= Biased-step entropy production model
{title2=$S_n=a(2K-n),\ a=\ln(p/q)$}
Take $n$ <independent> steps, each $+1$ with <probability> $p>1/2$ and $-1$ with <probability> $q=1-p>0$. Reverse a step sequence by reading it backward and changing each sign; use the same biased law for both experiments. With $K$ positive steps, the forward sequence weight is $p^Kq^{n-K}$ and its reversed weight is $p^{n-K}q^K$. Their <logarithm> ratio is
$$
S_n=(2K-n)a,\qquad a=\ln(p/q)>0,\qquad
K\sim\operatorname{Binomial}(n,p).
$$
Here $S_n$ is dimensionless <entropy production>, not an <entropy> in joules per kelvin. A <Markov chain> on a three-state ring gives a state-space realization: start uniformly, take a clockwise step with <probability> $p$ and a counterclockwise step with <probability> $q$. The uniform <stationary distribution> makes endpoint weights cancel. A physical interpretation as exchanged <heat> requires <local detailed balance> with bath <entropy> $\pm k_Ba$ per directed step; the abstract model alone specifies path <probabilities>.
For an allowed value $s=a(2k-n)$, the <binomial distribution> gives
$$
p_n(s)=\binom nkp^kq^{n-k},\qquad
\frac{p_n(s)}{p_n(-s)}
=\frac{\binom nkp^kq^{n-k}}{\binom n{n-k}p^{n-k}q^k}
=\left(\frac pq\right)^{2k-n}=e^s.
$$
This is an exact same-law finite <fluctuation theorem>, with no long-time approximation. For $p=3/4$, $n=9$ and $k=6$, $s=3\ln3$ and $p_n(s)/p_n(-s)=27$. The positive mean is $\langle S_n\rangle=na(p-q)$, although the all-negative path has nonzero <probability> $q^n$. Directly, $\langle e^{-S_n}\rangle=(pe^{-a}+qe^a)^n=(q+p)^n=1$.
\Image[/biased-step-entropy-probabilities.png]
{title=Exact biased-step model: discrete entropy-production probabilities and their log ratio at nine steps}
The left panel shows exact <probability masses> for $p=3/4$, with negative values highlighted; the right panel tests the <logarithm> ratio against $s$. These are model values, without experimental measurements or simulated measurement noise.
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