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The following is an example problem very common in a Computational Statistics course I have. I'm asked to comment the result of the following experiment:

A person has certain amount of money $C$ and participates in the following game: in each time unit, with probability $p$ wins a unit, with probability $q$, looses a unit and with probability $r$ remains equal. We assume that $p+q+r = 1$ and $C \ge 1$. The game continues until the player's ruin ($C \le 0$) and we are interested in knowing the time $t$ in which he ruins.

I'm asked to do a simulation of the process but my doubt here is a claim by my professor:

If there is probability to be ruined, then eventually one will ruin.

I googled Gambler's ruin problem instinctively and thought that my situation fits into that framework. The difference is that here I have just a conventional player and I think that the second player could be a player with an infinite number of coins so that he never ruins. In that sense, the Huygens' formulas that Wikipedia offer would confirm the claim of my professor.

So my question is, is my view correct or I should refer to another model for it? Also, the simulation we did was with $p = q$ (fair coin flipping), so, is the claim still valid with unfair flipping?

Related problems:

Here Gambler is saved when reaching some $N$

Here the idea of infinite capital comes in but with certain probability of avoiding ruin.

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    $\begingroup$ I would think the "effective" second player in this case would be the house, which, by definition, has infinite resources. $\endgroup$ – John Polcari Jul 7 '18 at 14:51
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The Huygens' formulas change the size of the bet as time goes on. This is not the best model for your description.

You should look into Random Walk. Your problem is directly related to a 1-D random walk and it is discussed very well in that article. This is also related to diffusion in physics because given enough time, a small particle will seemingly move in a still environment. In 1827, Robert Brown noticed that pollen 'danced around' and moved about while floating on completely still water. This is modeled with a 2-D random walk and it is called Brownian Motion.

A 1-D random walk algorithm should be quite easy to simulate to get your player to ruin. No matter how small the probability of loosing is, it will eventually 'diffuse' to ruin, it just may take some time.

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  • $\begingroup$ thank you very much for your answer, i need to read it carefully, but at first sight i wonder why the two answer differs in their conclusions $\endgroup$ – Javier Jul 8 '18 at 8:28
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You ref'ed the Wikipedia article, it has a section "Unfair coin flipping", which applies here. According to the formula there, Your player gets broke before reaching $C+n$ (i.e., before the opposing "bank" with initial payroll $n$ gets broke) with probability $$\tag1 p_{\text{ruin}}=\frac{1-a^n}{1-a^{C+n}}$$ with $a:=\frac pq$. Note that $r$ plays no roll for this - it only adds a delay to the events. If $a=1$ (i.e., $p=q$), formula $(1)$ does not apply, but rather its limit as $a\to 1$: $$\tag2 p_{\text{ruin}}=\lim_{a\to1}\frac{1-a^n}{1-a^{C+n}}=\lim_{a\to 1}\frac{-na^{n-1}}{-(C+n)a^{n+C-1}}=\frac n{C+n}$$ We are interested in the limit of $p_{\text{ruin}}$ as $n\to\infty$. If $p< q$, we have $0\le a<1$ and hence from $(1)$ $$ \lim_{n\to\infty}\frac{1-a^n}{1-a^{C+n}}=\frac{1-0}{1-0}=1.$$ If $p>q$, we have $a>1$ and $$ \lim_{n\to\infty}\frac{1-a^n}{1-a^{C+n}}=\lim_{n\to\infty}\frac{a^{-n}-1}{a^{-n}-a^{C}}=\frac1{a^C}.$$ If $p=q$, from $(2)$, $$ \lim_{n\to\infty}\frac{n}{C+n}=1.$$ We conclude that for $p\le q$, the probability of eventual ruin is $1$, but for $p>q$, the probability is strictly between $0$ and $1$, namely $\left(\frac qp\right)^C$ (i.e., there is a positive probability to be ruined, but also a positive probability not to eventually be ruined).


Remark: I suppose your professor was referring to Kolmogorov's zero-one law. But it seems that we are not dealing with a "tail event" here.

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  • $\begingroup$ thank you very much for your answer, i need to read it carefully, but at first sight i wonder why the two answer differs in their conclusions $\endgroup$ – Javier Jul 8 '18 at 8:28

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