Determining expected number of rounds I'm trying to determine a formulation for the expected number of rounds for a game of chance based on dice.
The details of the game are as follows:


*

*The game initially consists of $N$ players

*Each player has a fair die or access to one.

*For each round each player will roll their die once

*The mean of the rolls is determined

*Any player that rolled a value less than the mean is eliminated from the game

*The rounds are repeated until their is only player left.
My question is how does one determined the expected number of round in terms of $N$.
My thinking is that for any round the expectation is that half of the players will be eliminated. This leads to a continuous halving of the number of players, which seems to suggest that $\text{E}(\text{Rounds}) \approx \log_2 {N}$, or specifically:
$$ \lim_{N \to \infty}  \text{E}(\text{Rounds}) =\log_2 {N} $$
I wasn't able to formulate a definitive relation, so I ran a simple monte-carlo of the game, for numbers of players ranging from 2 players to 200 players, one million simulations per game size. The results can be found here:
https://gist.github.com/anonymous/f0db85f06343070045b78f7494f19565
The graph for the results and $\log_2 {N}$ is shown in the following as red and blue respectively:

Where I'm having difficulty is explaining the continuous and consistent over-estimate of the result curve (via simulation) when compared to the $\log_2{N}$ curve.
 A: Let e(n) be the expected number of rounds, starting with n players.
For a given positive integer n, the exact value of e(n) can be found via a recursive procedure.

I implemented such a procedure in Maple, and the results, for $2 \le n \le 8$, are shown below (the fractions are the exact values).
\begin{align*}
e(2) &= \dfrac {6} {5} \approx 1.200000000\\
e(3) &= \dfrac {303} {175}  \approx 1.731428571\\
e(4) &= \dfrac {81486} {37625} \approx 2.165740864\\
e(5) &= \dfrac {654386} {263375} \approx 2.484616991\\
e(6) &= \dfrac {1123369554} {409548125} \approx 2.742948839\\
e(7) &= \dfrac {1155681595606}  {389948321875} \approx 2.963678854\\
e(8) &= \dfrac {49218033279086166} {15594311926296875} \approx 3.156152930
\end{align*}
Here's my Maple implementation of e(n) ...


A: I've run a C++ program very close in algorithm to the Maple program by @quasi.  That is, it's based on
$$ e(n) = \frac{1+\sum_{0 < k < n}p_{k,n}\cdot e(k)}{1-p_{n,n}}\enspace,$$
where $p_{k,n}$ is the probability of there being $k$ survivors out of $n$ players.
Here's a snippet of the code, which shows how one works with the C++ bindings of the GMP bignum library.

static void print_expected_survivors(std::vector<mpz_class> const & counts,
                                     long i, mpz_class outcomes)
{
  mpq_class exp_surv = 0;
  for (long j = 0; j != i; ++j) {
    exp_surv += counts[j] * (j+1);
  }
  exp_surv /= outcomes;
  mpf_class ef = exp_surv;
  std::cout << "s(" << i << ") = " << ef << std::endl;
}

In the GMP library, mpz, mpq, and mpf stand for multiple precision integer, rational, and floating-point number, respectively.  This little function manipulates data of all three types as well as of type long. 
Here's a plot of $e(n) - \log_2(n)$:

Not obvious from the data whether there is a horizontal asymptote different from $y=0$.
