Determining expected number of rounds

Question

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:

1. The game initially consists of $N$ players

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

3. For each round each player will roll their die once

4. The mean of the rolls is determined

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

6. 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.

Answer 1

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) ...

Answer 2

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 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 $y=0$.