Asymmetric guessing games

Abstract

This paper theoretically and experimentally investigates the behavior of asymmetric players in guessing games. The asymmetry is created by introducing $r>1$ replicas of one of the players. Two-player and restricted N-player cases are examined in detail. Based on the model parameters, the equilibrium is either unique in which all players choose zero or mixed in which the weak player ($r=1$) imitates the strong player ($r>1$). A series of experiments involving two and three-player repeated guessing games with unique equilibrium is conducted. We find that equilibrium behavior is observed less frequently and overall choices are farther from the equilibrium in two-player asymmetric games in contrast to symmetric games, but this is not the case in three-player games. Convergence towards equilibrium exists in all cases but asymmetry slows down the speed of convergence to the equilibrium in two, but not in three-player games. Furthermore, the strong players have a slight earning advantage over the weak players, and asymmetry increases the discrepancy in choices (defined as the squared distance of choices from the winning number) in both games.

Appendices

Appendix A1: Proofs

Proof of Lemma 1

Assume $p<\frac{r+1}{2r}<1$ for any $r>1.$ Suppose $g_{\text {w}}<g_{\text {s}}$. Since $T=(\frac{g_{\text {w}}+r.g_{\text {s}}}{r+1})p$, T can never be greater than or equal to $g_{\text {s}}$. Then, we have two possible cases: 1) $T\le g_{\text {w}}<g_{\text {s}}$ and 2) $g_{\text {w}}\le T<g_{\text {s}}$. If $T\le g_{\text {w}}<g_{\text {s}}$, the weak player wins. If $g_{\text {w}}\le T<g_{\text {s}}$, the weak player wins again since $|g_{\text {w}}-T|-|g_{\text {s}}-T|=2T-g_{\text {w}}-g_{\text {s}}=\underset{<0}{\underbrace{\left( \frac{2p}{r+1}-1\right) }}g_{\text {w}}+\underset{<0}{\underbrace{\left( \frac{2pr}{r+1}-1\right) }}g_{\text {s}}<0.$ Now suppose that $g_{\text {s}}<g_{\text {w}}$. If $T\leqslant g_{\text {s}}<g_{\text {w}}$, the strong player wins. If $g_{\text {s}}\leqslant T<g_{\text {w}}$, the strong player wins again, since $|g_{\text {s}}-T|-|g_{\text {w}}-T|=2T-g_{\text {s}}-g_{\text {w}}=\underset{<0}{\underbrace{\left( \frac{2pr}{r+1}-1\right) }}g_{\text {s}}+\underset{<0}{\underbrace{\left( \frac{2p}{r+1}-1\right) }}g_{\text {w}}<0.$

Thus, this implies that if $p<\frac{r+1}{2r}$, the player who chooses a smaller number wins the game. This further implies that choosing zero is the weakly dominant strategy for each player because choosing zero guarantees a win or at least a draw. Thus, (0,0) is the unique Nash Equilibrium in weakly dominant strategies.$\square$

Proof of Lemma 2

Assume $1>p>\frac{r+1}{2r}$.

(i) Suppose $g_{\text {w}}>g_{\text {s}}$. Since $T=\left( \frac{g_{\text {w}}+rg_{\text {s}}}{r+1}\right) p,$ the weighted average is already closer to $g_{\text {s}}$ than $g_{\text {w}}$. It is easy to see that if the weighted average is multiplied by $p<1$, the resulting number, (T), will be even more closer to $g_{\text {s}}$ than $g_{\text {w}}$. Thus, for all $p<1$ and $r>1$, playing $g_{\text {w}}<g_{\text {s}}$ is necessary for the weak player to win. Now, suppose $g_{\text {w}}<g_{\text {s}}$. Then, we have two possible cases: 1) $T\le g_{\text {w}}<g_{\text {s}}$ and 2) $g_{\text {w}}\le T<g_{\text {s}}$.

If $T\le g_{\text {w}}<g_{\text {s}}$, clearly the weak player wins. This implies that $T\le g_{\text {w}}\Rightarrow \frac{g_{\text {w}}+rg_{\text {s}}}{r+1}p\le g_{\text {w}}\Rightarrow \left( \frac{pr}{r+1-p}\right) g_{\text {s}}<g_{\text {w}}\Rightarrow a_{1}g_{\text {s}}<g_{\text {w}}$ where $0<a_{1}=\frac{pr}{r+1-p}<1.$

If $g_{\text {w}}\le T<g_{\text {s}}$, the required condition for the weak player to win is: $|g_{\text {w}}-T|-|g_{\text {s}}-T|<0\Rightarrow 2T-g_{\text {w}}-g_{\text {s}}<0\Rightarrow \left( \frac{2p}{r+1}-1\right) g_{\text {w}}+(\frac{2pr}{r+1}-1)g_{\text {s}}<0\Rightarrow \left( \frac{2pr-r-1}{r+1-2p}\right) g_{\text {s}}<g_{\text {w}}\Rightarrow a_{2}g_{\text {s}}<g_{\text {w}}$ where $0<a_{2}=\frac{2pr-r-1}{r+1-2p}<1.$

Let $a(p,r)=\min \{a_{1},a_{2}\}$. Since $a_{2}<a_{1}$, $a(p,r)=a_{2}=\frac{2pr-r-1}{r+1-2p}$. For brevity, $a=a(p,r).$ Hence, for the weak player to win she should choose a number such that $ag_{\text {s}}<g_{\text {w}}<g_{\text {s}}$. In all other cases except $g_{\text {w}}=g_{\text {s}}$ that leads to a tie, strong player wins.

(ii) Suppose for a contradiction that $(g_{\text {s}},g_{\text {w}})$ is a PSNE.

(a) If a winner exists at $(g_{\text {s}},g_{\text {w}})$, the other player always has an incentive to deviate because she always has the chance to share the prize by choosing her opponent’s strategy at $(g_{\text {s}},g_{\text {w}})$. Hence, if one of the players is the winner at $(g_{\text {s}},g_{\text {w}})$, this point cannot be PSNE.

(b) If there is a tie at ($g_{\text {s}},g_{\text {w}}$) with $g_{\text {w}}\ne 0$, the strong player has an incentive to deviate because she can win by choosing a $g_{\text {s}}<$ $g_{\text {w}}$. Hence, such a point cannot be a PSNE.

(c) If there is a tie at ($g_{\text {s}},g_{\text {w}}$) with $g_{\text {w}}=0$, the strong player has an incentive to deviate because, given $p>\frac{r+1}{2r}$, choosing any $g_{\text {s}}>0$ guarantees winning. Thus, (0,0) cannot be a PSNE, either.

Hence, the game has no PSNE. Since the game is finite, we have at least one mixed-strategy Nash Equilibrium.

Now we show that the weak player imitates the strong player in equilibrium. Suppose that the strong player randomizes between $g_{{\text {s}}i}$ where $i=1,\ldots ,l_{\text {s}}$ and $l_{\text {s}}\le z+1$. Without loss of generality, we order pure strategies of the strong player such that $0\le g_{\text {s1}}<g_{\text {s2}}<\cdots <g_{sl_{\text {s}}}$. By part i, we know that the weak player cannot win by choosing a pure strategy $g_{\text {w}}>\max \{g_{\text {s1}},g_{\text {s2}},\ldots ,g_{sl_{\text {s}}}\}$.²⁸ In other words, all pure strategies that satisfy $g_{\text {w}}>\max \{g_{\text {s1}},g_{\text {s2}},\ldots ,g_{sl_{\text {s}}}\}$ are weakly dominated. Now suppose that the weak player randomizes between $g_{wi}$ where $i=1,\ldots ,l_{\text {w}}$ and $l_{\text {w}}\le z+1$ and $\max \{g_{\text {w1}},g_{\text {w2}},\ldots ,g_{wl_{\text {w}}}\}>\max \{g_{\text {s1}},g_{\text {s2}},\ldots ,g_{sl_{\text {s}}}\}$. In this case, the weak player can always increase her expected payoff by reducing the numbers over which she randomizes such that $\max \{g_{\text {w1}},g_{\text {w2}},\ldots ,g_{wl_{\text {w}}}\}\le \max \{g_{\text {s1}},g_{\text {s2}},\ldots ,g_{sl_{\text {s}}}\}$. The reason is that for no $g_{\text {w}}>\max \{g_{\text {s1}},g_{\text {s2}},\ldots ,g_{{\text {s}}l}\}$ that is assigned positive probability does the weak player have a chance of winning the game. Alternatively, she has a chance to win or attain a draw by playing smaller than or equal to the highest value of the mixed-strategy of the strong player. Thus, no $g_{\text {w}}>\max \{g_{\text {s1}},g_{\text {s2}},\ldots ,g_{{\text {s}}l}\}$ can be a part of a mixed-strategy for the weak player in equilibrium. In other words, any mixed-strategy that assigns positive probability to a $g_{\text {w}}>\max \{g_{\text {s1}},g_{\text {s2}},\ldots ,g_{sl_{\text {s}}}\}$ is weakly dominated and not part of the Nash equilibrium as the weak player can guarantee a positive payoff otherwise (see Appendix A2 for simple examples). This implies that in any equilibrium, the weak player should randomize between her strategies $g_{wi}$ where $i=1,\ldots ,l_{\text {w}}$ and $l_{\text {w}}<z+1$ such that $max\{g_{\text {w1}},g_{\text {w2}},\ldots ,g_{wl_{\text {w}}}\}\le \max \{g_{\text {s1}},g_{\text {s2}},\ldots ,g_{sl_{\text {s}}}\}$.

Furthermore, given any p and r such that $1>p>\frac{r+1}{2r}$, if a $g_{{\text {s}}i}$ satisfies $a\cdot g_{{\text {s}}i}>g_{{\text {s}}i}-1$, then for all pure strategies less than or equal to $g_{{\text {s}}i}$, the strong player wins unless there is a tie. This implies that the strong player randomizes over $0,1,\ldots ,g_{{\text {s}}i}$ equally (because any other randomization reduces her payoff), and possibly some other pure strategies in equilibrium. As the best response, the weak player must randomize over $0,1,\ldots ,g_{{\text {w}}j}=g_{{\text {s}}i}$ equally, and also some other pure strategies such that $\max \{g_{\text {w1}},g_{\text {w2}},\ldots ,g_{wl_{\text {w}}}\}\le \max \{g_{\text {s1}},g_{\text {s2}},\ldots ,g_{sl_{\text {s}}}\}$ based on the above argument. Note that the weak player does not randomize only over $0,1,\ldots ,g_{{\text {w}}j}=g_{{\text {s}}i}$ in equilibrium, because if she does then the strong player has an incentive to deviate (e.g., the strong player can win by choosing $g_{\text {s}}=z$).

Hence, the weak player never plays greater than the highest value of the mixed-strategy of the strong player and that the equilibrium does not involve the weak player always choosing small numbers. These together prove that the weak player imitates the strong player in equilibrium.$\square$

Proof of Lemma 3

Suppose $r>1$ and $p=\frac{r+1}{2r}$.

(i) Assume $g_{\text {w}}=0$. Then, $T=\frac{g_{\text {s}}}{2}.$ Thus, $|g_{\text {w}}-T|-|g_{\text {s}}-T|=0$ for any $g_{\text {s}}.$

(ii) If $g_{\text {w}}\ne 0$ and $g_{\text {s}}<g_{\text {w}}$, the strong player will be the winner because for all $p<1$ and $r>1$, the necessary condition for the weak player to win is $g_{\text {w}}<g_{\text {s}}$ by Lemma 2. If $g_{\text {w}}\ne 0$ and $g_{\text {w}}<g_{\text {s}}$, the weak player will be the winner, since $|g_{\text {w}}-T|-|g_{\text {s}}-T|=2T-g_{\text {w}}-g_{\text {s}}=(\frac{1}{r}-1)g_{\text {w}}<0$.

(iii) Firstly, we show that in any equilibrium, $g_{\text {w}}=0$. Suppose a pure strategy by the weak player where $g_{\text {w}}\ne 0$. Then, the strong player can always win the game by choosing $0\le g_{\text {s}}<g_{\text {w}}$. Now suppose that the weak player randomizes between $g_{wi}$ where $i=1,\ldots ,l$ and $l<z+1$. Again, the strong player can always win or guarantee a draw by choosing $0\le g_{\text {s}}\le \min \{g_{\text {w1}},g_{\text {w2}},\ldots ,g_{{\text {w}}l}\}$. Thus, there cannot be an equilibrium that includes a pure or mixed-strategy by the weak player with $g_{\text {w}}\ne 0$ and in any equilibrium, $g_{\text {w}}=0$.

Secondly, it is easy to show that $(g_{\text {s}},g_{\text {w}})=(0,0)$ and $(g_{\text {s}},g_{\text {w}})=(1,0)$ are PSNE because neither player has an incentive to deviate, given the other player’s strategy (the strong player is indifferent between choosing zero and one and in both cases, there is a tie). Moreover, by the proof of Lemma 2-ii a and b, there cannot be any other PSNE.

Finally, we show that there are infinitely many mixed-strategy Nash Equilibria including randomization between $g_{\text {s1}}=0$ and $2\le g_{\text {s2}}\le z$ equally and any randomization between $g_{\text {s1}}=0$ and $g_{\text {s2}}=1$. Note that if $g_{\text {w}}=0$, any pure strategy $g_{\text {s}}$ (part i) or any mixed-strategy of the strong player will result in a tie. However, $g_{\text {w}}=0$ is the best response of the weak player only for the mentioned mixed strategies. The second part is trivial: any (infinitely many) randomization between $g_{\text {s1}}=0$ and $g_{\text {s2}}=1$ will provide the same expected payoff since all payoffs are the same in both PSNE ($(g_{\text {s}},g_{\text {w}})=(0,0)$ and $(g_{\text {s}},g_{\text {w}})=(1,0)$). We now show the first part that $g_{\text {w}}=0$ and randomization between $g_{\text {s1}}=0$ and $2\le g_{\text {s2}}\le z$ equally are the best responses to each other. Given the randomization between $g_{\text {s1}}=0$ and $2\le g_{\text {s2}}\le z$ equally, any $0\le g_{\text {w}}<g_{\text {s2}}$ will give the same expected payoff (no $g_{\text {w}}\ge g_{\text {s2}}$ or no randomization by the weak player can provide a higher expected payoff) to the weak player. So, $g_{\text {w}}=0$ is a best response to the equal randomization between $g_{\text {s1}}=0$ and $2\le g_{\text {s2}}\le z$. Given $g_{\text {w}}=0$, randomizing equally between $g_{\text {s1}}=0$ and $2\le g_{\text {s2}}\le z$ will provide the exact same expected payoff to the strong player as any other pure or mixed-strategy. So this set of mixed strategies are best responses to $g_{\text {w}}=0$. Moreover, for any other mixed-strategy of the strong player that does not include zero, the weak player can strictly increase her expected payoff by deviating from $g_{\text {w}}=0$ and choosing less than or equal to the minimum of actions the strong player assigns positive probability. More precisely, suppose that the strong player randomizes between $g_{{\text {s}}i}\ne 0$ where $i=1,\ldots ,l$ and $l<z+1$. However, $g_{\text {w}}=0$ cannot be a best response to this mixed-strategy because the weak player can strictly increase her expected payoff by choosing any $1\le g_{\text {w}}\le \min \{g_{\text {s1}},g_{\text {s2}},\ldots ,g_{{\text {s}}l}\}$ (she even wins for sure if $1\le g_{\text {w}}<\min \{g_{\text {s1}},g_{\text {s2}},\ldots ,g_{{\text {s}}l}\}\ne 1)$.$\square$

Proof of Proposition 4

Suppose $1>p>\frac{r+1}{2r}$. Remember that $a=a(p,r)=\frac{2pr-r-1}{r+1-2p}$ from Lemma 2. Then, for all $g_{\text {s}}\in \{0,1,\ldots ,z\}$, $\underset{r\rightarrow \infty }{\lim }ag_{\text {s}}=\underset{r\rightarrow \infty }{\lim }\frac{2pr-r-1}{r+1-2p}g_{\text {s}}=(2p-1)g_{\text {s}}$. Thus, for each $g_{\text {s}}\in \{0,1,\ldots ,z\},\text { there exists }r_{i}^{*}>1,\text { }i\in \{0,1,\ldots ,z\}$, such that

$$\begin{aligned} \frac{2pr_{i}^{*}-r_{i}^{*}-1}{r_{i}^{*}+1-2p}\text { }g_{\text {s}}=a_{i}(p,r_{i}^{*})g_{\text {s}}=a_{i}^{*}g_{\text {s}}=\lceil (2p-1)g_{\text {s}}-1\rceil \end{aligned}$$

where $\frac{2pr_{i}^{*}-r_{i}^{*}-1}{r_{i}^{*}+1-2p}=a_{i}(p,r_{i}^{*})=a_{i}^{*}$. Since $\frac{{\text {d}}a(p,r)}{{\text {d}}r}>0,\text { for all }r>r_{i}^{*},\text { }ag_{\text {s}}\in (a_{i}^{*}g_{\text {s}},\text { }(2p-1)g_{\text {s}})$, which means that for each $g_{\text {s}}$, there exists $r_{i}^{*}>1$ such that for all $r>r_{i}^{*}$ the range in which the weak player wins does not change. Now, if we set $r^{*}=\max \{r_{0}^{*},r_{1}^{*},\ldots ,r_{z}^{*}\}$, then for all $g_{\text {s}}\in \{0,1,\ldots ,z\},$ there exists $r^{*}>1$ such that for all $r>r^{*}$ the game structure does not change. Since $\frac{{\text {d}}a(p,r)}{{\text {d}}r}>0$, it is clear that for $r<r^{*},$ as r increases the range in which the strong player wins gets larger.$\square$

Proof of Proposition 5

Suppose $r>1$, $p>\frac{r+1}{2r}$ and $g_{\text {w}}\ne g_{\text {s}}$. By Lemma 2, remember that if $p>\frac{r+1}{2r}$, for the weak player to win, she must choose her action ($g_{\text {w}}$) such that $ag_{\text {s}}<g_{\text {w}}<g_{\text {s}}$. Since $g_{\text {w}},g_{\text {s}}\in \{0,1,\ldots ,z\}$, if we have $ag_{\text {s}}>g_{\text {s}}-1$ for all $g_{\text {s}}$, the weak player can never win the game. To prove the result, it is sufficient to find $p^{*}<1$ such that for all $p>p^{*}$, $ag_{\text {s}}>g_{\text {s}}-1$ for all $g_{\text {s}}.$ Moreover, note that finding $p^{*}$ satisfying $az=z-1$ is sufficient:

$$\begin{aligned} ag_{\text {s}}=g_{\text {s}}-1\Rightarrow & {} ag_{\text {s}}-1=g_{\text {s}}-2\Rightarrow ag_{\text {s}}-a>g_{\text {s}}-2\Rightarrow a(g_{\text {s}}-1)>g_{\text {s}}-2 \end{aligned}$$

Then,

$$\begin{aligned} az=z-1\Rightarrow & {} \left( \frac{2p^{*}r-r-1}{r+1-2p^{*}}\right) z=z-1\Rightarrow p^{*}=1-\frac{1}{2}\left( \frac{r-1}{rz+z-1}\right) <1. \end{aligned}$$

Since $\frac{{\text {d}}a}{{\text {d}}p}>0,$ given any $r>1,$ for all $p>p^{*}=1-\frac{1}{2}\left( \frac{r-1}{rz+z-1}\right) ,\text { }ag_{\text {s}}>g_{\text {s}}-1$ is satisfied for all $g_{\text {s}}$.$\square$

Proof of Lemma 6

Assume $p<\min \left\{ \frac{\overset{N}{\underset{i=1}{\sum }}r_{i}}{2r^{*}},1\right\}$ where $r^{*}=max\{r_{1},r_{2},r_{3},\ldots ,r_{N}\}$ for any given set of r values, $\{r_{i}\}|_{i=1}^{N}$, $r_{i}\ge 1$ for all i. There are two cases: Case 1: $p<\frac{\overset{N}{\underset{i=1}{\sum }}r_{i}}{2r^{*}}\le 1$ and case 2: $p<1<\frac{\overset{N}{\underset{i=1}{\sum }}r_{i}}{2r^{*}}$. Without loss of generality, we can order the players such that $1\le r_{1}\le r_{2}\le \cdots \le r_{N-1}\le r_{N},$ implying that $r^{*}=r_{N}=max\{r_{1},r_{2},\ldots ,r_{N-1},r_{N}\}.$

Case 1: Suppose $p<\frac{\overset{N}{\underset{i=1}{\sum }}r_{i}}{2r^{*}}\le 1$. Note that this case implies a sufficiently high asymmetry. Specifically, the strongest player is assumed to be stronger than or equal to the sum of the strengths of all of the other players ($2r^{*}\ge \overset{N}{\underset{i=1}{\sum }}r_{i}$ or $r^{*}\ge \overset{N-1}{\underset{i=1}{\sum }}r_{i}$). To show that the strategy profile $(0,0,\ldots ,0)$ is the unique Nash equilibrium, we need to show that the strongest player cannot win the game by choosing the largest number among the players. In other words, for any given distribution of guesses of the other $N-1$ players, the strongest player does not have an incentive to guess a larger number than the maximum of the rest of the players. Moreover, it is trivial to show that if the strongest player does not have an incentive to guess the largest number, no other player does. If this is the case, then all players have an incentive to guess smaller numbers to win the game. Then, by iterated elimination of weakly dominated strategies, the strategy profile $(0,0,\ldots ,0)$ is obtained as the unique PSNE. Now, we show that the strongest player cannot win the game by choosing the largest number among the players.

Let $g^{*}=max\{g_{1},g_{2},g_{3},\ldots ,g_{N}\}$ be the maximum of the guesses of N players. To prove our claim, we suppose that $g^{*}=g_{N}$ and show that the strongest player N cannot win by guessing $g^{*}$. Let $\hat{g}=max\{g_{1},g_{2},g_{3},\ldots ,g_{N-1}\}$ and $\hat{r}$ be the coefficient of the player who guesses $\hat{g}$. Given that $T=\frac{\overset{N}{\underset{i=1}{\sum }}r_{i}.g_{i}}{\overset{N}{\underset{i=1}{\sum }}r_{i}}.p$ and $p<\frac{\overset{N}{\underset{i=1}{\sum }}r_{i}}{2r^{*}}\le 1$, we can write

$$\begin{aligned} T<\frac{\overset{N}{\underset{i=1}{\sum }}r_{i}.g_{i}}{\overset{N}{\underset{i=1}{\sum }}r_{i}}*\frac{\overset{N}{\underset{i=1}{\sum }}r_{i}}{2r^{*}}\Longrightarrow 2T<\frac{\overset{N}{\underset{i=1}{\sum }}r_{i}.g_{i}}{r^{*}}=\frac{\overset{N-1}{\underset{i=1}{\sum }}r_{i}.g_{i}}{r^{*}}+g^{*}. \end{aligned}$$

If $T\le \hat{g}<g^{*}$, then the strongest player cannot win anyways. If $\hat{g}<T<g^{*}$, then $g^{*}-T>T-\hat{g}\Longrightarrow \hat{g}+g^{*}>2T$ is sufficient to show that the strongest player N cannot win by guessing $g^{*}$. Note that if we can show $\hat{g}+g^{*}\ge \frac{\overset{N-1}{\underset{i=1}{\sum }}r_{i}.g_{i}}{r^{*}}+g^{*}$, this automatically implies $\hat{g}+g^{*}>2T$.

$$\begin{aligned} \hat{g}+g^{*}\ge \frac{\overset{N-1}{\underset{i=1}{\sum }}r_{i}.g_{i}}{r^{*}}+g^{*}\Longrightarrow r^{*}\hat{g}\ge \overset{N-1}{\underset{i=1}{\sum }}r_{i}.g_{i} \end{aligned}$$

The condition we supposed at the beginning, $p<\frac{\overset{N}{\underset{i=1}{\sum }}r_{i}}{2r^{*}}\le 1$, implies $r^{*}\ge \overset{N-1}{\underset{i=1}{\sum }}r_{i}$ or $r^{*}\hat{g}\ge \overset{N-1}{\underset{i=1}{\sum }}r_{i}.\hat{g}$. But since $\hat{g}=max\{g_{1},g_{2},g_{3},\ldots ,g_{N-1}\}$, $r^{*}\hat{g}\ge \overset{N-1}{\underset{i=1}{\sum }}r_{i}.\hat{g}\ge \overset{N-1}{\underset{i=1}{\sum }}r_{i}.g_{i}$ (The second inequality is not strict to include the extreme case where $\hat{g}=g_{1}=g_{2}=g_{3}=\cdots =g_{N-1}$). Thus, the above condition is satisfied.

Case 2: Suppose $p<1<\frac{\overset{N}{\underset{i=1}{\sum }}r_{i}}{2r^{*}}$. This implies $r^{*}<\overset{N-1}{\underset{i=1}{\sum }}r_{i}$. First, suppose the extreme case where $\hat{g}=g_{1}=g_{2}=g_{3}=\cdots =g_{N-1}$. Then, $T=\frac{\hat{g}.\overset{N-1}{\underset{i=1}{\sum }}r_{i}+g^{*}.r^{*}}{\overset{N}{\underset{i=1}{\sum }}r_{i}}.p$. But since $r^{*}<\overset{N-1}{\underset{i=1}{\sum }}r_{i}$, the weighted average $\frac{\hat{g}.\overset{N-1}{\underset{i=1}{\sum }}r_{i}+g^{*}.r^{*}}{\overset{N}{\underset{i=1}{\sum }}r_{i}}$ will be closer to $\hat{g}$ than $g^{*}$even without multiplying it with $p<1$. But since $\hat{g}=max\{g_{1},g_{2},g_{3},\ldots ,g_{N-1}\}$, the weighted average $\frac{\overset{N-1}{\underset{i=1}{\sum }}r_{i}.g_{i}+g^{*}.r^{*}}{\overset{N}{\underset{i=1}{\sum }}r_{i}}$ will be even smaller and will be farther away from $g^{*}$. Thus $g^{*}-T>T-\hat{g}$ is satisfied, which proves that the strongest player N cannot win by guessing $g^{*}$. Thus, $p<\min \{\frac{\overset{N}{\underset{i=1}{\sum }}r_{i}}{2r^{*}},1\}$ is a sufficient condition for all players choosing zero to be the unique PSNE of this game.$\square$

Appendix A2: Some Example Games and Their Solutions

Appendix B1: Instructions

(Translated from Turkish) Thank you for coming. You are participating in an experiment on decision making. You can earn substantial amounts of money in this experiment. Understanding the instructions correctly will help you in making better decisions to increase your earnings. You will also take a quiz about the instructions before proceeding to the actual experiment. Please follow the instructions carefully. From now on you are not allowed to talk with any other participant in the experiment. Whenever you have a question, raise your hand and wait for the experimenter.

The experiment will take approximately one hour. At the end of the experiment, you will be paid in cash whatever amount you have earned and an additional 10 Turkish Lira (TL) as a show-up fee. All the decisions you make during the experiment will be kept anonymous. They will be used for research purposes only and will not be shared with anyone.

Groups and Roles: At the beginning, we will randomly assign you into a 2-person (3-person) group. Each group member will be randomly assigned to either “Role A” or “Role B” (or “Role C”) and these assignments will stay constant over all 10 rounds of the experiment, but none of you will know with whom you are paired. You will play the game for 10 rounds.

Game: In each round, each participant must choose an integer number between 0 and 100 (including these boundaries). The winner of each group will be the person who selects the number that is closest to 1/2 of the mean of the numbers chosen in that group. In other words, there will be a target number and the participant who selects a closer number to this target will win the game in that round. The target number is calculated as follows:

$H=\frac{(g_{A}+g_{B})}{2}*\frac{1}{2}$ for S2 where $g_{i}$ is the number chosen by Player i.

($H=\frac{(g_{A}+9.g_{B})}{10}*\frac{1}{2}$, $H=\frac{(g_{A}+g_{B}+g_{C})}{3}*\frac{1}{2}$, and $H=\frac{(g_{A}+g_{B}+8.g_{C})}{10}*\frac{1}{2}$ for AS2, S3, and AS3, respectively.)

After each round, we will inform you about the number you have chosen, the number chosen by the other member(s) of the group, the target number, and the winning player.

Payment: The winner of each group receives 10 (15) TL for 3 randomly chosen rounds out of 10 rounds. In the case of a tie, this is divided between the winners (i.e., each winner gets 5 TL). The one(s) who loses receives nothing (zero).

In the decision screen, you will see the current round of the game and the remaining time to enter your number in that round. If the time has expired and you have not made your decision, a warning will be displayed in the upper right corner. The allotted time (45 s) is sufficient for you to choose your number. The game will not proceed to the next round unless the players in all groups enter their numbers and press the “ OK” button. Please manage your time accordingly. Please do not enter a number outside of the determined range, from 0 to 100 (including 0 and 100).

In each decision screen, you will see the target number (H) formula and your role, that will not change during the game. At the end of 10 rounds, you will be asked to answer a few short questions and complete a questionnaire. The last screen will show how much you will be paid in total.

To summarize, the experiment will proceed as follows:

• First, you will be asked to answer the multiple choice quiz regarding the rules of the game you will play and how you will be paid. The correct answer will appear on the screen after each question.

• You will see your randomly selected player role on the screen, then the first round of the game will start.

• In each round, after you choose your number and click the “OK” button, you will be given feedback on that round. This feedback will remain on the screen for 15 s and the next round will start.

• After 10 rounds are completed, you will be asked a few questions and asked to complete a questionnaire.

• Finally, the 3 rounds that have been randomly selected from 10, and your earnings in these rounds will be shown on screen. By adding the participation fee to your earnings, your total earnings will be determined. When you see this screen, remain seated. We will call and pay everyone individually. After the payment, the experiment will end.

Any questions?

Appendix B2: Examples of Subject Responses

Reports of some weak players are as follows:

“Although I had a lower chance, I tried to predict the other player’s choices from his/her previous choices,”

“I think I was disadvantaged, the other player was often the one who determined the target,”

“I was disadvantaged so I tried to mimic the other player,”

“I thought that my number was ineffective since it was multiplied by 1,”

“I was disadvantaged because my number was multiplied by 1,”

“Since player C (strong player) was decisive, I focused on her choices,”

“I wish I was player C (strong player) because (s)he had the advantage,”

“I especially focused on player C’s choices,”

“I tried to choose lower numbers to eliminate the advantage of player C,”

“Player C had the advantage by all means,”

“I was not able to affect the target since I was player A (weak player),”

“I chose based on Player C’s choices,”

“The outcome was mostly up to player C,”

“Player C was obviously favored,”

“Since I was disadvantaged I followed player C’s strategies.”

Reports of some strong players are as follows:

“I thought that whatever I chose, the target would be closer to me,”

“I thought that I could win since I was player B (strong player),”

“I had the highest impact on the target,”

“Since I was player B (strong player), I thought if I chose a big number, target would be closer to me”

“I had the advantage so I chose zero to get target closer to the smaller number,”

“Since I was player B (strong player), I thought I could get closer to the target by choosing a number as big as possible,”

“I had the advantage, so I chose small numbers,”

“I thought that the target would be closer to me since I was player B,”

“I thought that I had the advantage, and I could win by choosing big numbers”

“Since my number was multiplied by 8, I had the advantage,”

“I governed the game for the first couple periods since I was player C,”

“I tried to use my advantage but I failed,”

“I was player C and I thought I could govern the game,”

“Since my coefficient was higher, I thought I had the advantage,”

“Player B (strong player) had the advantage by all means,”

“I had the coefficient advantage.”