Continue on: Prove that $BPP^{BPP} = BPP$

Question

regarding the following post: Prove that $BPP^{BPP} = BPP$

Let $L\in BPP^{BPP}$, by definition $BPP^{BPP} =\bigcup_{O\in BPP}BPP^O $, therefore there exists a language $A\in BPP$ such that $L\in BPP^{A}$.
This means that there exists a $\mathsf{PTM} $ $M^{A}(x,r)$ such that $\mathbb{P}_{r}(M(x,r)\neq L(x))\leq \frac{1}{3}$, and $M^A$ runs in polinomial time.
In addition, since $A\in BPP_{[\frac{1}{3},\frac{2}{3}]} \subset BPP_{[2^{-n},1-2^{-n}]}$, there exists a $\mathsf{PTM} $ $M'(x,r)$ such that $\mathbb{P}_{r}(M'(x,r)\neq L(x))\leq 2^{-n}$, and $M'$ runs in polinomial time.
I defiened a $\mathsf{PTM}$ $N$, that on input $x$ does the following:
Simulate $M^A$ on $x$ and every time we need to do an orcale quarey $\alpha_i$, we would simulate $M'$ on $\alpha_i$ and the answer to the query would be the answer of $M'$.

I understand why $N$ runs in polynomial time, but I was stuck in the probabilistic analysis.
This is what I have so far:
If $x\in L$ then $N$ answer correctly iff $M^A$ answer correctly and $M'$ answer correctly about every query, therefore I got:
$\mathbb{P}_{r}(N(x,r)= 1)= \mathbb{P}_{r}(M(x,r)= 1)\mathbb{P}_{r}(\forall i, M'(\alpha_i,r)$is correct$)\geq \frac{2}{3} \prod_{i}(1-\frac{1}{2^{|\alpha_i|}}) $

How can I prove from here that $\mathbb{P}_{r}(N(x,r)= 1) \geq \frac{2}{3}$?
Would appreciate your help:)

Answer 1

Your BPP algorithm makes polynomially many queries to the BPP oracle, and assuming that all these queries are answered correctly, has error probability at most $1/3$, say.

In order to simulate this algorithm without appealing to the oracle, we replace each oracle call with the BPP algorithm itself. By repeating the BPP algorithm polynomially many times, we can reduce the error probability of each oracle call to $2^{-n}$, say. The union bound then implies that all oracle calls are answered correctly with probability $1-o(1)$, and so the entire algorithm has error probability at most $1/3+o(1)$, which suffices to put the language in BPP.