sniffnoy | Answers to and brief notes on the problem in last entry (Reply)

Real case: As the radius goes to infinity, the probability goes to 1. As the radius goes to 0, the probability goes to 1/2. At a radius of 1, the probability is 13/24.

p-adic case: As the radius goes to infinity, the probability goes to 1. As the radius goes to 0, there's no limit. At a radius of 1, the probability is p/(2p+2). (Even if p=2, yes, though the proof is different. Could there be one that unifies them? Dunno.)

For the probabilities in general, they are given as follows:

I'll use the notation from the problem as it was assigned; in the real case, P(B) will denote the probability for a radius of B; in the p-adic case, P(B_k) will denote the probability with a radius of p^-k.

Then we have P(B)=1/2+B/24 for B≤4 and P(B)=1-2/3√B for B≥4.

In the p-adic case, assuming p≠2, for k≥0 even, P(B_k)=p/(2p+2), while for k≥1 odd, it's 1/(2p+2). For k≤0 even, it's 1-(p^k/2+1+2p^k/2)/(2p+2), while for k≤1 odd, it's 1-(p^(k-1)/2+2p^(k+1)/2)/(2p+2).

In the 2-adic case, we actually get the same probabilities as in the odd case (though for a different reason) so long as k≤1. For k≥2, we get half that. We could also write this as follows: For k≥2 even, we get 1/6; for k≥3 odd, we get 1/12; for k<2 even, we get 1-2^k/2+1/3; and for k<3 odd, we get 1-(2^(k-3)/2+2^(k+1)/2)/3.

Is this an instance of some more general regular phenomenon? I have no idea, I just solved the problem as written... don't suppose any of you know about this sort of thing?