I have been looking into the Coupon Collector Problem and I can't find a formula to calculate the number of trials needed to collect all the coupons in X% of the attempts. Let me explain. The formula to calculate the expected number of trials needed to collect all coupons tells us the number of trials that in (approximately) 50% of the attempts will allow us to collect all coupons. Does anyone know how to calculate the number of trials needed for an arbitrary level of confidence e.g. the number of trials that in 75% of the attempts will allow us to collect all the coupons?
If we look at the formula below I would like to change $p$ (probability to collect all coupons) with an arbitrary probability e.g. 0.75 and given $m$ (e.g. 10) solve the equation for $k$.
$$p = \sum^{m}_{i=1} (-1)^{i+1} {m \choose i} \left(\frac{m-i}{m}\right)^k$$
Since you seem to be OK with the approximation that about $50\%$ of attempts succeed after the expected number of required coupons, you could take this approximation one step further and approximate the entire distribution by a normal distribution. As is well-known, the number of required coupons can be viewed as the sum of $m$ geometric variables with probabilities $\frac jm$ for $j=1,\ldots,m$, so its mean and variance are, respectively,
\begin{eqnarray} \mu &=&\sum_{j=1}^m\frac mj \\ &=& mH_m \\ &\approx& m(\log m+\gamma) \end{eqnarray}
(where $H_m$ is the $m$-th harmonic number and $\gamma$ is the Euler–Mascheroni constant) and
\begin{eqnarray} \sigma^2 &=& \sum_{j=1}^m\left(\left(\frac mj\right)^2-\frac mj\right) \\ &=& m^2\sum_{j=1}^m\frac1{j^2}-mH_m \\ &\approx& m^2\frac{\pi^2}6-m(\log m+\gamma) \\ &\approx& m^2\frac{\pi^2}6\;. \end{eqnarray}
To approximate the number of coupons required to reach a certain probability of completing the collection, we can invert the cumulative distribution function of the normal distribution with this mean and variance:
$$ \frac12\left(1+\operatorname{erf}\left(\frac{k-\mu}{\sqrt2\sigma}\right)\right)=p\;, $$
and thus
\begin{eqnarray} k&=&\mu+\sqrt2\sigma\operatorname{erf}^{-1}(2p-1) \\ &\approx& \left(\log m+\gamma+\frac\pi{\sqrt3}\operatorname{erf}^{-1}(2p-1)\right)m\;. \end{eqnarray}