Rolling dice depending on evens and odds Roll a standard $6$-sided die. If the result is even, replace it with a roll of two dice; if the new total is even, replace it with a roll of three dice, etc. Continue to roll one more die than previously as long as the total is even. When you finally roll an odd total, you score that many points (i.e. if your last throw is 1,2,2,6, you get 11 points) and the game ends. What is the expected number of points you would get? 
I don't have any significant ideas. So one can definitely notice that the contribution from a single turn is $\frac{1}{6}(1+3+5)$ but already for two turns it becomes complicated, as $9$ can be obtained in more ways than $11$. I expect there might be some clever partition of events to give $m = $ simple function$(m)$, but I can't find it.
Any help appreciated!
 A: Proposition: The expected value for an odd roll using $n\geq2$ dice is $3.5n$.
With this, the expected value of your game is
$$
\frac12\cdot3+\frac1{2^2}\cdot7+\frac1{2^3}\cdot10.5+\cdots+\frac1{2^n}\cdot3.5n+\cdots\\
=\frac{27}{4}
$$
as the probability of an odd roll using any number of dice is well known to be $\frac12$.
Proof: For two dice, the expected value of an odd roll is $7$ by symmetry (3 is as probable as 11, and so on), and since the expected value of two dice without restrictions is also $7$, the expected value for an even throw must also be $7$ by the law of total expectation.
For induction, assume the expected value for an odd roll using $k\geq2$ dice is $3.5k$, and the same for even. Now we throw $k+1$ dice. We look at the expected total value by separating the first die out from the rest.
If the first die is $1$, then the remaining dice must roll even. The expected total value for such a throw is $1+3.5k$, and the probability is $\frac16$.
If the first die is $2$, then the remaining dice must roll odd. The expected total value for this is $2+3.5k$, and the probability is $\frac16$.
Keep going, and we find by the law of total expectation that the expected value of an odd roll with $k+1$ dice is
$$
\frac16(1+2+3+4+5+6+6\cdot 3.5k)=3.5(k+1)
$$
It follows that an even roll must also have expected value $3.5(k+1)$.
A: The probability of getting odd (or even) is clearly $1/2$ for each round. The average odd value for $n$ dice is $n \cdot (1 + 3 + 5) / 3 = 3 n$. So your value for $n$ rounds is the probability of $n - 1$ failed rounds ($2^{- n + 1}$) and one successful round ($1/2$) times the average for the $n$-th round. Sum over all possible $n$:
$\begin{align*}
   \sum_{n \ge 0} \frac{1}{2^{n - 1}} \cdot \frac{1}{2} \cdot 3 n
      &= 3 \sum_{n \ge 0} \frac{n}{2^n} \\
      &= 6
\end{align*}$
For the sum, note that:
$\begin{align*}
   \frac{1}{1 - z}
     &= \sum_{n \ge 0} z^n \\
   z \frac{d}{d z} \frac{1}{1 - z}
     &= \sum_{n \ge 0} n z^n \\
   \frac{z}{(1 - z)^2}
     &=
\end{align*}$
Setting $z = 1/2$ (this is OK, the series converge for $\lvert z \rvert < 1$) gives the cited value.
