How to use a -coin so a TM can decide an undecidable language in polynomial time? In "Computational complexity- A modern approach" book (page 117) for the lemma 7.12 (following) the author mentioned that if the ρ is efficiently computable ρ-coin cannot give probabilistic algorithm a new power. Then he  states: "The exercises show that if ρ is not efficiently computable, then a ρ-coin can indeed provide additional power."

Lemma 7.12: A coin with Pr[Heads] = ρ can be simulated by a PTM in expected time O(1) provided the ith bit of ρ is computable in poly(i) time.

The exercise that supposed to show, if ρ is not efficiently computable, then a ρ-coin can indeed provide additional power is the following:

Exercise: Describe a real number ρ such that given a random coin that comes up “Heads” with probability ρ, a Turing machine can decide an undecidable language in polynomial time. 

Now, how can we show that if ρ is not efficiently computable, then a ρ-coin can indeed provide an additional power for a probabilistic algorithm to decide some undeniable language in polynomial time?

*I removed the hint because it made me more confused but you can find it at the end of the book.

I strongly believe that this is a research level problem and proper to be discussed in theoretical computer science society because this is a high level question about computational complexity and this question has not been discussed anywhere else.
I'll take as given the existence of Chaitin's constant $\Omega\in[0,1]$, and that knowing its first $k$ bits is equivalent to be able to decide the halting problem for all Turning machines of size up to $k$:
https://en.wikipedia.org/wiki/Chaitin%27s_constant

Given access to an $\Omega$-biased coin, one can use Chernoff bounds to compute $\hat\Omega_k$, which agrees with $\Omega$ up to $k$ bits (i.e., $|\Omega-\hat\Omega_k|&lt;1/2^{k+1}$), by sampling the coin $O(2^{2k})$ times. Now this is exponential in $k$ and is actually tight, so how does this square with the problem statement:
"Describe a real number $\rho$ such that given a random coin that comes up “Heads” with probability $\rho$, a Turing machine can decide an undecidable language in polynomial time" ?

The trick is to use a very inefficient encoding for the undecidable language.
Fix a universal encoding (the same one used to define $\Omega$) and consider the language $L$, which consists of all Turing machine descriptions $&lt;M&gt;$, such that $M$ halts on all inputs. Clearly, $L$ is undecidable. Now order $L$ in lexicographic order where $&lt;M_i&gt;$ is the $i$th word, and define $L'$ to consist of the words $&lt;M_i'&gt;$, where $M_i'$ is a TM equivalent to $M_i$ (the two halt on and accept the same set of inputs), but $M_i'$ contains an additional $2^i$ dummy states.

Thus, to determine whether a word $x$ belongs to the undecidable language $L'$, one only needs to know $O(\log|x|)$ bits of $\Omega$, which is feasible in expected poly$(|x|)$ time via the argument above.