BPP amplification to exponentially small error

Let $M$ be a BPP machine for $L$. On input $x$, run $M$ independently $k$ times and output the majority answer. For the correctness indicators $X_1,\dots ,X_k$, [L1] gives a Bernoulli parameter $p_x\ge 2\mathrm{/}3$.

If the majority vote is wrong, then $X_1+\cdots +X_k\le k\mathrm{/}2\le (3\mathrm{/}4)p_{x}k$ because $p_x\ge 2\mathrm{/}3$. Thus the bad event is contained in the lower-tail event from [L2] with $\alpha =1\mathrm{/}4$, and therefore $\mathbb{P}(\text{majority is wrong on }x)\le \exp (-(1\mathrm{/}4{)}^2{p}_{x}k\mathrm{/}2)\le \exp (-k\mathrm{/}48)$.

Choose $k$ to be a sufficiently large multiple of $r(\mathrm{\mid }x\mathrm{\mid })$, for example $k=\lceil 48(\log 2)r(\mathrm{\mid }x\mathrm{\mid })\rceil$. Then step 2.1 gives error at most $2^{-r(\mathrm{\mid }x\mathrm{\mid })}$. Since $k$ is polynomial in $\mathrm{\mid }x\mathrm{\mid }$, the amplified machine is still probabilistic polynomial-time.