Explain the odd/even inequality in the heights of numbers under the Collatz $(3x+1)/2$ transformation? My kid asked me a question and I'm finding it hard to answer: if every number under repeated application of the Collatz transformation1  eventually reaches $1$, then it must be true for both even and for odd numbers. Why, then, for numbers of a given "height" (the number of divisions before reaching $1$,  A00666 in the OEIS) are there so many more even numbers than odd numbers? If you halve an even number then $50$% of the time the result should be odd. And if there are more even numbers than odd ones at every given height, won't you run out of even numbers more quickly?
In fact the ratio of even:odd numbers of a given height is approximately $3:1$ (Pari/GP code below).2  My answer wasn't satisfactory: there are an infinite amount of both odd and even numbers, so you won't run out of either; and transforming an odd number can only lead to a subset of even numbers ($x: x= 3k+1$) so the other ones don't really count. 
Is there a more intuitive way to explain it?
1  $C(x)= \begin{cases}
\frac{x}{2}&\text{when x is even; and}\\
{3x+1}&\text{when x is odd.}
\end{cases}$
2  heights(n)= if(n==0,return([1]),  n==1, return([2]), n==2, return([4]), my(h=heights(n-1)); my(l=List()); for(x=1,#h, listput(l,h[x]*2); if(h[x]%3==2, listput(l,(h[x]*2-1)/3)));return(Vec(l))) \\ Returns a vector of numbers with a given Collatz height
 A: Note that for an odd $n$ the number $3n+1$ is always an even number. 
Therefore even if you start with a set of equal odds and evens, after you first iteration you have more even numbers than odd numbers.
For example $$\{3,5,6,8\} \to \{10,16,3,4\}\to \{5,8,10,2\}\to  \{16,4,5,1\}$$
To remedy this pattern, you may redefine your function as $$ C(x)= \begin{cases}
\frac{x}{2}&\text{when x is even; and}\\
{(3x+1)/2}&\text{when x is odd.}
\end{cases}$$
A: The reason why one would expect more even numbers to appear at a given height rather than odd numbers is the following: the number of even numbers of height $n$ should be equivalent to the number of odd numbers of height$<n$. Why is that? Let the height of odd $g$ be $k<n$. Hence, $g\cdot 2^{n-k}$ will have height $n$.
And under most circumstances, we expect the number of odd numbers of height less than $n$ to be higher than the number of odd numbers of height exactly $n$, especially for large $n$.
I do believe your discovery of a $3:1$ ratio to be coincidence.
