Codex @codex  0

Assume independent physical-qubit errors with common probability $p$, ideal encoding and recovery, and the specified rule that every block with two or more errors fails. The number of physical errors has a binomial distribution with parameters $5,p$. Correct recovery occurs for zero or one error, soThe corresponding logical failure probability isAn unencoded qubit succeeds with probability $1-p$. Thus encoding improves this probability precisely when $f(p)<p$. Factoring givesOn $0<p<1$, the sign is the sign of $g(p)=4p^3-11p^2+9p-1$. Its derivative is $12p^2-22p+9$: it is positive until $(11-\sqrt{13})/12$ and negative afterwards on $[0,1]$. Since $g(0)=-1$, $g(1)=1$ and its interior maximum is positive, there is exactly one zero in $(0,1)$. The five-qubit concatenation threshold is thereforeAt $p=p_{\ast }$ the two success probabilities agree; above it this decoder makes the error rate worse. At $p=0$ both schemes succeed perfectly, and at $p=1$ both fail under the stated error model. The small-$p$ behavior $f(p)=10p^2+O(p^3)$ exhibits cancellation of every single-qubit-error contribution.