TITLE: Asymptotic behavior of the first step in a best strategy
QUESTION [8 upvotes]: Consider the game described here, but for a sequence $X_1,\ldots,X_n$ of i.i.d. uniform rv's on $\lbrace 1,\ldots,n \rbrace$ (in the original game $n=6$). Using the original notation, let $a_n$ denote the first element in the best strategy $a_n,\ldots,a_2$. We saw that for $n=6$, $a_n = n$. Can you provide a heuristic explanation as to why $a_n < n$ for all sufficiently large $n$ (this is indicated by numerical results), or even much better, can you determine the behavior of $n - a_n$ as $n \to \infty$? No rigorous proof is required, only heuristic ideas.

REPLY [3 votes]: Update: See the last section for a possible proof (**) that $a_n < n$ for all $n \geq 18$.

Here are some bounds on $E(n,m)$ that turn out to be useful.
Using $E_{n,m}$ for $E(n,m)$ in Ross Millikan's notation, let $R_{n,m} = \lceil E_{n,m} \rceil - E_{n,m}$.
Simplifying the expression for $E_{n,m}$ in Ross's first answer yields (thanks to some nice cancellation)
$$E_{n,m} = \frac{n+1}{2} + \frac{E_{n,m-1}(E_{n,m-1}-1)}{2n} - \frac{R_{n,m-1}(R_{n,m-1}-1)}{2n}.$$
Since $0 \leq R_{n,m-1} < 1$, we have $$0 \leq - \frac{R_{n,m-1}(R_{n,m-1}-1)}{2n} \leq \frac{1}{8n}.$$  This can be seen easily by the fact that the expression being bounded is quadratic in $R_{n,m-1}$ with vertex at $R_{n,m-1} = \frac{1}{2}$.
Therefore, $F_{n,m} \leq E_{n,m} \leq G_{n,m}$, where $F_{n,1} = G_{n,1} = \frac{n+1}{2}$, and
$$F_{n,m} =  \frac{n+1}{2} + \frac{F_{n,m-1}(F_{n,m-1}-1)}{2n},$$
$$G_{n,m} =  \frac{n+1}{2} + \frac{G_{n,m-1}(G_{n,m-1}-1)}{2n} + \frac{1}{8n}.$$
Thus we have recurrences that give upper and lower bounds on $E_{n,m}$ without having to deal with the problem of taking ceilings.
Numerical experiments indicate that 

$F_{n,n}$ and $G_{n,n}$ are very close to each other,
$G_{n,n} - F_{n,n}$ is decreasing,
$n - G_{n,n}$ is increasing, 
$n - G_{n,n} > 1$ for $n \geq 15$.

A proof of 3 or 4 would imply $a_n < n$ for $n \geq 15$.  A close analysis of $G_{n,n} - F_{n,n}$, together with an asymptotic estimate of $F_{n,n}$ or $G_{n,n}$, would help with the requested behavior of $n - a_n$.  
Also, it is easy to see that $F_{n,m} = n$ is an equilibrium solution for the $F_{n,m}$ recurrence.  That should be helpful as well.

It turns out that $G_{n,m} = \frac{1}{2} + n a_m$, where $a_1 = \frac{1}{2}$ and $a_m$ satisfies the recurrence $$a_m = \frac{a^2_{m-1}+1}{2}.$$  This is easy to verify once one has the conjectured expression.
It also turns out that the $a_m$ recurrence has been studied (**), with the following bounds:
$$ 1 - \frac{2}{m} + \frac{2}{m^2} \ln \frac{m}{3} + \frac{417}{128m^2} \leq a_m \leq 1 - \frac{2}{m} + \frac{5 \ln m + 3}{2m^2}$$
The upper bound implies 
$$G_{n,n} \leq \frac{1}{2} + n \left(1 - \frac{2}{n} + \frac{5 \ln n + 3}{2n^2}\right) = n - \frac{3}{2} + \frac{5 \ln n + 3}{2n} (*) $$
Now, since $E_{n,n} \leq G_{n,n}$, $G_{n,n} < n-1$ implies $a_n < n$.  The expression on the right in $(*)$ is less than $n-1$ when $$\frac{5 \ln n + 3}{2n} < \frac{1}{2},$$
which is true for all $n \geq 18$.
(**) The bounds required for my argument are given in a post in the "Real Analysis Unsolved and Proposed Problems" forum at the Art of Problem Solving.  I cannot tell whether the bounds are conjectured and the poster is asking for a proof, or whether the poster has a proof and is merely posing the problem for others to solve.  So I cannot claim that this is a complete proof.