= Geometric success deadline
{title2=$N_{\alpha}=\lceil\log(1-\alpha)/\log(1-p)\rceil$}
How many attempts give at least a chosen success <probability> $0<\alpha<1$? For a positive-integer <geometric distribution> with $0<p<1$, completion by attempt $N$ means $T\leq N$, so
$$
\mathbb P(T\leq N)\geq\alpha
\quad\Longleftrightarrow\quad
q^N\leq1-\alpha
\quad\Longleftrightarrow\quad
N\geq\frac{\log(1-\alpha)}{\log q},\qquad q=1-p.
$$
The <logarithm> $\log q$ is negative, which reverses the inequality when dividing. The least <integer> meeting the target is therefore obtained with the <ceiling function>:
$$
N_{\alpha}=\left\lceil\frac{\log(1-\alpha)}{\log(1-p)}\right\rceil.
$$
For our constructed <independent> retry mechanism with $p=0.3$ and $\alpha=0.95$,
$$
\frac{\log(0.05)}{\log(0.7)}\approx8.3991,\qquad N_{0.95}=9.
$$
To verify minimality directly, eight attempts give $\mathbb P(T\leq8)=1-0.7^8=0.94235199<0.95$, whereas nine give $\mathbb P(T\leq9)=1-0.7^9=0.959646393\geq0.95$. Under the zero-based convention, the same completion <event> is $F\leq8$, with failure to complete given by $F\geq9$; saying $F\leq9$ would allow ten attempts. A 95% deadline controls a tail <probability>, while the <expected value> $10/3$ describes an average.
At $p=1$ success occurs on the first attempt, so $N_{\alpha}=1$; the <logarithm> formula is inapplicable. At $p=0$ no finite deadline achieves a positive target. For $0<p<1$, no finite deadline guarantees success with <probability> one because $q^N>0$. A retry cap at $N$ produces the <capped geometric waiting time> $\min(T,N)$ as the number of attempts actually spent; reaching the cap can mean either success on the last allowed attempt or failure of all $N$ attempts.
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