Same as BPP, except that the random bit source is biased as follows. Each bit could depend on all the previous bits in arbitrarily complicated ways; the only promise is that the bit is 1 with probability in the range [δ,1-δ], conditioned on all previous bits.
So clearly 0-BPP = P and 1/2-BPP = BPP.
It turns out that, for any δ>0, δ-BPP = BPP [VV85], [Zuc91].
Same as δ-BPP, but for RP instead of BPP.
For any δ>0, δ-RP = RP [VV85].