A Game of Nontransitive Dice

TL;DR

This paper analyzes a two-player game involving selecting n-sided dice, revealing a unique pure strategy Nash Equilibrium for n>3 where players choose standard dice, and provides an algorithm to improve nonstandard dice.

Contribution

It proves the uniqueness of the Nash Equilibrium in pure strategies for n>3 and introduces an algorithm to generate dice that beat any given nonstandard die.

Findings

Unique pure strategy Nash Equilibrium for n>3 is the standard n-sided die.

An algorithm can generate a die that beats any nonstandard die.

The equilibrium strategy involves both players choosing the standard n-sided die.

Abstract

We consider a two player simultaneous-move game where the two players each select any permissible $n$-sided die for a fixed integer $n$. A player wins if the outcome of his roll is greater than that of his opponent. Remarkably, for $n>3$, there is a unique Nash Equilibrium in pure strategies. The unique Nash Equilibrium is for each player to throw the Standard $n$-sided die, where each side has a different number. Our proof of uniqueness is constructive. We introduce an algorithm with which, for any nonstandard die, we may generate another die that beats it.

Equations

$${\mathbb{E}}_p(D_n)=\sum _{i=1}^n{d}_i{p}_{d_i}=\frac{n+1}{2}$$

$$\begin{split}D_{n}&=\big{\{}[d_{1},d_{2},\dots,d_{n}]\big{\}}\end{split}$$

$$U_A(A_n,B_n)=\mathrm{Pr}(A_n>B_n)+\frac{1}{2}\mathrm{Pr}(A_n=B_n)$$

$$U_{A}(D_{n},S_{n})=\sum_{i=1}^{n}\bigg{(}\frac{1}{n}\bigg{)}\bigg{(}\frac{d_{i}-1}{n}+\frac{1}{n}\cdot\frac{1}{2}\bigg{)}=\bigg{(}\frac{1}{n}\bigg{)}\bigg{(}\mathbb{E}[D_{n}\big{]}-\frac{1}{2}\bigg{)}=\frac{1}{2}$$

$$\Phi_{i,j}(D_{n})=\big{\{}[d_{1},\dots,d_{i-1},d_{i}-1,d_{i+1},\dots,d_{j-1},d_{j}+1,d_{j+1},\dots,d_{n}]\big{\}}$$

$$\begin{split}\gamma_{k}&=|\big{\{}b_{i}|b_{i}=k\big{\}}|\end{split}$$

$$\begin{array}{rl}\sum _{k=1}^n{\gamma }_k& =n\end{array}$$

$$\begin{array}{rl}\sum _{k=1}^{n}k{\gamma }_k& =\frac{n(n+1)}{2}\end{array}$$

$${\xi }_k={\gamma }_k+{\gamma }_{k+1}$$

$${\gamma }_1+{\gamma }_2={\gamma }_2+{\gamma }_3={\gamma }_3+{\gamma }_4=\cdots ={\gamma }_{n-1}+{\gamma }_n$$

$$\begin{split}\gamma_{1}&=\gamma_{3}=\cdots=\gamma_{k}\hskip 28.45274pt\forall\hskip 5.69046pt\text{odd integers}\hskip 2.84544ptk\in\big{\{}1,2,\dots,n\big{\}}\\ \gamma_{2}&=\gamma_{4}=\cdots=\gamma_{j}\hskip 28.45274pt\forall\hskip 5.69046pt\text{even integers}\hskip 2.84544ptj\in\big{\{}1,2,\dots,n\big{\}}\\ \end{split}$$

$$\begin{array}{rl}\sum _{k\text{odd}}^n{\gamma }_k+\sum _{j\text{even}}^n{\gamma }_j& =n\end{array}$$

$$\begin{array}{rl}\sum _{k\text{odd}}^{n}k{\gamma }_k+\sum _{j\text{even}}^{n}j{\gamma }_j& =\frac{n(n+1)}{2}\end{array}$$

$$\begin{array}{rl}\frac{n+1}{2}{\gamma }_1+\frac{n-1}{2}{\gamma }_2& =n\end{array}$$

$$\frac{n}{2}{\gamma }_1+\frac{n}{2}{\gamma }_2=n$$

$$\begin{array}{rl}{\gamma }_1+{\gamma }_2& =2\end{array}$$

$$\frac{n^2}{4}{\gamma }_1+\frac{n(n+2)}{4}{\gamma }_2=\frac{n(n+1)}{2}$$

$$\begin{array}{rl}\frac{n}{2}{\gamma }_1+\frac{n+2}{2}{\gamma }_2& =n+1\end{array}$$

Full text

Keywords: Efron’s Dice, Nontransitive Dice, Probability Paradoxes, Game Theory

1 Introduction

If Hercules and Lychas play at dice

Which is the better man, the greater throw

May turn by fortune from the weaker hand.

William Shakespeare,

The Merchant of Venice

Nontransitive dice are a fascinating topic in applied probability. They first came into the limelight as a result of a paper by Martin Gardner [8] and are one of a larger class of nontransitivity “paradoxes” (see [16], [2]), which also include the well-known Condorcet Voting Paradox, as described in [7].

Recent papers published concerning nontransitive dice include [1], [4], and [15]. Another recent interpretation of the situation is [10], which instead reinterprets the scenario through throws of unfair coins. Additionally, the issue of nontransitive dice is the subject of a recent polymath project paper (see [12]), and indeed we borrow some terminology from that paper.

One problem that may be considered is how nontransitive dice should be played in a strategic interaction of two or more players. [14] was the first to explore this: it explores the two player game where each player could choose one of four specific111[14] formulates the problem using the four six-sided “Efron Dice”. non-transitive symmetric dice and finds the set of equilibria. It then extends the analysis to cover the situation in which the players each choose two dice.

Here, we investigate a broader problem: we consider a two-player, one-shot, simultaneous move game in which each player selects a general $n$-sided die and rolls it. The player with the highest face showing wins a reward, which we may normalize to $1$. The solution concept that we use for the game is that of a (pure strategy) Nash Equilibrium.222Note that this game is a constant-sum game; therefore equivalent to a zero sum game, and for two players the concept of a Nash Equilibrium is equivalent to that of a saddle point. We show that for $n>3$ there is a single, unique333There may be additional mixed strategy equilibria; however, in this analysis we focus exclusively on pure strategy equilibria. Henceforth, by Nash Equilibrium or equilibrium, we mean only those in pure strategies. Nash Equilibrium in which both players play the standard die. Moreover, our proof of uniqueness is constructive and contains an algorithm that, for any non-standard die, generates a die that beats it. One notable implication of this is that for any non-standard die, there is at least one die that is the result of a one-step, the most elementary step of our algorithm, applied to the standard die, that beats it. That is, for any die that is not the standard die, there is a die very similar to the standard die that beats it.

The closest paper to this one, [6], considers the same problem, where for some fixed integer $n$, two players each choose a die and roll against each other. It also shows that the standard die ties any other die in expectation, and that every nonstandard die loses to some other die. [5] explores dice games as well in a slightly more general setting, and the existence and uniqueness of the the equilibrium in which both players each roll the standard die follows from Propositions 6 and 8 in that paper.

Our paper differs from [6] and [5] in the following key ways. We provide different proofs of the existence and uniqueness of the Nash Equilibrium in the game444[6] do not explicitly show this, but it clearly follows from their Lemma 1 and Theorem 2., and we are able to do so using exclusively elementary mathematics. Additionally, our proof is constructive and we formulate a simple algorithm that allows us, for any non-standard die, to generate a die that beats it. Moreover, our last result–that for any non standard die, there is a die that beats it that is merely a one-step away from the standard die–is also novel.

Finally, dice games can be placed in a more general context, as a member of the family of Colonel Blotto games. First developed by E. Borel in 1921 (see [3]), a burgeoning literature has resulted, due to the game’s general applications in economics, operations research, political science, and other areas. Some recent papers include [13] and [9]. In addition, recently, [11] explores a $n$ player continuous version of this game played on the interval $[0,1]$. For two players, the unique equilibrium is the continuous analog of the unique equilibrium here, the uniform distribution.

2 The Basic Game

Define a general $n$-sided die (henceforth just “die”) as an integer-valued random variable $D_{n}$ that takes values in the finite set \big{\{}1,2,...,n\big{\}}, where the distribution must satisfy the following conditions:

For each possible value of $D_{n}$, $d_{i}$, the probability that it occurs, $p_{d_{i}}$, is a multiple of $1/n$. 2. 2.

[TABLE]

For a given $n$, denote the set of all $n$-sided dice by $\mathcal{D}_{n}$.

Then, a standard $n$ -sided die, $S_{n}$, is simply a die where each value occurs with equal probability, $1/n$. Naturally $S_{n}\in\mathcal{D}_{n}$.

We can represent any $n$-sided die, $D_{n}\in\mathcal{D}_{n}$, as a (discrete) uniformly distributed random-variable that takes values in the multiset of size $n$, with elements in \big{\{}1,2,...,n\big{\}} and sum equal to $\frac{n(n+1)}{2}$:

[TABLE]

Example 1**.**

The five $4$-sided dice are S_{4}=\big{\{}[1,2,3,4]\big{\}}, X_{4}=\big{\{}[1,1,4,4]\big{\}}, Y_{4}=\big{\{}[2,2,2,4]\big{\}}, \big{\{}[1,3,3,3\big{\}}, \big{\{}[2,2,3,3]\big{\}}.

2.1 The Game

Consider two players, Amy ($A$) and Bob ($B$). They play the following one shot game. Fix $n$, Amy and Bob each independently select any $n$-sided die, $A_{n},B_{n}\in\mathcal{D}_{n}$ and then roll them against each other. Their expected payoffs are the probability that the realization of their roll is higher than the realization of their opponents roll555If the realizations of the rolls are the same, the winner is decided by a (fair) coin flip..

A Strategy for Amy, (and analogously for Bob), is simply a choice of die $A_{n}\in\mathcal{D}_{n}$. For any pair of strategies, $(A_{n},B_{n})$, Amy’s expected payoff, $U_{A}(A_{n},B_{n})$, is

[TABLE]

Example 2**.**

Suppose $n=4$ and let Amy and Bob choose dice $X_{4}$ and $Y_{4}$ from Example 1, respectively. Then $U_{A}(X_{4},Y_{4})=7/16$ and $U_{B}(X_{4},Y_{4})=9/16$.

The main result of this paper is the following theorem.

Theorem 1**.**

For any $n$, the unique Nash Equilibrium of the two player game is where both players play the standard die $S_{n}$. That is, the unique Nash Equilibrium is the strategy pair $(S_{n},S_{n})$.

We shall prove this theorem by proving two propositions: Proposition 1, that $(S_{n},S_{n})$ is a Nash Equilibrium; and Proposition 2, that $(S_{n},S_{n})$ is the unique equilibrium (for $n\geq 4$).

Proposition 1**.**

$(S_{n},S_{n})$* is a Nash Equilibrium.*

Proof.

First, we show that for all i\in\big{\{}A,B\big{\}} and for all $D_{n}\in\mathcal{D}_{n}$, $U_{i}(S_{n},D_{n})=U_{i}(D_{n},S_{n})=\frac{1}{2}$.

Suppose player $B$ chooses the standard die, $S_{n}$ and player $A$ chooses an arbitrary die D_{n}=\big{\{}[d_{1},d_{2},\dots,d_{n}]\big{\}}. If the realization of $D_{n}$ is $d_{i}$; that is, dice $D_{n}$ “lands” showing face $d_{i}$, then with probability $(d_{i}-1)/n$, $D_{n}$ beats the standard die, and with probability $1/n$, $D_{n}$ ties the standard die. Hence,

[TABLE]

Since $(S_{n},S_{n})$ yields to player $B$ a payoff of $1/2$, there is no profitable deviation. ∎

For any die $D_{n}\in\mathcal{D}_{n}$ and any $i$ and $j$ such that $i\neq j$, $d_{i}\neq 1$ and $d_{j}\neq n$, a One-step from $D_{n}$ is $\Phi_{i,j}(D_{n})$ defined by

[TABLE]

Evidently, $\Phi_{i,j}(D_{n})$ satisfies the conditions in Equation 1 and so $\Phi_{i,j}(D_{n})$ is a die. We may make the following remark:

Remark 1**.**

Any die can be reached in a sequence of one-steps from any other die. This of course implies that any die can be reached in a sequence of one-steps from the standard die, $S_{n}$.

For any two dice $A_{n},B_{n}\in\mathcal{D}_{n}$, we say $A_{n}$ Beats $B_{n}$ if the number of pairs $(a_{i},b_{j})$ with $a_{i}>b_{j}$ exceeds the number of pairs with $a_{i}<b_{j}$.

It remains to show uniqueness, which we accomplish in the following lemma.

Proposition 2**.**

The Nash Equilibrium, $(S_{n},S_{n})$, is unique for $n\geq 4$.

Proof.

Clearly, for any strategy pair, player $i$ has a profitable deviation if and only if there is a die that beats her opponents die. Our proof is constructive and we show that for any die $B_{n}\neq S_{n}$, we can construct a die, ${G}_{n}$, that beats it.

To that end, let B_{n}=\big{\{}[b_{1},b_{2},\dots,b_{n}]\big{\}}, and (recall) S_{n}=\big{\{}[1,2,\dots,n]\big{\}}. We introduce $\gamma_{k}$, defined as

[TABLE]

for $k=1,2,\dots,n$. Note that each $\gamma$ is a non-negative integer and that each satisfies

[TABLE]

[TABLE]

Next, for $k=1,2,\dots,n-1$ define $\xi_{k}$ as

[TABLE]

To construct a die $G_{n}$ that beats $B_{n}$, we need simply find a pair $(\xi_{i},\xi_{j})$ with $\xi_{i}>\xi_{j}$ (and so clearly $j\neq i$) and $i\neq j+1$. Then, simply add $1$ to $s_{i}$ and take $1$ away from $s_{j+1}$ (hence the need for $i\neq j+1$), and the resulting die will beat $B_{n}$.

Example 3**.**

Suppose player $A$ chooses die $X_{4}$ from our previous examples, X_{4}=\big{\{}[1,1,4,4]\big{\}}. We have $\gamma_{1}=2$, $\gamma_{2}=\gamma_{3}=0$, and $\gamma_{4}=2$, and so $\xi_{1}=2$, $\xi_{2}=0$, $\xi_{3}=2$, and $\xi_{4}=2$. Evidently $\xi_{1}>\xi_{2}$ and $1=i\neq j+1=2+1=3$. Hence, we add $1$ to $s_{1}$ and subtract $1$ from $s_{3}$, yielding the die Y_{4}=\big{\{}[2,2,2,4]\big{\}}, which beats $X_{4}$. Indeed, should player $B$ choose $Y_{4}$ she would achieve a payoff of $9/16>1/2$.

Evidently, if $\xi_{a}\neq\xi_{b}$ for some $a$, $b$, then there must be some $i$, $j$ with $\xi_{i}>\xi_{j}$. Thus, we establish the following lemma:

Lemma 1**.**

If $n\geq 4$, then for any non-standard $n-$sided die $\exists$ a pair a,b\in\big{\{}1,2,...,n\big{\}}, for which $\xi_{a}\neq\xi_{b}$.

Proof.

Evidently \xi_{a}=\xi_{b}\hskip 14.22636pt\forall a,b\in\big{\{}1,2,\dots,n\big{\}} if and only if

[TABLE]

which holds if and only if

[TABLE]

We also have the following two relationships:

[TABLE]

and

[TABLE]

We can combine equations 2 and 3 to obtain

[TABLE]

for odd $n$. For even $n$, equations 2 and 3 yield

[TABLE]

or

[TABLE]

Furthermore, from equations 2 and 4, we have

[TABLE]

or

[TABLE]

for even $n$. Now observe, we cannot have both $\gamma_{1}\geq 1$ and $\gamma_{2}\geq 1$ since if one were equal to $1$ and the other were greater than $1$, this would violate 3; and if they were both equal to $1$, then $B_{n}=S_{n}$, a contradiction. Thus, either $\gamma_{1}$ or $\gamma_{2}$ must be equal to [math].

Suppose $n$ is odd and that $\gamma_{1}=0$. From equation 5 we have $(n-1)\gamma_{2}=2n$, which does not have a solution in integers $n$, $\gamma_{2}$ for $n>3$. Next, suppose $n$ is odd and that $\gamma_{2}=0$. From equation 5 we have $(n+1)\gamma_{1}=2n$, which does not have a solution in integers $n$, $\gamma_{1}$ for $n>1$. Thus, we conclude that $n$ cannot be odd.

Suppose $n$ is even and that $\gamma_{1}=0$. From equations 6 and 7 we must have $\gamma_{2}=2$ and $n+2=n+1$, which is obviously a contradiction. Finally, suppose $n$ is even and that $\gamma_{2}=0$. From equations 6 and 7 we must have $\gamma_{1}=2$ and $n=n+1$, which is also a contradiction. Thus we have proved Lemma 1. ∎

To wrap up the proof of Proposition 2 we need to verify that we cannot have the situation where the only pair $\xi_{i},\xi_{j}$ that satisfies $\xi_{i}>\xi_{j}$ occurs when $i=j+1$. To that end, suppose $\xi_{i}>\xi_{j}$ for $i=j+1$. First, let $j\neq 1$. Then, if $\xi_{j-1}\leq\xi_{j}$, relabel $j-1$ as $j^{\prime}$, which yields $\xi_{i}>\xi_{j^{\prime}}$ for $i\neq j^{\prime}+1$. On the other hand, if $\xi_{j-1}>\xi_{j}$, relabel $j-1$ as $i^{\prime}$, implying $\xi_{i^{\prime}}>\xi_{j}$ for $i^{\prime}\neq j+1$. Next, let $j=1$. If $\xi_{i+1}\geq\xi_{i}$, relabel $i+1$ as $i^{\prime}$, which yields $\xi_{i^{\prime}}>\xi_{j}$ for $i^{\prime}\neq j+1$. If, instead, $\xi_{i+1}<\xi_{i}$, relabel $i+1$ as $j^{\prime}$, and thus we have $\xi_{i}>\xi_{j^{\prime}}$ for $i\neq j^{\prime}+1$. ∎

We may also write the following Corollary, which we have proved along the way.

Corollary 1**.**

Let $n\geq 4$. Then, for any die $B_{n}\neq S_{n}$, $\exists$ a die $G_{n}$ that is the image of a one-step $\Phi_{i,j}$ on the standard die $S_{n}$ that beats $B_{n}$.

Note that given some die $B_{n}\neq S_{n}$, the algorithm developed in our proof yields every winning die (i.e. a die that beats $B_{n}$) that is a one-step away from $S_{n}$. Moreover, it is easy to see how by “flipping” the algorithm we could also obtain the set of losing dice that are a one-step away from $S_{n}$. Furthermore, the algorithm also enables us to find the “best” (and “worst”) die/dice a one-step away from $S_{n}$ to play versus $B_{n}$. The die or dice with the greatest net gain from the one-step would have the highest likelihood of beating $S_{n}$. In the same manner, the die or dice with the greatest (in absolute value) net loss from the one-step would have the lowest likelihood of beating $S_{n}$.