If someone wins 3000 rounds of a game out of 100000 numbered rounds of a game, what is the expected no. of last-3-digits of a round that they solved? I participate in r/picturegame on Reddit. Each round is numbered from Round 0 (1 actually, but let's say it was 0) up to Round 111400 at the moment.
Suppose the top player of this game had won 3000 rounds out of all the rounds up to and including round 99999 and has been playing since the beginning (this is roughly true and keeps the numbers convenient I think.) Let's also suppose that he isn't especially trying to win rounds of a certain number, and the round numbers don't follow any particular pattern relative to his sleep schedule. (This is also realistic.)
Now let's take the last 3 digits of each round that he won, so round 89345 would be "345", round 40080 would be "080" and (for the sake of argument) round 66 would be "066".
Out of the 3000 rounds that he won, what is the expected number of round-endings, from "000" to "999", that are not represented amongst those rounds?
 A: (Note: close but not quite correct approach) A simple way to look at this problem is to say that wins happen $3\%$ of the time, so on any given round a particular set of ending digits has a $97\%$ chance of not being a winner.  Each set of ending digits occurs $100$ times among the full set of rounds, giving $.97^{100}$ as the chance that set will never be a winner.  Apply this as the average across all the sets of digits to get $.04755...$ or $4.755\%$ of the $1000$ sets of digits that will not be winners on average.
Edit:
As noted in comments below, the simplistic approach detailed above gives an approximation but is not quite correct.  In particular, for a given set of digits which occurs $100$ out of $100000$ rounds, there is the chance that one of the $999$ other digit sets will be winners in each of $3000$ wins.  This happens
$$\frac{\binom{99900}{3000}}{\binom{100000}{3000}}$$
percent of the time when considering the count of such events among the total set of $3000$ wins in $100000$ rounds.  That the numbers work out pretty close to equal is a misleading coincidence.
A: (I had several false starts on this problem due to wrongly reading the problem statement.  Here's hoping I finally have it right.)
Suppose we write each of the integers from $0$ to $999$ on $m$ balls each, so there are $1000m$ balls in all.  We draw $n$ balls from the mix without replacement, with each ball equally likely to be selected. Winning $3000$ games out of $100000$ corresponds to this model with $n=3000$, $m=100$. Define
$$X_i = \begin{cases}
1 \qquad \text{if no ball labelled i is drawn} \\
0 \qquad \text{otherwise}
\end{cases}$$
for $0 \le i \le 999$. Then
$$P(X_i = 1) = \frac{\binom{999m}{n}}{\binom{1000m}{n}}$$
The total count of numbers not found is $\sum_{i=0}^{999} X_i$, and by linearity of expectation,
$$E \left( \sum_{i=0}^{999} X_i \right) = \sum_{i=0}^{999} E(X_i) = 1000 \left( \frac{\binom{999m}{n}}{\binom{1000m}{n}} \right)$$
For $m=100$ and $n=3000$, the result is $47.480$.
