Extreme bionomal distribution, finding at least $x$ successes amongst many trials Sorry, I don't know much about advanced probabilities. I've learned most of what I know in the last few hours. So be gentle.
Essentially, I want to calculate the odds of dropping at least 80 swords in an idle RPG game. We kill 100 monsters per minute, and the drop chance of 1 is 1/20000 (0.005%). So, I want to know what the probability is for at least 80 to drop after 1 day, 10 days, 30 days and 365 days. Any of these will suffice for this question though, just to understand how it works.
As I understand it, 10 days is 1,440,000 trials:
$$10 days \times 24 hours \times 60 minutes \times 100 kills/minute$$
I've tried an online calculator which was limited to 1000 trials, and also manually calculating how likely exactly 80 swords is after 10 days:
$$\frac{n!}{x!\times(n-x)!} \times p^x \times q^{(n-x)}$$
n = number of trials
x = number of successes
p = probability of success in any one trial
q = probability of failure in any one trial (q = 1 - p)

$$\frac{1.44e6!}{80!\cdot(1.44e6-80)!} \times (1/20000)^{80} \times (19999/20000)^{(1.44e6-80)} \approx 2.9\% $$
But how do I find the chance that at least 80 drops will have occurred after $X$ minutes?
 A: Let $X_n$ count the number of collected swords after $n$ minutes.
Note that $X_n$ follows a Binomial Distribution, $X_n\sim B(100n,\frac{1}{20000})$. The formula for calculating when a Binomial is lower or equal than a number is easily defined, so we will use it.
$\begin{align}
P(X_n\geq 80)&=1-P(X_n\leq 79)\\
&=1-\sum_{i=0}^{79}\binom{100n}{i}\frac{1}{20000^i}\frac{19999^{n-i}}{20000^{n-i}}\\
&=1-\frac{19999^n}{20000^n}\sum_{i=0}^{79}\binom{100n}{i}\frac{1}{19999^i}\\
&=1-\frac{19999^n}{20000^n}\sum_{i=0}^{79}\frac{100n!}{i!(100n-i)!19999^i}
\end{align}$
Now you simply have some calculations to do and a sumatory
A: Let $p = \frac{1}{20000},$ 
that is, $p$ is the probability to drop a sword in each trial.
The expected value of the number of swords dropped per trial is $p$,
and the variance in the number of swords dropped per trial is
$p - p^2$ (since each trial is a Bernoulli random variable).
The expected value $\mu$ of a sum of random variables is the sum of
their expected values, so in $N$ trials, the expected number
of dropped swords is $\mu = Np$.
Assuming each trial is a random variable independent of all the other trials,
the variance $\sigma^2$ of the sum of the variables 
is the sum of their variances,
so in $N$ trials the variance in the number of swords dropped
is $\sigma^2 = N(p - p^2).$
The standard deviation $\sigma$ of a probability distribution is the
square root of its variance.
Since $p$ is so small in this case, 
$\sigma = \sqrt{N(p-p^2)} \approx \sqrt{Np}$.
If you add together $N$ independent Bernoulli variables,
when $N$ is very large the resulting binomial distribution starts to look like a
Gaussian distribution (also called a normal distribution).
You can then use the mean and standard deviation of that normal distribution,
along with tables that tell you what percentage of the time a normal 
random variable will be less than $m$ standard deviations above the mean,
to estimate the probability that your binomial variable will be
less than $m$ standard deviations above its mean.
$80$ is not a particularly large number in this context, so the
approximation may be somewhat rough.  But for $N = 1.44 \times 10^6$ trials,
the expected number of dropped swords is $\mu = 72$ and the
standard deviation is $\sigma = 8.485$. That is, $80$ swords is just about
$\mu + 0.94 \sigma$.
We look up $0.94$ in a table for area under the normal curve
(such as the table on this page),
or we have a computer calculate it for us, and we find that
nearly $83\%$ of the area under the normal curve is to the left of that point,
giving us about a $17\%$ chance to exceed $80$ swords
(but recalling that that is a rough approximation,
I'd just say it's somewhere around $15$ or $20$ percent).
The normal distribution tables tell us there is about a $0.01$
probability to exceed $\mu + 2.33\sigma$,
so suppose $N$ is just large enough that 
$\mu + 2.33\sigma \approx Np + 2.33\sqrt{Np} = 80$.
Letting $y = \sqrt{Np}$, this means we want to solve the quadratic equation
$$y^2 + 2.33y - 80 = 0.$$
Once we find $y$, we can work backwards to find that $N = \frac{y^2}{p}$.
The only positive root of the equation above is $y \approx 7.85$,
so in that case $N \approx 1.23 \times 10^6.$
In other words, you have to do something like $1.23 \times 10^6$
trials in order to have at least a $1\%$ chance of dropping $80$ or more swords.
The tables also say that there is 
about a $0.0001$ chance to exceed $\mu + 3.73\sigma$, 
which implies a $0.9999$ chance to exceed $\mu - 3.73\sigma$.
Solving for $y$ in
$$y^2 + 3.73y - 80 = 0,$$
the only positive root is $y \approx 11.0016$, which implies
$N \approx 2.42 \times 10^6$.
So the number of trials needed in order to have a $99.99\%$ chance
of $80$ or more dropped swords is just about twice the number of trials
that give you a $1\%$ chance of $80$ or more dropped swords.
Or in other words, once you get to $1\%$, it takes only about the
same amount of time again to reach $99.99\%$ probability.
Is that what you meant by a "narrow timespan"?
To make this better than just an educated guess, however, you would still need
to try calculating the exact binomial probabilities for each number of swords
from $0$ to $79$ inclusive to find an accurate probability that you drop
fewer than $80$ swords.
By the way, it may help in  your calculations
to use the fact that
$$\begin{eqnarray}
\frac{n!}{x!(n-x)!} p^x q^{n-x}
  &=& \frac{n(n-1)(n-2)\cdots(n-x+1)}{x!} \left(\frac pq\right)^x q^n \\
  &=& \left(\frac{n}{1}\cdot\frac pq\right)
\left(\frac{n-1}{2}\cdot\frac pq\right)
\left(\frac{n-2}{3}\cdot\frac pq\right)\cdots
\left(\frac{n-x+1}{x}\cdot\frac pq\right) q^n.
\end{eqnarray}$$
In particular, once you have the probability of dropping $x$ swords,
call this $P(X = x)$,
the probability of dropping $x+1$ swords is
$$\begin{eqnarray}
P(X = x+1) &=& \frac{n!}{(x+1)!(n-x-1)!} p^{x+1} q^{n-x-1} \\
&=& \frac{n-x}{x+1} \cdot \frac pq \cdot \frac{n!}{x!(n-x)!} p^x q^{n-x} \\
&=& \frac{n-x}{x+1} \cdot \frac pq \cdot P(X = x).
\end{eqnarray}$$
So computing all the probabilities from $x = 0$ to $x = 79$ inclusive
is not terribly hard to do with the help of a computer.
(If you are not using software that can deal with very large numbers,
you may need to do the calculations using logarithms in order to avoid
exceeding the maximum size of a floating-point number.
But with that warning in mind, I think
even a decent spreadsheet program can be made to do the calculation.)
