Hello Sjoerd,
Thanks for your answer. Your proposal is interesting. However, I don't see how this problem can be solved with Markov chains as I am not familiar with this theory, but I will read more about it. The distribution depends strongly on m. Take the case where m=0, then P(1)=1/a^2 and P(v)=0 for every v>1, or if m=1 then P(v) is as in my first post. The solution I found has a similar form:
P[v_] := If[v == 1, 1/a^2,
Sum[(G.MatrixPower[B, v - 2].A)[[i]], {i, 1, 3}]];
The matrices G and B are as follows (stand for good and bad results):
B = {{1 - m + ((-1 + a) m^2)/a^2, ((-1 + a) m^2)/a^2, ((a - m) m)/a^2},
{((-1 + a) m^2)/a^2, 1 - m + ((-1 + a) m^2)/a^2, ((a - m) m)/a^2},
{((-1 + a) (a - m) m)/a^2, ((-1 + a) (a - m) m)/a^2, (a - m)^2/
a^2}};
G = {{((1 - m) m)/a + m^2/a^2, 0, 0},
{0, ((1 - m) m)/a + m^2/a^2, 0},
{0, 0, m^2/a^2}};
And the vector A is
A = {(-1 + a)/a^2, (-1 + a)/a^2, (-1 + a)^2/a^2};
As you can see, the dimension could be reduced to 2x2 matrices as there are equal elements, but for the different events it was easier to "visualize" the problem in this manner. It is quite difficult to explain, how I found this result. I would have liked a more straightforward method as I would like to calculate the distribution for a target of an arbitrary length (so not just for a pair of numbers). With the method I used, this will probably be a difficult task.