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I realized that an earlier question I'd posted was actually different than what I was actually asking.

My question is: say we have a game where you win 2 dollars, or lose 1 dollar, both with probability 1/2. What is the probability that you end up with exactly the same amount of money that you started with?

Is this equivalent to a ruin problem against an infinitely rich adversary?

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It rather depends on when you decide to end the game. When you run out of money? When you next have what you started with (and what happens if you lose 1 then gain 2)? When you have a million times what you started with? The first of these to happen? – Henry May 3 '11 at 22:56
say it functions exactly as a random walk on the real number line. It ends when you next have what you started with, and you are allowed to go into the negative. – gentisse May 3 '11 at 23:07
Random walks do not usually jump over integer points, but I will take that as "don't stop". – Henry May 3 '11 at 23:08

To stop on your original position you need a multiple of three steps (twice as many $-1$ as $+2$ steps).

The probability of stopping after the first three steps is $\frac{3}{8} = 0.375$: eight possibilities for the steps, of which three end up at a net zero gain. In general the probability of first stopping after $3n$ steps is

$$\frac{{3n \choose n}}{2^{3n-1}(3n-1)}. $$

So the probability of ever stopping, rather than heading off to $+\infty$, is

$$\sum_{n=1}^{\infty} \frac{{3n \choose n}}{2^{3n-1}(3n-1)} =\frac{3}{4} \left(3-\sqrt{5}\right) \approx 0.572949\ldots$$

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Thank you, that is very helpful! So the probabilities do converge. – gentisse May 4 '11 at 0:00
how did you get the denominator in your probability? I'm not really seeing where the 3n-1 comes from – gentisse May 4 '11 at 0:13
It gives the wrong answer otherwise, in the same way the Catalan numbers ${2n \choose n}/(n+1)$ have $n+1$ in the denominator: I arrived at it trying to simplify something ugly. There are $15$ paths which arrive back at zero after six steps but $3\times 3$ first arrived after three, leaving $6$; there are $84$ paths which arrive back at zero after nine steps but $3\times 15$ first arrived after three and $6\times 3$ first arrived after six, leaving $21$; and so on. – Henry May 4 '11 at 1:29

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