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Assume that each physical qubit undergoes an independent quantum bit-flip channel with probability $p$ during one storage interval, and that encoding, error syndrome extraction, recovery and decoding are all ideal. Use the three-qubit bit-flip syndrome table to apply the indicated minimum-weight recovery. Zero or one bit flips are corrected. Two bit flips have the same error syndrome as the complementary one-site error: for example, $X_1{X}_2$ gives $01$, so recovery $X_3$ leaves $\overline{X}=X_1{X}_2{X}_3$. Three bit flips give $00$ and also leave $\overline{X}$. This exchanges $|000⟩$ and $|111⟩$, and therefore acts as a logical Pauli X gate after decoding.

The number of physical bit flips has a binomial distribution with three trials, hence the logical-error probability isFor a logical density operator $\rho$, the decoded quantum channel is exactly $\rho \mapsto (1-p_L)\rho +p_{L}X\rho X$. Here logical failure means a nontrivial logical Pauli X gate error; a particular input such as $|+⟩$ can be invariant under that error. At $p=0.1$, exactly two flips have probability $3(0.1{)}^2(0.9)=0.027$ and three flips have probability $(0.1{)}^3=0.001$, giving $p_L=0.028$. The block succeeds with probability $0.972$ compared with $0.9$ for an unencoded qubit. At $p=0.01$, $p_L=0.000298$, illustrating the quadratic suppression of rare independent bit flips.

Factoring the improvement givesThus one level reduces logical-error probability for $0<p<1/2$, is equal at $p=0,1/2,1$, and makes it worse for $1/2<p<1$ under this same decoder. The suppression arises because a logical error requires at least two physical bit flips. With independent disjoint inner blocks, the same noise model and ideal correction at every level, a concatenated quantum error-correcting code has $p_{\ell +1}=3p_{\ell }^2-2p_{\ell }^3$, with $p_0=p$. At $p_0=0.1$, two levels use nine data qubits and give $p_2=3(0.028{)}^2-2(0.028{)}^3=0.002308096$.

This calculation describes ideal correction of stored information with the specified quantum bit-flip channel. Repeated correction in time additionally requires a fresh specified noise model for each interval; imperfect quantum gates, faulty measurements and correlated errors change the calculation. The timed bit flip during repetition-code syndrome extraction and exact encoded-state failure with a faulty parity check show concretely why this storage calculation alone is not a fault-tolerance result.

The PDF places the extra $X$ on $q_2$ after that qubit has interacted with $a_1$ but before it interacts with $a_2$. Its timing is essential. With no incoming error, the first extracted parity is zero. The fault then changes the data toThe remaining parity measurement gives $a_2=1$ on both branches. Thus the measured error syndrome is $01$, which the rule in the preceding part interprets as an error on $q_1$. Applying $X_1$ leavesThis is outside the original encoded subspace: neither of its supporting basis strings is $000$ or $111$. The incorrect syndrome causes a second data error rather than restoring the input. This timed bit flip during repetition-code syndrome extraction demonstrates why a simple parity circuit need not tolerate its own faults.

The same failure can be checked for every incoming channel error. Let $f\in \{0,1\}$ indicate the internal fault and let $c(s)$ be the correction bit pattern from the table. The observed syndrome and the final pre-correction data pattern areBecause $s$ is linear over binary addition and $s(c(s_{\mathrm{obs}}))=s_{\mathrm{obs}}$, the residual pattern $r=d\oplus c(s_{\mathrm{obs}})$ satisfiesWhenever the fault occurs, the residual syndrome is $10$, so recovery never returns the state to the encoded subspace, even if a channel error also occurred. The diagram therefore must not be treated as an extra independent pre-extraction data flip that would always be diagnosed normally.