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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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Astronomically low :-) – joriki Aug 28 '12 at 7:57
Now that you know the answer, you might want to make this an acceptable question, by adding what you know, what you tried, and so on. – Did Aug 28 '12 at 7:59
Actually the probability that the game ever ends is 1. However the most likely endings are: (a) the player refuses to continue to play after he noticed he's losing. (b) the player is bankrupt and cannot continue playing. (c) the player dies of old age. – celtschk Aug 28 '12 at 8:30
I don't understand any of the assumptions you've added. Are they meant to change the question? It seemed clear enough to me before you added them. a) What's the relevance of starting with €0 when the terminating condition refers only to profit? b) What does it mean that the losses can't be negative -- does that imply that the profit can't be positive? c) What's $P(n)$? And in case it's meant to be the probability of the game ending, i) how you can you make assumptions on that when it's what you're trying to calculate, and ii) that inequality is the wrong way around. – joriki Aug 28 '12 at 9:08
You've now substantially changed the question and then accepted my answer, which was an answer to the original question and isn't a correct answer to the new question. Also, the change in the question has now made it relevant how much money you have to begin with, but you've removed that information, so the question is now incomplete. – joriki Aug 28 '12 at 15:35
up vote 7 down vote accepted

[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




with characteristic equation


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


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



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

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@celtschk: I think that comment was based on a misunderstanding; I've tried to clarify the answer. – joriki Aug 28 '12 at 8:56
Ah, indeed, now I understand what you mean, thanks for the clarification. However I think you've exchanged "increases" and "decreases" in your second sentence (the formula gets it right, though). – celtschk Aug 28 '12 at 10:38
@celtschk: Indeed, thanks; fixed. – joriki Aug 28 '12 at 11:00

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 $500+x$.)

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