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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.

Use the probability of failing to restore an arbitrary unknown input in the original encoded subspace as the error criterion. With no extraction fault, the bit-flip repetition code succeeds exactly on channel-error patterns of weight zero or one. Independent physical flips giveThe probability $B$ is the logical bit-flip probability for the ideal recovery. The preceding calculation shows that every occurrence of the specified internal fault leaves the code space. Independence of that fault therefore gives exact encoded-state failure with a faulty parity check:A single unencoded qubit has bit-flip error probability $p$. Thus the encoded scheme improves this error criterion precisely whenFor $0<p<1/2$, $p-B=p(1-p)(1-2p)>0$ and $A=(1-p{)}^2(1+2p)>0$. Rearranging givesEquality gives no improvement. At $p=0$ the unencoded channel is already perfect; at $p=1/2$ ideal recovery only ties it. For $1/2<p\le 1$, the stipulated majority correction is no better even without the internal fault, and this fault cannot improve it. Hence there is no nonnegative $q$ giving a strict improvement in those ranges.

The criterion concerns restoration of every possible logical state: a logical bit flip can accidentally preserve a particular logical $X$ eigenstate, but that is not protection of an unknown input. The threshold assumes the displayed single fault location and ideal encoding, measurement and conditional correction. Adding another recovery circuit or comparing a different average-fidelity metric would define a different noise model or performance criterion.