The "expected value" in a heuristic argument about the Collatz conjecture

In a popular heuristic argument in favor of the Collatz conjecture, one calculates a kind of "geometric expected value" for the ratio $C(x)/x$, where $x$ is odd, and

\begin{align*} C(x) := \frac{3x+1}{2^{\upsilon_2 (3x+1) }} \end{align*}

is next odd number after $x$ in the sequence obtained by repeatedly applying the Collatz function

\begin{align*} T(n) := \begin{cases} n/2 &\text{if $n$ is even,}\\ 3n+1 &\text{if $n$ is odd} \end{cases} \end{align*}

to $x$. The argument is that the quantity

\begin{align*} \frac{C(x)}{x} = \frac{3+\frac{1}{x}}{2^{\upsilon_2 (3x+1) }} \approx \frac{3}{2^{\upsilon_2 (3x+1)}} \end{align*} will equal $3/2$ with probability $1/2$, will equal $3/4$ with probability $1/4$, and will - generally - equal $3/2^k$ with probability $1/2^k$, and so the "expected value" of $C(x)/x$ is calculated as the following infinite product, which converges to $3/4 < 1$:

\begin{align*} \prod_{k=1}^\infty \left(\frac{3}{2^k}\right)^{\frac{1}{2^k}}. \end{align*}

My question: Why don't we calculate the expected value in the "usual" (ie. arithmetic, as opposed to geometric) way, ie. as

\begin{align*} \mathbb{E}\left(\frac{C(x)}{x}\right) = \sum \hspace{0.1cm} \text{"outcome"} \cdot \text{"probability of this outcome"} = \sum_{k=1}^\infty \frac{3}{2^k} \cdot \frac{1}{2^k}? \end{align*}

If we did this, the expected value would be different, namely equal to $1$.

• A "geometric" approach may be more attractive since it helps explain why the Collatz Conjecture seems to converge to 1 and it takes into account the incremental consequences of $n/2$. The value 3/4 more clearly explains why on average the Conjecture converges, where as the value 1 does not give that sort of insight. Commented Aug 25, 2017 at 14:48

Also, when using ratios and geometric arguments, there is the useful property that fixed size components such as $+1$ shrink to zero as you move into greater numbers, while they can accumulate when using sums.