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.
