Here is one way of interpreting your question. The main theorem of my joint paper:

• Joel David Hamkins and Alexei Miasnikov, The halting problem is decidable on a set of asymptotic probability one, Notre Dame J. Formal Logic 47 (2006), no. 4, 515--524. blog post

is that for some of the standard models of computation, the halting problem is decidable with probability one. Specifically, we prove that for the usual one-way-infinite Turing machine model, there is a set $A$ of Turing machine programs such that almost every program is in $A$, in the sense that the proportion of all $n$-state programs in $A$ goes to $1$ as $n$ goes to infinity, and furthermore $A$ is decidable and the halting problem is decidable for programs in $A$. So this is a sense in which the halting problem is decidable with probability one.

The argument is sensitive, however, to the computational model, and for some other models of computation, the best currently known is that we can decide the halting problem on a set of probability $\frac 1{e^2}$, which is about 13.5%.

Here is one way of interpreting your question. In my joint paper:

• Joel David Hamkins and Alexei Miasnikov, The halting problem is decidable on a set of asymptotic probability one, Notre Dame J. Formal Logic 47 (2006), no. 4, 515--524. blog post,

the main theorem is that for some of the standard models of computation, the halting problem is decidable with probability one. Specifically, we prove that for the usual one-way-infinite Turing machine model, there is a set $A$ of Turing machine programs, such that:

• Almost every program is in $A$, in the sense that the proportion of all $n$-state programs in $A$ goes to $1$ as $n$ goes to infinity;
• it is decidable whether a given program is in $A$; and
• the halting problem is decidable for programs in $A$.

So this is a sense in which the halting problem is decidable with probability one.

The argument is sensitive, however, to the computational model, and for some other models of computation, the best currently known is that we can decide the halting problem on a set of probability $\frac 1{e^2}$, which is about 13.5%.

Here is one way of interpreting your question. The main theorem of my joint paper:

• Joel David Hamkins and Alexei Miasnikov, The halting problem is decidable on a set of asymptotic probability one, Notre Dame J. Formal Logic 47 (2006), no. 4, 515--524. blog post

is that for some of the standard models of computation, the halting problem is decidable with probability one. Specifically, we prove that for the usual one-way-infinite Turing machine model, there is a set $A$ of Turing machine programs such that almost every program is in $A$, in the sense that the proportion of all $n$-state programs in $A$ goes to $1$ as $n$ goes to infinty, and furthermore $A$ is decidable and the halting problem is decidable for programs in $A$. So this is a sense in which the halting problem is decidable with probability one.

The argument is sensitive, however, to the computational model, and for some other models of computation, the best currently known is that we can decide the halting problem on a set of probability $\frac 1{e^2}$, which is about 13.5%.

Here is one way of interpreting your question. The main theorem of my joint paper:

• Joel David Hamkins and Alexei Miasnikov, The halting problem is decidable on a set of asymptotic probability one, Notre Dame J. Formal Logic 47 (2006), no. 4, 515--524. blog post

is that for some of the standard models of computation, the halting problem is decidable with probability one. Specifically, we prove that for the usual one-way-infinite Turing machine model, there is a set $A$ of Turing machine programs such that almost every program is in $A$, in the sense that the proportion of all $n$-state programs in $A$ goes to $1$ as $n$ goes to infinity, and furthermore $A$ is decidable and the halting problem is decidable for programs in $A$. So this is a sense in which the halting problem is decidable with probability one.

The argument is sensitive, however, to the computational model, and for some other models of computation, the best currently known is that we can decide the halting problem on a set of probability $\frac 1{e^2}$, which is about 13.5%.