Regarding Collatz, ratio of odd steps to even steps? If $b > 1,\ \ b  \ \to_c \ 1$ in $m + n$ steps ($m$ through odd numbers, $n$ through even) then $2^A + b \ \to_c  \ {3^m \over  2^n} 2^A + 1$.  
If we imagine that ${3^m \over 2^n} 2^A + 1 < 2^A + b$, then we obtain the inequality $ {3^m \over 2^n} + 2^{-A} < 1 + b \cdot 2^{-A}$. We could choose $A$ to be some massively huge integer, then $2^{-A}$ would be a massively small number, and it seems that we can conclude ${3^m \over 2^n} < 1$, leading to ${m \over n} < {\ln2 \over \ln3}$, an upper bound on the ratio of odd steps to even steps.
I wrote a program to check if the inequality holds up to $10000000$ and it checks out, unless I'm having hidden overflow errors! Is there any reason why this upper bound ought to be correct, or a better way to check the inequality for larger integers?
Edit: here is a plot of $m/n$ for $2 \le b \le 10000$ versus ${\ln2 \over \ln3}$

 A: I found an explicit proof for the claim - I thought to add it in the question as an edit but I mean it's a final answer anyway,
Suppose $b \ \to_c \ 1 < b$ in $m+n$ steps, $m$ through odd and etc
then $b \ \to_c \ {3^m \over 2^n} b + {O \over 2^n} = 1$ for some integer $O \ge 0$, then we have 
${3^m \over 2^n} b \le {3^m \over 2^n} b + {O \over 2^n} = 1 < b \\ \to {3^m \over 2^n} < 1 \\ \to {m \over n} < {\ln2 \over \ln3}$
A: Let us -what is in your notation $b \to_c 1$ with $b \gt 1$ - denote as
$$ 1 = T(b;B_1,B_2,...,B_N) \tag 1$$  with the exponents $B_k$ indicating the powers of $2$ representing the collected $x/2$-steps between two $3x+1$-steps.
Let us also denote the number of $3x+1$-steps by $N$ and the sum-of-exponents (which are also the number of $x/2$ - steps ) by $S = B_1+B_2+...+B_N$ .
Then we can rearrange in (1) and as well write 
$$ 1 = {3^N \over 2^S} b + T(0;B_1,B_2,...,B_N) \tag 2$$.
Let us now denote ${3^N \over 2^S} = q$ for shortness. then we have, because always $T(0;B_1,...,B_N) \gt 0$ from $ 1 = q \cdot b + T(0;B_1,B_2,...,B_N) $ that $$q = {3^N \over 2^S}\lt \frac 1b \lt 1 \tag 3$$ 

I think the following is superfluous, but because you fiddled with this:
If we introduce some $2^A$ then for all $A>S$ we'll have with valid implicite Collatz-transformations
$$ q \cdot 2^A + 1 = q \cdot (2^A+b) + T(0;B_1,B_2,...,B_N) \tag {4.1}$$
Of course immediately follows 
$$ q \cdot 2^A  + 1 \lt q (2^A+b) \tag {4.2}$$ and also, because $q \lt 1$
$$ q \cdot 2^A  + 1 \lt 1 (2^A+b) = 2^A+b  \tag {4.3}$$ and
$$ q  \lt 1+{b-1\over  2^A}  \tag {4.4}$$
But this last formulation is overly weak, because we have already from (3) that $(q=) {3^N \over 2^S} \lt \frac 1b \lt 1 $ .
Remark: that superfluous manipulations irritated me in catching the sense of your question, but well, I think I've got your intention correctly now
From this we get also your correct result: 
$${3^N \over 2^S} \lt \frac 1b \lt 1  \\ 
   3^N \lt \frac {2^S} b \lt 2^S \\ 
  N \ln3 \lt S \ln2 - \ln b \lt S \ln2 $$ $$ 
  {N \over S}  \lt {\ln2 \over \ln 3 } - {\ln b \over S  \ln3} \lt { \ln2 \over \ln3} \tag 5 
$$ where no additional numerical evidence is needed.

I've looked at some pictures, too. To make some possibly relevant patterns better visible I use only odd $b$. Also often a logarithmic scaling of the$b$-axis makes things clearer in exponential diophantine problems.
Caution-remark: I'm used to letters $a$ and $A$ for this problems instead of your $b$ here and didn't think about this while generating the pictures. So you need to mentally translate the notations, sorry...
So here are the number of 3x+1-steps (=$N$) in the trajectories of $b \to 1$:

Besides the expectable increase of $N$ with increasing $b$ I find it a remarkable observation, that seemingly (at least) 2 patterns are overlaid here: something from bottom upwards and something from a first staircase above the bottom expanding upwards and downwards. Similarly this looks with $S$, the number of even steps:

The pattern-overlay seems even clearer here.
After that the picture for the ratio, corresponding to your own picture but showing a bit more detail/pattern:

Of course, because I've omitted the even numbers in $b$, we also do not have the perfect powers of $2$ in $b$, and thus not the zero-ratios which are visible in your plot.               

After that I looked at the Collatz-modification $5x+1$ . Here we know, that trajectories do not - as in the $3x+1$-version - end in the cycle $1 \to 1 \to \cdots$ - but in at least 3 cycles ($1 \to 3 \to \cdots$, $13 \to 33 \to 83 \to \cdots$ and $17 \to 27 \to 43 \to \cdots$) and very likely diverge to infinity (should we introduce the concept of "the cycle at infinity" here?) and so I computed the ratios for each group of values of $b$ depending on their final characteristic: for the $b$ which enter a cycle, the ratio of $N/S$ before they enter the cycle, and for the diverging $b$ I stopped the trajectories at $x \ge 1e48$ and used $N$ and $S$ taken so far. Here is the picture:

Of course, the diverging $b$ have a ration above $\ln2 / \ln5$ and the converging 
$b$ smaller (except the value $b=5$ which gives a larger ratio because it is smaller than the smallest value of the final cycle (which is $13$))
A: I made a few graphs of this and my even over odd ratio seems to tend passed ln(3)/ln(2).
Even/Odd ratio for n<100000
