Looking for help calculating expectation of a “modified” geometric distribution - 3 bit signals

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hey everyone I was hoping I could get a hint about how to solve the following question - it’s a previous years test question so I only have the final answer

The Question:
3 bit signals are transmitted through a noisy signal that flips every bit $(1 \to 0 \ ,0 \to 1)$ with a probability of 1/3, independent of the other bits. What’s the expectation of the number of signals transmitted wholly transmitted before the first one with a flipped bit

My Thought process is this: I wanted to define two indicator series: $X =$ the number of bits unflipped. $$X = \sum_{i=1}^{\infty} X_i \quad X_i = \begin{cases} 1 & \text{the ith bit hasn’t flipped nor have the j<i bits.} \\ 0 & \text{otherwise}\end{cases}$$ and $Y=$ the number of signals that have been transmitted - what I need to find. $$Y = \left\lfloor \frac{X}{3} \right\rfloor$$ $$E(Y) = E(E(Y|X)) = E(E(\lfloor X/3 \rfloor |X))$$ I hit a snag trying to calculate this. is this the best approach or even a correct one? if it is how do I deal with the resulting sum? how would you do it instead? $$\sum_{i=1}^{N} \left\lfloor \frac{i}{3} \right\rfloor \frac{1}{3} \left(\frac{2}{3}\right)^{i-1}$$ I know that once I find this sum the question is easy since I know $E(X)$ from $X \sim \text{Geom}(1/3)$

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The probability of a signal getting transmitted correctly is $(2/3)^3 = 8/27$. What is asked can be calculated by thinking of it as a geometric random variable with $p = 1-8/27 = 19/27$. Thus, the expected value of the number of signals transmitted correctly is $(1-p)/p = 8/19$.

If you want to do it your way, then note that the sum should instead be

$\begin{align*} \frac{1}{3}\sum_{i=1}^\infty \left\lfloor \frac{i}{3}\right\rfloor\left(\frac23\right)^i &= \frac{1}{3}\sum_{n=0}^\infty \sum_{k=0}^2 \left\lfloor\frac{3n+k}{3}\right\rfloor\left(\frac{2}{3}\right)^{3n+k}\\ &= \frac{1}{3}\sum_{n=0}^\infty n\left(\frac{2}{3}\right)^{3n}\left(1+\frac{2}{3}+\left(\frac{2}{3}\right)^2\right)\\ &= \frac{19}{27}\sum_{n=0}^\infty n\left(\frac{8}{27}\right)^n\\ &= \frac{8}{19}. \end{align*}$

You can evaluate the last sum by differentiating the geometric series.