When does BPP with a biased coin equal standard BPP? Let a probabilistic Turing machine have access to an unfair coin that comes up heads with probability $p$ (flips are independent). Define $BPP_p$ as the class of languages recognizable by such a machine in polynomial time. It is a standard exercise to prove that:

A) If $p$ is rational or even $BPP$-computable then $BPP_p=BPP$. (By $BPP$-computable I mean: there is a randomized polynomial algorithm that being fed $n$ in unary returns w.h.p. the binary rational with denominator $2^n$ that lies within $2^{-n-1}$ of $p$.)

B) For some uncomputable $p$ the class $BPP_p$ contains an undecidable language and hence is larger than $BPP$. Such values of $p$ form a dense set in $(0,1)$.

My question is the following: what happens in between? Is there a criterion for $BPP_p=BPP$? In particular:

1) Do uncomputable in $BPP$ probabilities $p$ exist such that $BPP_p=BPP$? (They may be computable in some higher classes).

2) Is $BPP_p$ wider than $BPP$ for all uncomputable $p$? (The parameters in question are those whose binary expansion contains very long sequences of zeros and/or ones. In this case computing bits by random sampling may take very long, even uncomputable time, and the problem cannot be rescaled to polynomial time. Sometimes the difficulty can be overcome by another base of expansion, but certain $p$ may fool all bases).
1) Yes, but only because of your definition. Take a unary language $L\in EXP\setminus BPP$ (yeah, I know this might be empty, in that case just take something even bigger than $EXP$), that is very sparse in the sense that $n\notin L$ if $n$ is not a tower of $2's$, i.e., of the form $2^{2^{2^\ldots}}$. Define $p=\sum_{n\in L} 1/n$. This $p$ is not $BPP$-computable, but $p$ can be approximated in $P$ up to a small enough additive error that allows the simulation of a $BPP_p$ machine.

Had you defined $BPP$-computable such that you want to approximate $p$ up to an additive error of $1/n$ (instead of $1/2^n$) in polynomial time, things would be different.

Update. The below answer is for the case when the additive error we allow is $2^{-n}$ instead of $2^{-n-1}$.

2) Yes, because here you can forget about the polynomial restriction on the classes and by sampling $2^n$ times you can get the $n$-th bit of $p$ in $BPP_p$.