I came to the same answer as you, but by a different means. If you have a computer, it's easy to do these problems by just looking at all possible rolls. For three rounds, there's 27 possible combinations.deathdonut said:Instead, the question I tried to answer is: What is the chance of dying on the THIRD round assuming you have been unconcious for TWO.
Code:
(X = Strike; - = nothing; O = stabilize)
--> X X X 729 / 8000
--> X X - 810 / 8000
--> X X O 81 / 8000
--> X - X 810 / 8000
--> X - - 900 / 8000
--> X - O 90 / 8000
X O X 81 / 8000
X O - 90 / 8000
X O O 9 / 8000
--> - X X 810 / 8000
--> - X - 900 / 8000
--> - X O 90 / 8000
--> - - X 900 / 8000
--> - - - 1000 / 8000
--> - - O 100 / 8000
- O X 90 / 8000
- O - 100 / 8000
- O O 10 / 8000
O X X 81 / 8000
O X - 90 / 8000
O X O 9 / 8000
O - X 90 / 8000
O - - 100 / 8000
O - O 10 / 8000
O O X 9 / 8000
O O - 10 / 8000
O O O 1 / 8000
729 / 7220 = 10.1%
Which matches what you came up with. You're solution was more elegant. But with this technique, you can extend it to any number of rounds, do any amount of conditional probability you want, and come up with precisely the right answer.