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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 $p=\frac{\beta}{\alpha+\beta}$, $\alpha,\beta\geq 0$
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    Is the value of $s$ chosen by the player? Is $s$ constant once the game starts (i.e., all bets are the same amount)? – 2017-02-07
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    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. – 2017-02-07
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    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. – 2017-02-07
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    $p(s)$ 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. – 2017-02-07
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    You mean $s$ doesn't change. – 2017-02-07
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    $p(s)$ is the probability; $s$ is the amount of the bet. – 2017-02-07
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    No, I mean $p(s)$ does not change. Let's say that $p(\pm 1)=\frac{1}{2}$ and $p(s)=0$ otherwise. Then no matter what the current fortune of the gambler is or what happened before, in every round he will either gain $+1$ or lose $-1$ with probability $\frac{1}{2}$ each. So $s$ is not always the same, but the probability with which $s$ is chosen is. So the gambler cannot adapt and choose $s=\min(x/2,(T-x)/2)$ where $x$ is his current fortune since that would require adapting the probability each round. – 2017-02-07
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    Ok,I follow. The gains and losses in each round are not up to the player. They depend on the probability distribution, based on the function $p(s)$. – 2017-02-07
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    I'm fairly sure that your claim that the expected duration is always maximized at $M= T/2$ is false, so I wouldn't even consider trying for a proof. Since the calculations are tedious, it won't be so easy to get a counterexample by hand, and writing a program to find a counterexample is a nontrivial task in its own right. Still, in my opinion, it's a counterexample or nothing. – 2017-02-07
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    Can you elaborate on where that intuition comes from so that I may look for counterexamples in a more guided direction? – 2017-02-07
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    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. – 2017-02-07
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    You could use $3$ values (as in the examples you gave), instead of $4$. Then the search space is reduced to testing triples $(a,b,c)$ instead of quadruples $(a,b,c,d)$. – 2017-02-07

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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[1] &= 164984/46004 \approx 3.586296844\\[6pt] E[2] &= 252681/46004 \approx 5.492587601\\[6pt] E[3] &= 317280/46004 \approx 6.896791583\\[6pt] E[4] &= 331160/46004 \approx 7.198504478\\[6pt] E[5] &= 331500/46004 \approx 7.205895140\\[6pt] E[6] &= 253149/46004 \approx 5.502760630\\[6pt] E[7] &= 214016/46004 \approx 4.652117207 \end{align*}

In particular, since $E[5] > E[4]$, the expected duration is not maximized at $M=T/2$.

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    Thanks, that is a very interesting example. I will implement the numerics myself to confirm and then accept your answer. – 2017-02-07