= Binary symmetric channel capacity
{title2=$C=1-h_2(p)$}
A <binary symmetric channel> independently flips each transmitted <bit> with crossover <probability> $p$. Its <channel capacity> is
$$
C=1-h_2(p),\qquad 0\leq p\leq1,
$$
in <bits> per use, where $h_2$ is <binary entropy> and the endpoint convention is $0\log_2 0=0$. Knowing the input leaves just the uncertainty of the flip, so $H(Y\mid X)=h_2(p)$ for every input <probability distribution>. Since the output is a <bit>, $H(Y)\leq1$, and thus $I(X;Y)\leq1-h_2(p)$. A fair input produces a fair output and reaches the bound. This is the two-symbol case of <weakly symmetric channel capacity>; https://ocw.mit.edu/courses/16-36-communication-systems-engineering-spring-2009/a49d9e2954b440def3794fe73e79756c_MIT16_36s09_lec13_14.pdf[Eytan Modiano, MIT 16.36, Channel Capacity and Coding, slides 4–5] gives this calculation.
For crossover <probabilities> $p=0.01,0.1,0.25$, the resulting <channel capacities> are approximately $0.91920686$, $0.53100441$, and $0.18872188$ <bits> per use. At $p=0$, the <communication channel> is noiseless; at $p=1/2$, the output is <independent> of the input and the capacity is zero. At $p=1$, every <bit> is inverted, a known reversible rule, and capacity is one again. More generally $C(p)=C(1-p)$, because inverting every received <bit> exchanges the two crossover <probabilities>. The existing <mutual information> figure shows this full symmetric curve.
An uncoded transmission illustrates why the raw bit-error <probability> is not itself <channel capacity>. At $p=0.1$, transmitting 100 uncoded <bits> succeeds on all of them with <probability> $0.9^{100}\approx0.00002656$. The <probability> of at least one error is therefore approximately $0.99997344$. A suitable <block code> changes which received words count as recoverable evidence of a message; simply repeating uncoded transmission at rate one does not obtain the reliability promised at rates below $C$.
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