Limit of optimal strategy of Ebert's hat problem Regarding this question https://puzzling.stackexchange.com/questions/100023/guess-your-hat-color-but-you-dont-have-to
What I have understood is that in general with $n$ logicians the optimal strategy, i.e. the optimal probablity $p(n)$, is that one described by tehtmi,  but we don't know the size $g(n)$ of the minimal dominating set for $n \geq 10$, i.e. we don't know the value of $p(n) = 1 - \frac{g(n)}{2^n}$
I think that $1/2 \leq p(n) \leq p(n+1) \leq 1 $ for all $n\geq 1$ since if we have that $p(n+1) < 1 - \frac{g(n)}{2^n} $ then we can forget the $n+1$-th logician, and the other $n$ logicians act as the are $n$ logicians. Hence we have that
$ \lim_{n\to \infty} p(n) $ exists and we have that $$p(9) = 225/256 \leq \lim_{n\to \infty} p(n) \leq 1$$
My question is there is a way to understand this limit without knowing the value of $p(n)$ ? It is true that the limit is equal to $1$? If not what is the limit?
Edit:
Or equivalently, what is the limit
$$ \lim_{n \to \infty} \frac{g(n)}{2^n} $$ ??
there is an asymptotic expression of $g(n)$ ? It is true that $g(n) =o(2^n) $? Or we only have that $g(n) = O(2^n) $ ?
 A: In the $n$-dimensional hypercube, a $k$-vertex set can only dominate at most $nk$ other vertices, so it cannot be dominating unless $k(n+1) \ge 2^n$. Therefore $g(n) \ge \frac{2^n}{n+1}$ and $p(n) \le 1 - \frac1{n+1}$.
Whenever $n = 2^k - 1$, we can achieve this value of $p(n)$ with a strategy based on the Hamming code.
I will assume the two hat colors are red and blue. Number all logicians from $1$ to $n$. For $i=1,\dots,k$, let $S_i$ be the set of logicians such that the $i^{\text{th}}$ binary digit of their number is $1$. Every logician is in some set $S_i$, and for every choice of one or more of these sets, there is a logician who is in exactly those sets and no others.
If we knew all the hat colors, then for every set $S_i$ we could compute a parity bit $p_i$ which is $1$ if an odd number of the logicians in set $S_i$ have red hats, and $0$ if an even number of them do. A logician does not know the value of these parity bits, because they do not know the color of their own hat. However, by considering both choices, they know that the sequence of parity bits has one of two values: $(p_1, \dots, p_k)$ or $(p_1', \dots, p_k')$. The rule a logician follows is:
If one of these sequences is $(0,0,\dots,0)$, assume the other one is correct, and guess a hat color accordingly. Otherwise, do not guess.
Whenever the parity sequence is in fact $(0,0,\dots,0)$, every single logician guesses wrong. In all other cases, there is exactly one logician contained in exactly those sets $S_i$ for which $p_i = 1$. That logician will guess the correct option, and no other logician will guess anything.
The dominating set here is the set of sequences with parity bits $(0,0,\dots,0)$ (these are the code words of the appropriate Hamming code). It includes exactly a $\frac1{2^k} = \frac1{n+1}$ fraction of all sequences, so for $n = 2^k-1$ we achieve $g(n) = \frac{2^n}{n+1}$ and $p(n) = 1 - \frac1{n+1}$.
For values of $n$ that are not of this form, we can let $k$ be the largest power of $2$ less than $n+1$, use the strategy above for $n' = 2^k-1$, and just have the first $n'$ logicians ignore all the others. This results in $p(n) = 1 - \frac1{n'+1} \ge 1 - \frac 1{2n+1}$. (In particular, the limit of $p(n)$ as $n \to \infty$ is equal to $1$.)
