1
$\begingroup$

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?

  • 0
    R$a$ndom walks do not usually jump over integer points, $b$ut I will take that as "$d$on't stop".2011-05-03

1 Answers 1

1

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

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

  • 0
    It gives the wrong answer otherwise, in the same way the [Catalan numbers](http://en.wikipedia.org/wiki/Catalan_number) ${2n \choose n}/(n+1)$ 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.2011-05-04