# Random walk in a sphere

Given a sphere of radius $R$, divided in cubic cells of size $l$, the probability for a particle to jump from a cube to another adiacent is: $P=\frac{1}{6}$. If we define the probability to exit from the sphere as: $P_0(l,R,t)$ what is the expression for this probability vs. time $t$, radius $R$ and cell's size $l$? We can consider for the particle to be outside the sphere when it occupies a cell with the center greater than $R$. The starting point of the process is the center of the sphere and the first cubic cell is located such that its center coincides with the center of the sphere itself. Thanks.

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Are we assuming that the number of cubes is large? If so I think this problem is not too hard and I can try to write an answer. – DanielSank Jan 2 '15 at 5:56
@DanielSank: Yes, the number of the cubes is large. – Riccardo.Alestra Jan 7 '15 at 10:08
If the number of cubes is large can't you just use a continuum approximation? Do you need an exact result? – DanielSank Jan 7 '15 at 16:16
@DanielSank: I can use a continuum approximation, if it is more suitable for calculation. – Riccardo.Alestra Jan 8 '15 at 9:25