# Duration of generalized Gambler's ruin game with average payoff 0

A gambler has a starting fortune of $M$. He will repeatedly play a game the game and his bets are such that he gains/loses $s$ with probability $p(s)$ (gains if $s\geq 0$, loses if $s<0$), where $p(s)$ is the same every game. If his fortune ever goes below 0 or above $T>M$ then he stops playing. Additionally, the expected return $\int_{-\infty}^\infty sp(s)ds=0$.

I am only interested in the duration of the game. I believe that the game will last the longest if $M=\frac{T}{2}$, i.e. the game starts in the middle. I am looking for help with the proof or a counterexample.

It maybe useful to consider that if $D_M$ is the expected duration of the game when the game starts at $M$, then we have the constraints

$$D_M=1+\int_{-\infty}^\infty p(s)D_{M+s}ds$$ $$D_0=D_T=0$$

For the standard Gambler's ruin with $p(+1)=p(-1)=\frac{1}{2}$ and $p(s)=0$ for $s\neq\pm 1$, the statement is true since the expected duration is then given by $D_M=M(T-M)$, which is maximal for $M=\frac{T}{2}$. I do not know however how to prove this for the general game.

Examples that do not appear to be counterexamples:

• $p(2)=\frac{1}{2}$, $p(-1)=p(-3)=\frac{1}{4}$
• $p(-\alpha)=p$, $p(\beta)=(1-p)$ with $p=\frac{\beta}{\alpha+\beta}$, $\alpha,\beta\geq 0$
• Is the value of $s$ chosen by the player? Is $s$ constant once the game starts (i.e., all bets are the same amount)? – quasi Feb 7 '17 at 13:15
• No, every turn the gambler bets such that the return is $s$ with probability $p(s)$. Alternatively this can be seen as a 1D random walk where we take a step of $s$ with probability $p(s)$ and if we ever reach $0$ or $T$ we stop. – Shinja Feb 7 '17 at 13:26
• If the gambler's wealth at a given turn is $x$, and if he always chooses $s = \text{min}(x/2,(T-x)/2)$, the game will never end. – quasi Feb 7 '17 at 13:34
• $p(s)$ does not change during the game. In other words the gambler does not adapt his strategy based on how much he currently has, nor on anything that happened previously in the game. – Shinja Feb 7 '17 at 13:35
• The expectation per round is required to be zero, but that doesn't force any kind of symmetry on $p(s)$. I would use $T = n$, where $n$ is a large even integer, and choose $p(-2) = a, p(-1) = b, p(1) = c, p(2) = d$, with $a,b,c,d$ all distinct, summing to 1 (hence $p(s) = 0$ for all other values of $s$), and such that $0 \le a,b,c,d \le 1$ and $-2a -b +c + 2d = 0$, It seems very doubtful that, over all such values of $a,b,c,d$, the duration is always maximized for $M=T/2$. A simulation with random choices of $a,b,c,d$ satisfying the given conditions could support my claim, or support yours. – quasi Feb 7 '17 at 14:36

Let $T = 8$, and let $p(s)$ be given by

\begin{align*} p(-2) &= 1/8\\[6pt] p(-1) &= 1/2\\[6pt] p(2) &= 3/8 \end{align*}

For $M$ from $1$ to $7$, let $E[M]$ denote the expected number of rounds until the player's remaining wealth is is no longer in the open interval $(0,T)$, given a starting wealth of $M$.

Then, assuming my Maple program is correct,

\begin{align*} E &= 164984/46004 \approx 3.586296844\\[6pt] E &= 252681/46004 \approx 5.492587601\\[6pt] E &= 317280/46004 \approx 6.896791583\\[6pt] E &= 331160/46004 \approx 7.198504478\\[6pt] E &= 331500/46004 \approx 7.205895140\\[6pt] E &= 253149/46004 \approx 5.502760630\\[6pt] E &= 214016/46004 \approx 4.652117207 \end{align*}

In particular, since $E > E$, the expected duration is not maximized at $M=T/2$.

• Thanks, that is a very interesting example. I will implement the numerics myself to confirm and then accept your answer. – Shinja Feb 7 '17 at 15:52