}$, is the set of agents with demands higher than $d_{i}$ and $\mathbf I _{\{\varphi >x\}}$ is the indicator function, which takes the value $1$ if the sum of demands higher than $d_{i}$ is less than $x$ and $0$ otherwise.¹⁴

Observe that if we take the derivative of $\pi _{i}(d_{i})$ with respect to $d_{i}$, we get $\pi _{i}^{\prime }(d_{i})=(1-F(d_{i}))^{k-1}-f(d_{i})\Phi +(1-F(d_{i}))\Psi +\mathop {\textstyle \sum }_{i=1}^{k}\gamma _{i}f(d_{i})(1-F(d_{i}))^{k}$ , where $\Phi$,$\Psi$, and $\gamma$ are some positive constants. Obviously, if $F(d_{i})<1$, this derivative is greater than zero. This implies that player $i$ can receive a share higher than $1/n$ just by marginally increasing his demand. This means that there must exist some player $j$ with $\pi _{j}<1/n$, which is in contradiction with findings from Steps 1, 2, and 3. Hence, a positive probability cannot be assigned to any demand above $1/n$ in equilibrium.

Therefore, the result follows from Steps 1, 2, 3, and 4. $\square$