One answer to your question is that it depends on the underlying model of computability you are using.
Specifically, although this is a little different than your set-up, but for one of the standard models of computability---Turing machines with a one-way infinite tape and single halt state---Alexei Miasnikov and I proved that the halting problem is decidable with asymptotic probability one. See also this MO
answer and also this MO answer, in which I mention similar ideas, reproduced in part below.
Our idea was to use asymptotic density. For any natural number
$n$, there are only finitely many Turing machine program
using $n$ states. The asymptotic density or asymptotic probability
of a set $A$ of Turing machine programs is the limit (if it
exists)
- $\lim_{n\to\infty} \frac{|A\cap P_n|}{|P_n|}$,
where $P_n$ is finite the set of Turing machine programs
with exactly $n$ states. Thus, the asymptotic probability
of a set $A$ of Turing machine programs is simply the limit
of the proportion of $n$-state programs in $A$. In
particular, if a set $A$ has asymptotic density $1$, then
it means that more than $99\%$, more than $99.9\%$, of
Turing machine programs are in $A$, as close to $1$ as
desired as the number of states increases. In this case, we
would seem to be justified in saying that almost every
Turing machine program is in $A$.
To give an elementary sample calculation, a Turing machine
program $p$ in finite alphabet $\Sigma$ with states $S$
(not counting the halt state) is a function
$\Sigma\times S\to \Sigma\times
(S\cup\{halt\})\times\{L,R\}$. For example, if the alphabet
has $2$ symbols and there are $n$ states, then there are
$(4(n+1))^{2n}$ many programs. The number of programs that
never transition to the halt state, however, is
$(4n)^{2n}$, which has proportion $(\frac{n}{n+1})^{2n}$,
which goes to $\frac{1}{e^2}$ as $n\to\infty$. Thus, the
density of programs that never halt at all, because they
can never transition to the halt state, is $\frac{1}{e^2}$,
or about $13.5\%$.
This way of thinking is the foundation of the topic of
generic case
complexity.
A central concern of this topic is the fact that many
undecidable or unfeasible decision problems admit a black
hole, a very small region where the problem is difficult,
outside of which it is easy. For example, it is not good to
base a financial encryption scheme on a problem whose
difficulty is confined to a black hole---a robber is after
all satisfied to rob the bank even only $90\%$ of the time,
or even only $1\%$ of the time. Alexei Miasnikov inquired
whether the halting problem itself admits a black hole, and
it turned out that for one of the standard models of
computability, the answer is yes:
Theorem.([1]) For the Turing machine
model with one-way infinite tapes, there is a set of Turing
machine programs $A$ such that
- $A$ has asymptotic density $1$, so almost every program
is in $A$.
- $A$ is polynomial time decidable.
- The halting problem is polynomial time decidable for
programs in $A$.
Thus, for this model of computation, the halting problem is
decidable with probability $1$. The reason has to do with
the fact that for the one-way infinite tape Turing machine
model, it turns out that almost every Turing machine
program, like Polya's drunken man, falls off the tape
before repeating a state. And this is something that can be
detected in linear time. It follows that with asymptotic
probability one, a Turing machine program computes a finite
set.
[1] J. D. Hamkins, A. Miasnikov, The halting problem is
decidable on a set of asymptotic probability
one, Notre Dame Journal
of Formal Logic, Notre Dame J. Formal Logic 47 (2006),
515–524.
