Balls are thrown randomly and uniformly into $n$ bins up until one bin has 3 balls. Let there be $T=T(n)$ be the number of throws we made until the occasion occurs. I am to assume $n$ is large and requested to find $f(n)$ so that the probability:
$$ P(T > 0.1f(n))$$ Is close to 1 (for example bigger than 0.9999) and the probability: $$ P(T > 10f(n))$$ Is close to 0 (for example smaller than 0.001)
Our hint was to investigate the random variable, when throwing $m$ balls $X(m)$ is the number of triplets $[i,j,k] \subset [1,2,...,m]$ so that the $i,j,k$ balls thrown fall in the same bin.
I have tried doing something similar to the birthday problem by defining a uniform random variable $X_i$ over $i \in [1,...,n]$ so its basically if there is a ball in the $i$ bin. then I made another variable $$Y_i,_j,_k =\begin{cases} 1, & \text{$X_i = X_j = X_k$} \\0, & \text{otherwise}\end{cases}$$ And then went on to calculate the expected value. Overall I received the result ${m \choose 3}*\frac{1}{n^2}$ but the results don't match the definition of $X(m)$.
I am really kind of lost as this was the main way I felt might work, where am I wrong in this, whether its my way of thought or just my math, all help would be appreciated.
p.s. this is a homework question but the due date already passed
You are asked to get the number of balls right within a factor of $10$ so we can be rather rough. If $n$ is fairly large you will have a Poisson distribution of the number of balls in each bin. If we throw $k$ balls the parameter in the Poisson distribution is $\lambda=\frac kn$. We want to choose this so there is a reasonable chance that at least one bin has three balls. The chance a given bin has three balls is $\frac {\lambda^3e^{-\lambda}}{3!}$. Since there are $n$ bins, we want this to be about $\frac 1n$ So (using $=$ instead of $\approx$) we have $$\frac {\lambda^3e^{-\lambda}}{3!}=\frac 1n\\ \frac {(\frac kn)^3e^{-\frac kn}}{3!}=\frac 1n\\ e^{-\frac kn}=\frac {6n^2}{k^3}$$ I did an approximate solution of this for $n$ from $5$ to $65$ in steps of $5$ in a spreadsheet. The graph is below.