Probabilistic halting problem

I'm a physics and math student working through Nielsen & Chuang's text on quantum computation and information. I don't have much experience in CS theory, so some of these exercises are confusing to me:

Suppose we number the probabilistic Turing machines and define the probabilistic halting function $h_p(x)$ to be 1 if machine $x$ halts on input of $x$ with probability at least 1/2 and 0 if machine $x$ halts on input of $x$ with probability less than 1/2. Show that there is no probabilistic Turing machine which can output $h_p(x)$ with probability of correctness strictly greater than 1/2 for all $x$.

I'm sure we're supposed to assume that there is a probabilistic Turing machine that does this, and then show that we can solve the deterministic halting problem using this machine, thus obtaining a contradiction. I don't have a clue how to go from this probabilistic Turing machine to a deterministic one, though.

Let $A$ be a randomized procedure which outputs $h_p$ with success probability larger than $1/2$.

Consider the following code, with input $x$:

• Let $y$ be the machine corresponding to machine $x$ running on itself as input.

• If $A(y) = 1$, go into an infinite loop, otherwise halt.

Denote this program by $B$.
What happens if we run $B$ on itself? Let us consider two cases. In both cases, let $y$ be the machine corresponding to $B$ running on itself.

Case 1: $B$ halts on $B$ with probability at least $1/2$. In this case, $h_p(y) = 1$, and so with probability more than $1/2$, $A(y) = 1$. Therefore $B$ goes into an infinite loop with probability more than $1/2$, contradiction.

Case 2: $B$ halts on $B$ with probability smaller than $1/2$. In this case, $h_p(y) = 0$, and so with probability more than $1/2$, $A(y) = 0$. Therefore $B$ halts with probability more than $1/2$, contradiction.

(This generalizes the usual proof of uncomputability of the halting problem via diagonalization.)

StackExchange Computer Science Q#112095, answer score: 3