09 Dec 2025
Note to self on learning and taking actions. Will rewrite it when I understand it deeper, one day.
Attempt process.
\[\begin{align*} X_t &\in \{0,1\}, \quad t = 1,2,\dots \\ \mathbb{P}(X_t = 1 \mid p) &= p, \\ \mathbb{P}(X_t = 0 \mid p) &= 1-p, \\ X_1, X_2, \dots &\text{ i.i.d.} \end{align*}\]Aggregates after $n$ attempts.
\[\begin{align*} S_n &= \sum_{t=1}^n X_t, \\ F_n &= n - S_n, \\ \hat p_n &= \frac{S_n}{n}. \end{align*}\]Probability of at least one success in $n$ attempts.
\[\begin{align*} P_n(p) &= \mathbb{P}(S_n \ge 1 \mid p) = 1 - (1-p)^n. \end{align*}\]Bayesian update.
\[\begin{align*} p &\sim \mathrm{Beta}(\alpha_0,\beta_0), \\ p \mid X_{1:n} &\sim \mathrm{Beta}(\alpha_0 + S_n,\; \beta_0 + F_n), \\ \tilde p_n &:= \mathbb{E}[p \mid X_{1:n}] = \frac{\alpha_0 + S_n}{\alpha_0 + \beta_0 + n}. \end{align*}\]Target planning.
\[\begin{align*} q &\in (0,1), \\ n^*(p,q) &:= \min\left\{n \in \mathbb{N} : 1 - (1-p)^n \ge q \right\} \\ &= \left\lceil \frac{\log(1-q)}{\log(1-p)} \right\rceil. \end{align*}\]Plug-in rule for next block size.
\[\begin{align*} \tilde p_n &\in \{\hat p_n,\; \mathbb{E}[p \mid X_{1:n}]\}, \\ n_{n+1}^{\text{plan}} &= n^*(\tilde p_n, q). \end{align*}\]Change-of-strategy indicator.
\[\begin{align*} p_{\min} &\in (0,1), \\ I_n &:= \mathbf{1}\{\tilde p_n < p_{\min}\}. \end{align*}\]