 Geometric success deadline

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 $T\le N$, soThe logarithm $\log q$ is negative, which reverses the inequality when dividing. The least integer meeting the target is therefore obtained with the ceiling function:For our constructed independent retry mechanism with $p=0.3$ and $\alpha =0.95$,To verify minimality directly, eight attempts give $\mathbb{P}(T\le 8)=1-0.7^8=0.94235199<0.95$, whereas nine give $\mathbb{P}(T\le 9)=1-0.7^9=0.959646393\ge 0.95$. Under the zero-based convention, the same completion event is $F\le 8$, with failure to complete given by $F\ge 9$; saying $F\le 9$ 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.