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Let's consider this simple dice game: A coin is faked so it has p chance to land on heads, and 1-p chance to land on tails. Every round costs 1, and gives you 2 if you win (for a total of +1).

Assume you're starting with n. What are your odds to "go infinite" - be able to play the game forever? This sounds like Markov Chains 101, it's just been ages since I read anything about Markov Chains.

Also - given any constant m, what are the odds of ever reaching $m in this game?

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    @Yuval: The probability is most certainly not 1. Imagine you start out with $1 and lose the first round...2011-06-24

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Define $f(n)$ as the probability of playing forever when starting out with n coins. Also, assume that the probability p of winning a round is bigger than $\frac{1}{2}$ (otherwise, the probability of playing forever is 0). Then, we get the recurrence relation $ f(n) = p f(n + 1) + (1-p) f(n-1) $ with the boundary conditions $f(0) = 0$ and $\lim_{n \to \infty} f(n) = 1$.

The general solution of the recurrence relation is $ f(n) = a + b \left( \frac{1-p}{p} \right)^n $ and from the boundary conditions we get $a = -b = 1$. So, the probability to play infinitely is $ f(n) = 1 - \left( \frac{1-p}{p} \right)^n . $

The second question can be answered just as the first one, but with different boundary conditions: $f(0) = 0$ and $f(m) = 1$. This leads to the probability of reaching m as $ \frac{1 - \left( \frac{1-p}{p} \right)^n}{1 - \left( \frac{1-p}{p} \right)^m} . $

By the way, the keyword for this kind of problems is 'Random Walk'.