How to calculate the average number of turns until going bust for a simplified blackjack style game? If for each turn, you draw a random number from a uniform distribution with a range between 1 and 20 inclusive and you add this number to your total sum, what is the average number of turns it would take to go bust (total sum over 21)?
I'm wondering if there's a nicer way to think about this problem, because eventually I had to result to using a computer to figure it out.
Before that, all I could figure out is simple things like the average must be greater than 2, and I tried scaling down the problem to drawing numbers of 1-2, and bust if over 3, in which case the average turns I think was 2.125. But this seemed to provide not very actionable insights. I only had a hunch it would between 2 and 3 turns, and that if the problem with scaled up to larger numbers that it would converge to some value between there.
From using a computer I found that it is true and it converges to

 e

So now I'm curious as to why that is but don't know the exact place I should look to. Could someone please help me understand why that is, or point me towards a resource which will help me learn about this thing in particular? It seems like a simple problem so I wonder if there's a more general concept here that has a name and that's commonly taught.
 A: We can alternatively look at this as $n$ drawing numbers from $0$ to $N-1$, and seeing whether their sum is larger than $N-n$. The chance for having a total of $m$ after $n$ draws is
$$
\frac{1}{N^n}\binom{n-1+m}{m}
$$
provided that $m < N$. Since that's always true if we're not bust, the probability of not being bust on turn $n$ is
$$
\frac{1}{N^n}\sum_{m=0}^{N-n}\binom{n-1+m}{m} = \frac{1}{N^n}\binom{N}{n}.
$$
Thus, if $P(n)$ is the probability of going bust on exactly turn $n$, then the probability of going bust by turn $n$ is the sum of those. This is the complement of not being bust on turn $n$, so
$$
\sum_{i =1}^nP(i) = 1 - \frac{1}{N^n}\binom{N}{n}.
$$
We can then find the expected value through a bit of a sneaky trick:
\begin{multline}
\sum_{n=1}^{N+1}nP(n) = \sum_{n=1}^{N+1}\sum_{k=1}^{n}P(k) = \sum_{n=1}^{N+1}\sum_{k=n}^{N+1}P(k) =\sum_{n=1}^{N+1}\left[1-\sum_{k=1}^{n-1}P(k)\right]\\= \sum_{n=1}^{N+1}\frac{1}{N^{n-1}}\binom{N}{n-1} = \sum_{n=0}^{N}\binom{N}{n}\frac{1^{N-n}}{N^n} = \left(1+\frac{1}{N}\right)^N.
\end{multline}
For large $N$, this clearly approaches $e$.
