# BPP clarification

I know that $\sf BPP[2/3,1/3]= BPP[\alpha,\beta]$ when $\alpha\lt\beta$, but I read something on Wikipedia which got me confused:

In practice, an error probability of $1/3$ might not be acceptable, however, the choice of $1/3$ in the definition is arbitrary. It can be any constant between $0$ and $1/2$ (exclusive) and the set $\mathsf{BPP}$ will be unchanged.

The reason for my question is the this question that I'm trying to answer:

We define the class $PP_{\frac{7}{8}}$: $L \in PP_{\frac{7}{8}}$. There's a probabilistic Turing machine that for $x \in L$ accepts $x$ with probability $>$ than $\frac{7}{8}$ and for $x \notin L$ it accepts $x$ with probabilty $\leq \frac{7}{8}$.

So by the $\alpha, \beta$ first definition I can conclude that $PP_{\frac{7}{8}}$ which equals to $\sf BPP[7/8,7/8+\epsilon]$ also equals to $\sf BPP[2/3,1/3]$ but the I am asked to prove that $\sf NP \subseteq BPP$ which we don't know yet.

For what value of $\epsilon$ does PP$_{7/8}$ equal BPP$[7/8,7/8+\epsilon]$? All you're given is that if $x \in L$ then the probability is strictly larger than $7/8$, while if $x \notin L$ then the probability is at most $7/8$. It could be that when $x \in L$, the machine accepts with probability $7/8 + 2^{-n}$. In fact, PP contains NP, while BPP is conjectured (by some) to equal P.
• I see, so if it was a constant it was in $\sf BPP$ but since it is not, we can't tell? – Jozef Jan 22 '13 at 11:07