Reverse Bernoulli Trial? I'm struggling to figure out how to do what I think would be called a reverse Bernoulli trial, essentially:

How many coin flips must I make to have a 75% change of getting three heads?

First of all, does this even make sense as a thing to try to work out? As for my attempts so far, I've tried expanding the binomial ${n \choose j}$, taking the log of the whole thing and using Stirling's approximation on the $ln(n!)$-like terms, but I get horribly tangled up and can't see a way round.
Does anyone have a method to do what I'm trying to? Does it have a name?
 A: You could try to use the normal approximation of the binomial distribution, which states that the binomial distribution with parameters $n$ and $p$ converges to a normal distribution with mean $np$ and variance $np(1-p)$.
This approximation is "good" if $n$ is large, so either you ask for many heads, or a probability close to $1$.
Now let's say we can make this approximation: the number of tails is $N\sim\mathcal{N}(np,np(1-p))$. So 
\begin{align}
P(N\geq 3)&=0.75\\ &\iff P\left(\frac{N-np}{\sqrt{np(1-p)}}<\frac{3-np}{\sqrt{np(1-p)}}\right)=0.25\\&\iff \frac{3-np}{\sqrt{np(1-p)}}=\Phi^{-1}(0.25)\end{align}
Where $\Phi$ is the cdf of the standard Gaussian distribution. Now you have to solve this in $n$.
A: Equivalently, how many coin flips to have less than 25% chance of getting $0,1,2$ heads.
For n flips, probability of $X<3$ heads is:
$\Pr[X<3] = \Pr[X=0]+\Pr[X=1]+\Pr[X=2]= {n \choose 0}2^{-n}+{n \choose 1}2^{-n}+{n \choose 2}2^{-n} = \left[1+n+\dfrac{n(n-1)}{2}\right]\dfrac{1}{2^n} = \dfrac{n^2+n+2}{2^{n+1}} < \dfrac{1}{4}$
By trial and error, for $n=6,\Pr[X<3]=\dfrac{44}{128}>\dfrac{1}{4}$ while for $n=7,\Pr[X<3]=\dfrac{58}{256}<\dfrac{1}{4}$, so you will need 7 flips.
