\section{Introduction}

The red-and-black game problem goes back to 1965, when Lester E. Dubins and Leonard J. Savage wrote their seminal book \textit{How to gamble if you must} \cite{DS}. They presented a problem, in which player wanted to gain certain amount of money, and any other lower prize was not sufficient for their purpose. Having some initial capital and a goal, authors give us an answer, how to play - not whether to play at all - to increase the chances of winning. Moreover, a model was  presented, in which winning probability depends on the player's bet amount. Additionally, Dubins and Savage brought a Nash equilibrium for two players --  when your opponent bets his all fortune, you should play as economical as possible (such strategy is called \emph{timid}); when your opponent plays the lowest bet, you should play all your money (this is termed \emph{bold} strategy).

The work of Dubins and Savage provides a motivation for another ground-breaking book in the field, namely \cite{MS} by Ashok P. Maitra and William D. Sudderth. They continued the topic of red-and-black, having introduced new problems for the player, such as how harmful would it be to ignore previous part of the game when choosing the strategy at the current moment. Inspired on these two works, a number of articles has been published, which focused on several variations on the game, but still authors managed with bold-timid strategy. In 2005 L. Pontiggia in \cite{P1} presented two-person's game models, where the aim of both players is to get all of the opponent's money. Both models based on winning probabilities - in the first they were proportional to the player's bets, in the second they were estimated with additional weight parameter $\omega$. In 2006 M. Chen and S. Hsiau in \cite{Ch1} presented winning probability function $f$ as a function of one variable. They got two results concerning the best strategy for a player, when their opponent's strategy was known and when $f$ has certain properties. What is more, Chen and Hsiau gave a counterexample for Pontiggia's assumption for $N$-person model, $N \geq 2$, that was introduced in \cite{P1}. A year later, in 2007 L. Pontiggia in \cite{P2} presented new model for $N$-person game, including gambling house, which on every step of the game had the positive probability of winning all the players' bets. This time optimal strategies for all players is to use bold strategy. 


In this paper we will explore further the last discussed game, namely $N$-person game, where the winning sum in non-constant. Based also on \cite{W}, where a generalization for results of \cite{Ch1} has been given, we will show a less restrictive model. Then, we present examples of the functions that meet assumptions of our main theorem and thus we show that we extend the main result of \cite{P2}.
 


\section{Rules of the game}


Let $N\geq 2$ be a a fixed integer, which denotes number of the players in the game. Next, assume that $(x_1^0, \ldots x_N^0)$ is a vector of positive integers, where $x_j^0$ denotes the initial fortune of $j$-th player. Put $M := \sum_{i=1}^N x_i^0$ (the total amount of money at the beginning of the game) and let $G$ be a positive integer equal to a fixed goal the players aim to reach. We assume that the goal is the same for all players, only one player can win and moreover is is possible to win by one of the players. Therefore, we impose the following natural double inequality:
\begin{equation}
G \leq M < 2G.
\label{MG}
\end{equation}
Denote $S=\{0, 1, \ldots , M\}$.
We can define the state space for the game as
$$ P = \{(x_1, \ldots, x_N):   x_j \in S, j=1, \ldots, N , \, \sum_{i=1}^N x_i \leq M\}.$$
The absorbing states of the game consist of all vectors with one of the coordinates greater or equal to $G$ (when one of the players wins) and also of all vectors for which $\sum_{i=1}^N x_i<G$ (when it is no longer possible to win by any of the players). 

Now, we can define an action set of player $j$ when current fortunes of the players are equal to $(x_1, \ldots, x_j, \ldots , x_N)$:
$$A_j(x_1, \ldots, x_N) = \begin{cases} 
       \{1, \ldots, x_j\}, & \text{if } x_j \in \{1, \ldots, G-1\}, \\
      \{0\}, & \text{if } x_j \in \{0, G\},
   \end{cases}$$
and his payoff function:
\begin{equation}
W_j(x_1, \ldots, x_N) = \begin{cases} 
       1, & \text{if } x_j \geq G, \\
      0, & \text{if } x_j < G. 
   \end{cases}
\label{payoff}
\end{equation}

Note that the game is non-cooperative and each player has no knowledge of the actions simultaneously taken by the others.

Now, assume that we are given functions $F_j\colon P \to [0, 1]$, which represents probabilities of winning for the players. More precisely, if bets of the players are equal to $(a_1, \ldots, a_N)$, then the number  $F_j(a_1, \ldots, a_N)$ is the probability of victory of $j$-th player.
The following condition is natural and guarantees that the total probability does not exceed one:
\begin{equation}\label{1}
    \sum_{i=1}^N F_i(a_1, \ldots, a_{j}, \ldots, a_N) \leq 1.
\end{equation}
For technical reasons we will need a natural and by no means restrictive condition that a probability of victory of a given player is equal to zero if his bet is equal to zero (which, according to the rules of the game, is possible only he lost all his capital at an earlier stage of the game):

\begin{equation}
\label{00}
F_j(a_1, \ldots, a_{j}, \ldots, a_N) =0 \quad \textrm{if} \quad a_j = 0.
\end{equation}

Now, we are at the point to define precisely law of motion of the game. Let us fix a positive integer $m$, a current stage of the game. Let $X_{m,j}$ be a random variable which is equal to the fortune of $j$-th player at time $m$. By $a_{m,j}$ we denote an amount which he bids at this stage of the game. By the casino rules $1 \leq a_{m,j} \leq X_{m,j}$ for $j= 1, 2, \ldots, N$. The law of motion are:
\begin{equation}\label{r1}
     X_{1,1} = x_1^0, \qquad X_{1,2} = x_2^0, \qquad \ldots \qquad X_{1,N} = x_N^0  
\end{equation}
\begin{multline}\label{r2}
    (X_{m+1, 1}, \ldots, X_{m+1, N}) =\\ \begin{cases} 
       (X_{m,1} - a_{m, 1}, \ldots, X_{m,N} - a_{m, N}) , \qquad \text{w.p. } 1 - \sum_{i=1}^N F_i(a_1, \ldots, a_{N}), \\
       (X_{m,1} - a_{m, 1}, \ldots, X_{m,j} + \sum_{i\neq j} a_{m, i}, \ldots, X_{m,N} - a_{m, N}) , \qquad \text{w.p. } F_j(a_1, \ldots, a_{N})
   \end{cases}
\end{multline}
(here ''w.p.'' is an abbreviation of ''with probability'').

Note that inequality \eqref{1} together with the rules \eqref{r1}, \eqref{r2} implies that
\begin{equation}\label{r3}
    \sum_{i=1}^N X_{m+1, i} \leq \sum_{i=1}^N X_{m, i}.  
\end{equation}


\section{The main result}


Our main result is a generalization of a theorem of L. Pontiggia \cite{P2}*{Theorem 3.1}. We replace assumptions of super-additivity and super-multiplicativity by a much less restrictive functional inequality, which will be then discussed in the next section.

According to the definition of payoff function \eqref{payoff}, 
the players want to choose a strategy which maximizes probability of reaching the goal $G$. We provide conditions upon which bold strategies form a Nash equilibrium.


\begin{thm}\label{t1}
We consider a $N$-person red-and-black game with $N \geq 2$ and with the law of motion described by formulas \eqref{r1} and \eqref{r2}, with probability functions $F_j\colon S^N \to [0, 1]$ satisfying condition \eqref{1}.

Assume that for every $j \in\{1, \dots , N\}$ and every $X_{m,1}, \dots , X_{m, N}\in S$ such that $X_{m,1}+ \dots + X_{m, N}\geq G$ functions $f, g \colon S \to [0,1]$ given by
\begin{align}
f &= F_j(X_{m, 1}, \ldots, X_{m, j-1}, \cdot , X_{m, j+1}, \ldots, X_{m, N})
\label{f}\\
g &= \sum_{i \neq j} F_i(X_{m, 1}, \ldots, X_{m, j-1}, \cdot , X_{m, j+1}, \ldots, X_{m, N}),
\label{g}
\end{align}
satisfy functional inequality
\begin{equation}
f(x) - f(a) \geq g(a)f(x-a),
\label{main}
\end{equation}
for all $a, x \in S$ such that $a \leq x$. Then, a Nash equilibrium for all players is to play boldly.
\end{thm}
\begin{proof}
We will follow the idea of the proof of \cite{P2}*{Theorem 3.1}.

Fix $j \in\{1, \dots , N\}$ and assume that all players play boldly. Denote
\begin{equation}
    Q_j(X_1, \ldots, X_N) = \mathbb{P}[\text{Player }j\text{ reaches } G, \text{when the game starts at } (X_1, \ldots X_N)]
\end{equation}
Law of motion of the game at time $m$ is
\begin{multline}\label{r4}
    (X_{m+1, 1}, \ldots, X_{m+1, N}) = \\ \begin{cases} 
       (0, \ldots, 0) , & \text{w.p. } 1 - \sum_{i=1}^N F_i(a_1, \ldots, a_{N}), \\
       (0, \ldots, 0, \sum_{i=1}^N X_{m, i}, 0, \ldots, 0) , & \text{w.p. } F_j(a_1, \ldots, a_{N}).
   \end{cases}
\end{multline}
One can see that when all players adopt bold strategies, then the game terminates after the first round, i.e. $m=1$. Clearly, since all the players bet their entire fortunes, then either one of them wins and thus reaches his goal, or all players go bankrupt and the casino collects their bets. Moreover, since $X_{1,1}+ \dots + X_{1, N}\geq G$, then obviously 
$$ Q_j(0, \dots , 0, \sum_{i =1}^N X_{1,i}, 0, \ldots, 0)=1.$$
Thus we see that the expected return to player $j$ equals to
\begin{align*}
Q_j(X_1, \ldots, X_N) &= F_j(a_{1, 1}, \ldots, a_{1, N})\cdot Q_j(0, \ldots, \sum_{i=1}^N X_{1, i}, \ldots, 0) \\&= F_j(a_{1, 1}, \ldots, a_{1, N}).
\end{align*} 
We will show that $Q_j$ is excessive. Then \cite{MS}*{Theorem 3.3.10} implies that a bold strategy is optimal for player $j$ if all remaining players play boldly. This will complete the proof.
Therefore, we need to prove that if at first stage of the game player $j$ bets an amount $a_j<X_1,j$, i.e. less than his entire fortune, and then plays boldly for the rest of the game (if the game lasts till second round), then the expected return for him is less or equal than would be his expected return if he played boldly at the first stage as well.  
Denote the expected return for player $j$ who adopts this strategy by $\sigma_j(X_{1, 1}, \ldots, X_{1, j}, \ldots, X_{1, N})$.
Staking an amount $1\leq a_j\leq X_1,j$ means for him that 

\begin{align*}
    \sigma_j(X&_{1, 1}, \ldots, X_{1, j}, \ldots, X_{1, N}) \\ &=
    F_1(X_{1, 1}, \ldots, a_{1, j}, \ldots, X_{1, N})\cdot Q_j( a_{1, j} + \sum_{i\neq j} X_{1, i}, 0, \ldots, X_{1, j} - a_{1, j}, \ldots, 0) \\ &+ \ldots + F_j(X_{1, 1}, \ldots, a_{1, j}, \ldots, X_{1, N})\cdot Q_j(0, \ldots, \sum_{i=1}^N X_{1, i}, \ldots, 0) + \ldots + \\&+ F_N(X_{1, 1}, \ldots, a_{1, j}, \ldots, X_{1, N})\cdot Q_j(0, \ldots, X_{1, j} - a_{1, j}, \ldots, 0, a_{1, j} + \sum_{i\neq j} X_{1, i}) \\&+ \bigg[1- \sum_{i=1}^N F_i(X_{1, 1}, \ldots, a_{1, j}, \ldots, X_{1, N})\bigg]\cdot Q_j(0, \ldots, X_{1, j} - a_{1, j}, \ldots, 0).
    \end{align*}
Observe that
$$  Q_j(0, \ldots, 0, \sum_{i=1}^N X_{1, i}, 0, \ldots, 0) = 1 $$
\noindent
and
$$ Q_j(0, \ldots, 0, X_{1, j} - a_{1, j}, 0, \ldots, 0) = 0,$$
since $X_{1, j} - a_{1, j} < G$ and the payer will not be able to increase his fortune and, as a consequence to reach his goal $G$.

Moreover, for all $i\neq j$, $i = 1, \ldots, N$ we have
\begin{align*}
    Q_j( a_{1, j} + \sum_{i\neq j} X_{1, i}, 0,& \ldots, X_{1, j} - a_{1, j}, \ldots, 0) \\&= \ldots =  Q_j(0, \ldots, X_{1, j} - a_{1, j}, \ldots, 0, a_{1, j} + \sum_{i\neq j} X_{1, i}) \\&= F_j(X_{1, 1}, \ldots, X_{1, j} - a_{1, j}, \ldots, X_{1, N}).
    \end{align*}
\noindent
Finally,
\begin{align*}
    \sigma_j(X_{1, 1}, \ldots, &X_{1, j}, \ldots, X_{1, N}) =
    F_j(X_{1, 1}, \ldots, a_{1, j}, \ldots, X_{1, N})\cdot 1 \\&+ \sum_{i \neq j} F_i(X_{1, 1}, \ldots, a_{1, j}, \ldots, X_{1, N})\cdot F_j(X_{1, 1}, \ldots, X_{1, j} - a_{1, j}, \ldots, X_{1, N})).
\end{align*}

To show that $Q_j$ is excessive we need to verify the inequality
$$ Q_j(X_{1, 1}, \ldots, X_{1, N}) \geq \sigma_j(X_{1, 1}, \ldots, X_{1, N}).
$$
Note that it is equivalent to
\begin{align*}
F_j(X_{1, 1}, \ldots, &X_{1, N}) \geq F_j(X_{1, 1}, \ldots, a_{1, j}, \ldots, X_{1, N}) \\&+ \sum_{i \neq j} F_i(X_{1, 1}, \ldots, a_{1, j}, \ldots, X_{1, N})\cdot F_j(X_{1, 1}, \ldots, X_{1, j} - a_{1, j}, \ldots, X_{1, N})).
\end{align*}
Now, introduce functions $f, g$ as in the statement of the theorem, i.e.
$$ f(X_{1, j}) = F_j(X_{1, 1}, \ldots, X_{1, j-1}, X_{1, j}, X_{1, j+1}, \ldots, X_{1, N})$$
and
$$g(a_{1, j}) = \sum_{i \neq j} F_i(X_{1, 1}, \ldots, X_{1, j-1}, a_{1, j}, X_{1, j+1}, \ldots, X_{1, N}).$$
The inequality in question takes the form
$$ f(X_{1, j})- f(a_{1, j}) \geq g(a_{1, j}) \cdot f(X_{1, j} - a_{1, j}). $$
Note also that $f(0)=0$ by \eqref{00}.
Therefore, we reduced the problem to the inequality \eqref{main} and now the proof is completed. 
\end{proof}

\begin{rem}
A special case of the above theorem is when $N=2$ and with no possibility of winning by casino. In this situation condition \eqref{1} takes the form $F_1+F_2 = 1$. This, joined with inequality \eqref{main} applied for functions $f$ and $g$ for both players, after an easy calculation leads to the same functional inequality that appeared in \cite{W} in a bit different situation, where bold strategy was shown to be best response to the timid strategy (\cite{W}*{Theorem 1 and inequality (7) therein}).
\end{rem}


\section{Functional inequality}


We will begin this section with an easy but useful observation about solutions of inequality \eqref{main}, which in view of Theorem \ref{t1} is crucial for the present work.

\begin{prop}\label{p1}
Assume that we are given two functions $f, g \colon S \to [0,1]$ and $f$ is positive on $S$. Then $f, g$ satisfy inequality \eqref{main} for all $a, x \in S$ such that $a \leq x$  if and only if
\begin{equation}
g(y) \leq \min \left\{ \frac{f(x) - f(y)}{f(x-y)} : x \in \{ y+1, \ldots , M \} \right\}, \quad y \in S.
\label{min}
\end{equation}
\end{prop}
\begin{proof}
Straightforward.
\end{proof}

From the above proposition we have an immediate corollary.

\begin{cor}
Assume that $f \colon S \to (0,1]$ is a non-decreasing function and $g \colon S \to [0,1]$ is defined by the formula
\begin{equation}
g(y) = \min \left\{ \frac{f(x) - f(y)}{f(x-y)} : x \in \{ y+1, \ldots , M \} \right\}, \quad y \in S.
\label{g_min}
\end{equation}
Then $f, g$ satisfy inequality \eqref{main} for all $a, x \in S$ such that $a \leq x$.
\end{cor}

One can ask about solutions of a corresponding functional equation, when one replace inequality sign in \eqref{main} by equality, i.e.
\begin{equation}
f(x) - f(a) = g(a)f(x-a),
\label{eq}
\end{equation}
for all $a, x \in S$ such that $a \leq x$. However, it is not difficult to  find all solutions of  \eqref{eq}. Let us note the following observation.

\begin{prop}\label{pf}
Assume that we are given two functions $f, g \colon S \to \R$. Then $f, g$ satisfy functional equation \eqref{eq} for all $a, x \in S$ such that $a \leq x$  if and only if, either $g=0$ and $f$ is constant on $S$, or $f=0$ and $g$ is arbitrary on $S$,
or $f(x) = f(1)x$ for all $x \in S$ and $g=1$ on $S$.
\end{prop}
\begin{proof}
We will justify the "only if" implication only. Apply \eqref{eq} with $x=a$ to deduce that $g(a)f(0) = 0$ for all $a \in S$. Thus, either $g=0$ on $S$, or $f(0)=0$. We will discuss the second case. Put $a=0$ in \eqref{eq} to get that either $f=0$ on $S$, or $g(0)=1$. So, assume that $g(0)=1$. Next, observe that if for some $b \in S$, $b>0$ we have $f = 0$ on $\{1, 2, \dots , b\}$, then by \eqref{eq} we obtain $f(b+1)-f(1) = g(1)f(b)=0$, so $f(b+1)=0$. Therefore, we can assume that $f(1)\neq 0$. Without loss of generality assume that $f(1)=1$, by multiplying equation \eqref{eq} by a constant and replacing $f$ by $f(1)^{-1}f$, if necessary. Denote $c:=g(1)$ and use \eqref{eq} to get
$$f(x+1) - 1 = c f(x), \quad x \in S, \, x+1 \in S.$$
A straightforward induction leads to
$$f(y) = c^{y-1}+\dots +1, \quad y \in S, \, y>0.$$
Thus, if $c=1$, then $f(y) = y$ and $f(y) =  \frac{c^y-1}{c-1}$ if $c \neq 1$. It remains to insert the already obtained formulas back to our equation \eqref{eq}. In the first case we have $g=1$ on $S$, whereas second case leads us after some reductions to a contradiction.
\end{proof}

\begin{rem}
Proposition \ref{pf} and functional equation \eqref{eq} correspond to rather uninteresting possibilities in the game. First is when casino newer wins and each players wins with probability $1/N$ regardless the bets. Second is when casino collects all the bets with probability one. Note that solutions of equation \eqref{eq} are not necessarily probability functions. For example, in the last possibility spoken of in Proposition \ref{pf}, if $f(1)\neq 0$, then condition \eqref{1}, which guarantees that the sum of probabilities of winning does not exceed one, is not satisfied. Therefore, in this case solutions provided by this statement are of no use from the perspective of the game.
\end{rem}


\section{Examples and final remarks}


We will begin with a concrete example which shows that our approach is more general than the known works in the literature. It can be justified directly via Proposition \ref{p1}.

\begin{ex}\label{e1}
Let $M=1000$, $\varepsilon = 1/100$ and define $f \colon S \to (0,1]$ by the formula
$$ f(t) = 1 - e^{-\varepsilon t}, \quad t \in S.$$
Further, let $g \colon S \to (0,1]$ be defined as
$$ g(t) = \left\{  \begin{array}{ll}
	1-\varepsilon t,  & \textrm{if } t < 100, \\
	0,  & \textrm{if } t\geq 100.
	\end{array}            
\right. $$
One can calculate that inequality \eqref{main} is satisfied (for all $a, x$, in particular without restriction $a\leq x$). Moreover, one can see that $f+g \leq 1$ (and thus condition \eqref{1} is not violated).
\end{ex}

A calculation on \cite{P2}*{p. 552, before Remarks 3.1} leads us to the following corollary.

\begin{cor}
Assume that $\f\colon [0,1] \to [0,1]$ is a super-multiplicative and super-additive function and $x_1, \dots , x_n \in S$ are fixed. Then functions $F_i\colon S^N \to [0, 1]$ given by
$$F_i(a_1, \dots a_n ) := \f\(\frac{a_i}{a_i + \sum_{j\neq i}x_j } \)$$
satisfy assumptions of Theorem \ref{t1}.
\end{cor}

Two particular examples of function $\f$ which fulfils assumptions of the above corollary are given in  \cite{P2}*{Examples 3.1 and 3.2} as $\f(s) = w s$ with some $w \leq 1$ and $f(s) = s^p$ with $p \geq 1$.

\medskip 

Two further easy examples show that in context of non-constant sum games with bet-dependent probabilities it is no longer relevant whether the game is sub-fair or super-fair. Namely, it is possible to construct a super-fair game in which it is best for a player to adopt a bold strategy and a sub-fair game with timid strategy being optimal.

\begin{ex}
Assume that the win probability function for the first player is equal to 
$$ F_1(a_1, \ldots , a_N) = \left\{  \begin{array}{ll}
	0,  & \textrm{if } a_1 < x_1^0, \\
	1,  & \textrm{if } a_1 \geq x_1^0.
	\end{array}            
\right. $$
Thus, without any additional restrictions upon the remaining probability functions we see that the bold strategy is optimal for player one, in fact it is a dominant strategy guaranteeing her a sure victory. Clearly, here equilibrium is not unique (provided the game lasts more than one turn), since any bet of player 1 greater than or equal to $x_1^0$ is an optimal strategy.
\end{ex}

\begin{ex}
Now assume that for every $j \in\{1, \ldots,  N\}$ and every choice of bets $(a_1, \ldots , a_N)$ we have
$$ F_j(a_1, \ldots , a_N)=\frac{1}{N}.$$
In this situation a timid strategy is always optimal for each player (again, it is clearly a dominant strategy). 

One can modify this example by allowing the casino a positive probability of collecting all the bets, for example as follows:
$$ F_j(a_1, \ldots , a_N)=\frac{1}{2N}.$$
Now the situation changes and with probability $1/2$ all the bets are taken by casino. Still it is unwise for a single player to deviate from a dominant timid strategy and increase his bet. Consequently, with timid strategies of all players the probability that any of them will win decreases as the initial sum is large.

Note however that if the players were allowed to cooperate, then they would adopt bold strategies to increase their expected payoffs. 
\end{ex}






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   issn={0021-9002},
 
 
}

		




\end{biblist}
\end{bibdiv}

\end{document}


We will terminate the paper with results of a numerical experiment. We simulated the red-and-black game with the settings provided in Example \ref{e1}. Then, we run the game repeatedly 1000 times and counted the number of victories of Player 1 in three cases: 
\begin{itemize}
	\item[A.] she adopts bold strategy, 
	\item[B.] she adopts timid strategy, 
	\item[C.] she bets random amounts of her capital.
\end{itemize} 
The results are given in Table \ref{tab:numberOfVictories}.

\begin{table}
	\centering
		\begin{tabular}{|l||c|c|c|}	
		\hline
		&bold strategy & timid strategy & random bets  \\
		\hline\hline
		M=1000 & xx & yy & zz \\\hline
		M=???? & XX & YY & ZZ \\
		\hline
		\end{tabular}
	\caption{number of victories}
	\label{tab:numberOfVictories}
\end{table}






