# Drunk problem involving probability of being in a circle.

This is the typical drunk problem wherein the person is confined to moving either to the North, South, East, or West but never diagonally with just one step. A step has a length $L$. What is the probability that the drunk will never leave a circle of radius $2L$ after $N$ steps?

Obviously, the probability is zero for $N=1$ and $N=2$. For $N=3$, I got it to be $3/4$ although I am not sure whether this is correct. For $N>3$, I am just lost.

• This is a Markov process with $13+1$ states. – Hagen von Eitzen Jan 28 '14 at 7:48
• @Hagen von Eitzen: You can cut it down to 5 states by symmetry, and to 2 states (one of which is absorbing) if you look at the position after an odd number of steps. – Henry Jan 28 '14 at 7:59

After an odd number of steps, you are either at a point $(0,\pm 1)$, $(\pm 1,0)$, or outside the circle. You can only cross the circle on an odd step. If you start at one of those $(0,\pm 1)$, $(\pm 1,0)$ points, then you have a probability $\frac14 \times 1 + \frac12 \times \frac12 + \frac14 \times \frac14 = \frac{9}{16}$ of not leaving the circle in the next two steps.
After $N=1$ step, the probability of not having left the circle is, as you say, $1$. So for $N=3$ it is $\frac{9}{16}$, and you then keep multiplying by $\frac{9}{16}$ every two steps.
So after $N$ steps the probability of not having left the circle is $\left(\frac{9}{16}\right)^{\lfloor(N-1)/2\rfloor}$ for positive $N$.