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A fair dice roll: On a 6 you win €1 on everything else you lose €1. The game continues until you profit €500.

What are the odds of the game ever ending?

Update: While initially I hadn't clarified as to whether the player can enter debt in fairness to @joriki I've marked his answer as correct as it answers the initial problem.

Updated - Assumptions:

  • One cannot enter debt. Eg) If you are at €0 and roll a 1 you remain at €0.
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    Thanks @joriki. I've updated the problem to reflect this.2012-08-29

2 Answers 2

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[Note:] This is an answer to the original question. The question has been changed in the meantime, and the answer no longer answers the question in its current form. The answer to the current question is trivially that the game ends with probability $1$.


Consider the same game with a profit target of €$n$. If you roll a $6$ (why is it always a $6$?), your target effectively decreases by $1$; otherwise it increases by $1$. Thus the recurrence relation for the probability $p(n)$ of the game with target €$n$ ending is

$p(n)=\frac16p(n-1)+\frac56p(n+1)\;,$

or

$p(n+1)-\frac65p(n)+\frac15p(n-1)=0$

with characteristic equation

$\lambda^2-\frac65\lambda+\frac15=0$

with solutions $\frac15$ and $1$. Thus the general solution of the recurrence is

$p(n)=c_1+c_2\left(\frac15\right)^n\;,$

and the conditions $p(0)=1$ and $p(\infty)=0$ determine $c_1=0$ and $c_2=1$, so

$p(n)=\left(\frac15\right)^n$

and

$ p(500)=\left(\frac15\right)^{500}\approx3\cdot10^{-350}\;. $

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    @celtschk: Indeed, thanks; fixed.2012-08-28
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As joriki notes, the probability that the new version of the game (where the player cannot go into debt) ends is 1.

An easy way to see this is to note that, at any point in the game, the probability that you will win within the next $500$ turns is at least $(1/6)^{500}$. Now, if you play for $500$ turns and don't win, the probability of winning within the next $500$ turns is again at least $(1/6)^{500}$, and so on.

Thus, for any positive integer $k$, the probability of winning within $500k$ turns is at least $1 - (1 - (1/6)^{500})^k$. It's then an easy exercise in algebra to show that this probability tends to 1 as $k$ tends to infinity. (Hint: use the lemma that $\lim_{k \to \infty} a^k = 0$ for all $0 \le a < 1$.)

(That's assuming that you start with no money and need to reach $500$€. If you start with some finite amount $x$€ and need to reach $500+x$€, then just replace all instances of $500$ above with $500+x$.)