TITLE: Bank vaults and probability
QUESTION [10 upvotes]: A burglar breaks into a bank with the intention to open the vaults and steal some gold coins.
  He knows that each vault contains a number of 1 to 100 such coins with equal probability for each number. Since he is what we call an “ethical burglar”, he will only get 100 coins to help some people in need. How many vaults must he breach on average? 

This is really confusing because we are not given how many vaults we have. Is it safe to say that since the numbers from 1 to 100 have equal probability, then we have 100 vaults? 
Then, if vault Number 1 has 1 coin, 2 has 2 etc, we need to open as many that will give sum of 100? And then get the average? 
I need help with the interpretation and solution please!
Ok after having asked several people, here is my interpretation:
We may have any number of vaults, not necessarily 100 (maybe more, maybe less) but the number of coins in them has been placed randomly, with equal probability for each number from 1 to 100. 
The burglar will stop once he has collected 100 or more coins: That is, if, after the last vault, he has 99, he may open another one, with 23 coins, so he will have a total of 122 and this is OK.
Any ideas for the solution?
Thank you!

REPLY [3 votes]: Let $E(n)$ be the expected number of vaults needed to obtain at least $n$ coins.  By conditioning on the number $k$ of coins in the next vault, we have
$$E(n)=
\begin{cases}
0 &\text{if $n\le 0$}\\
1+\displaystyle{\frac{1}{100}\sum_{k=1}^{100} E(n-k)} &\text{if $n > 0$}
\end{cases}
$$
To make this derivation explicit for $n>0$, let random variable $V_n$ be the number of vaults needed to obtain at least $n$ coins, and let (discrete uniform) random variable $C$ be the number of coins in a vault.  By the law of total expectation, conditioning on $C$ yields
\begin{align}
\mathbb{E}(V_n)&=\sum_{k=1}^{100}\mathbb{E}(V_n|C=k)\mathbb{P}(C=k)\\
&=\sum_{k=1}^{100}(1+\mathbb{E}(V_{n-k}))\frac{1}{100}\\
&=1+\frac{1}{100}\sum_{k=1}^{100}\mathbb{E}(V_{n-k})
\end{align}
We want to compute $E(100)$, which turns out to be approximately 2.678.  You can verify that $E(n)=1.01^{n-1}$ satisfies the recursion if $n>0$, so the exact value is $E(100)=1.01^{99}$.