In the shooting room paradox $x$ people enter a room, one of them rolls two dice and if they both land on 6 everyone in the room is shot and killed. If not, everyone in the room is taken out and let go on their merry way and $10x$ people are then placed in the room and the experiment continues on like this until one group rolls a double six at which point it is complete. The usual question is "Knowing that you are entering the room what is the chance that you will get shot?". It would initially seem like the chance is $1/36$ but over $90\%$ of people that enter the room in any given run of the experiment are actually killed so this is the chance of you getting killed.
Taking an example of the experiment starting with one person: If it ends on the first go $1/1$ people involved have been killed giving $100\%$, if it ends on the second $10/11$ have been killed (the first didn't roll double sixes and was set free) giving approximately $90.91\%$ and $\lim_{n\to \infty} \frac{9\times10^n}{10^{n+1}-1}=0.9$ which is where the answer of $90\%$ comes from. But as this is the absolute minimum proportion of people killed, the actual expected value must be higher than this. If rather than rolling two sixes, the criterion for getting shot was rolling any number at all then the death rate would be $100\%$ and if it was to get any number except 1 on a 100-sided die would we not expect the death rate to be a fair bit higher than $90\%$?
My question is what is the exact expected value? I think it would be the answer to this difficult summation that I have had no luck conquering:
$\frac{1}{36}\sum_{i=0}^\infty\frac{35}{36}^i\times\frac{9\times10^i}{10^{i+1}-1}$ which simplifies a bit to $\frac{9}{36}\sum_{i=0}^\infty\frac{1}{36}^i\times\frac{350^i}{10^{i+1}-1}$
Any help would be greatly appreciated! Thanks :)