How many draws to pass 0.95 · Problem 193

s Shown after a correct answer. Revealing it first marks the problem as seen, not solved.

Idea. The maximum is below $0.95$ only if every draw is, which happens with probability $0.9 5^{n}$ . We need that below $\frac{1}{2}$ .

Step 1. $P (max \geq 0.95) = 1 - 0.9 5^{n},$ which is at least $\frac{1}{2}$ exactly when $0.9 5^{n} \leq \frac{1}{2}$ .

Step 2. Taking logarithms, $n \geq \frac{l n 2}{- l n 0.95} \approx \frac{0.6931}{0.0513} \approx 13.5$ .

Step 3: check. $0.9 5^{13} \approx 0.513$ and $0.9 5^{14} \approx 0.488$ . So the answer is $n = 14$ .

A rule of thumb. For a small probability $p$ per try, about $\frac{0.69}{p}$ tries give an even chance of at least one success: here $\frac{0.69}{0.05} \approx 14$ .