Background
In Dungeons and Dragons we (typically) roll $20$ sided dice. In my hypothetical situation on my roll I add a +11 to the outcome and if it's not at least a $20$ then I get to reroll the $20$ sided dice once again.
The question follows: What is the probability of me getting at least a $20$ on my turn (the original role, and the secondary roll if needed).
My logic
We are basically looking for $> 8$ on the roll since a $9$ would meet the criteria with the $+11$ bonus. A normal D20 would have an average of a $10.5$ on any particular role. On the first role I have a $12/20$ chance. I'm not exactly sure why but I feel like I can substuite any number lower than the needed $8$ with the average on the dice of a 10.5 So it would be:
$(8*10.5)$ + ${\sum_{i=9}^{20} i}$ = 258
so the odds of a success should be:
$1 -\frac{20}{258}$ or 92.2%
If this is correct, why? It feels correct but I can't prove to myself why it would be.
Edit
I fixed some silly grammatical and calculation errors.
Best Answer
First, to calculate the chance of getting at least a 20 on a single 1d20+11 roll, you count the number of "successful" possibilities and divide it by the total possibilities. Since you have the +11 bonus, it would suffice to roll an 9. In which case, there are 12 distinct "successful" cases (namely rolling a natural 10, 11, ..., 20) and 20 total cases. So the probability of rolling at least a 20 on a single 1d20+9 roll would simply be $\frac{12}{20} = \frac{3}{5}$.
Now, since you can re-roll a 1d20+11 if the first throw was unsuccessful, we split into multiple "scenarios". The probability of failure on the first throw would be $1-\frac{3}{5} = \frac{2}{5}$. The probability of success and failure of the second throw would similarly be $\frac{3}{5}$ and $\frac{2}{5}$, respectively. Since the first and the second throw are independent, we can multiply probabilities together to get the compound probability. So we have 3 basic cases:
$$\begin{cases} \frac{3}{5}, & \text{if the first throw succeeds} \\ \frac{2}{5}*\frac{3}{5} = \frac{6}{25}, & \text{if the first throw fails, but the second one succeeds}\\ \frac{2}{5}*\frac{2}{5} = \frac{4}{25}, & \text{if both throws fail.} \end{cases}$$
So, to find the total probability of "success" (i.e. at least one of the throws is successful), simply add up the probabilities of the successful cases. $$\frac{3}{5}+\frac{6}{25} = \frac{21}{25} = 84\% \text{ chance of success.}$$