Known:
BPP vs NEXP is open. BPP is strict subset of EXP^EXP.
Question:
Is BPP vs EXP^NP open?
If so, is there any class between EXP^NP, EXP^EXP concerning which vs BPP it's still open?
Thanks!
Context: Slaughtering open problems.
Known: BPP vs NEXP is open. BPP is strict subset of EXP^EXP. Question: Is BPP vs EXP^NP open? If so, is there any class between EXP^NP, EXP^EXP concerning which vs BPP it's still open? Thanks! Context: Slaughtering open problems. 


As far as I can see, BPP is not known to be distinct from $\mathrm{EXP}^\mathrm{NP}$. A standard argument shows that $\mathrm{BPP}\ne\mathrm{BPEXP}$. (If $\mathrm{BPP}=\mathrm{BPEXP}$, then in particular $\mathrm{BPP}=\mathrm{EXP}$, hence $\mathrm{BPEXP}=\mathrm{EEXP}$ by padding, hence $\mathrm{EXP}=\mathrm{EEXP}$, contradicting the deterministic time hierarchy theorem.) We have (by padding, using the corresponding results from the polynomial hierarchy) $\mathrm{BPEXP}\subseteq\mathrm{MAEXP}\subseteq\mathrm S_2^\mathrm{EXP}\subseteq\mathrm{ZPEXP}^\mathrm{NP}\subseteq\mathrm{NEXP}^\mathrm{NP}\cap\mathrm{coNEXP}^\mathrm{NP}$, so $\mathrm{BPEXP}$ is very close to being contained in $\mathrm{EXP}^\mathrm{NP}$, but not quite so. There is also the related result by Buhrman, Fortnow, and Thierauf that $\mathrm{MAEXP}\nsubseteq\mathrm P/\mathrm{poly}$ (note that $\mathrm{BPP}\subseteq\mathrm P/\mathrm{poly}$). 

