I need help understanding the solution to the following puzzle:
Puzzle: There is a "triangular" duel between the three shooters. Everyone shoots one by one, can shoot once and any person he wants. Smith always hits the target (100%) and shoots last. Brown hits the target 80% of the time and shoots second. Jones is the worst, hitting the target only 50% of the time and he shoots first. Who should Jones shoot in order to increase his chance of surviving?
Answer: Jones should shoot in the air. The worst shooter has the best chance of surviving in the triangular duel, which is Jones. After him goes Smith who never misses the target. Since Smith and Brown will be shooting each other when their turn comes, Jones must shoot no one until one of his enemies dies.
After that he shoots at his enemy, again having an advantage. First, it's easier to calculate the probability of surviving for Smith. In the duel with Brown he shoots first with a probability of 1/2. In this case he kills Brown. Brown also shoots first with a probability of 1/2.
Smith survives with a probability of 1/5. Thus, Smith will live over Brown with a probability of 1/2 + 1/2 * 1/5 = 3/5. With a probability of 1/2, Smith survives the duel with Jones. All in all: the probability to survive for Smith is 3/5 * 1/2 = 3/10.
The probability for Brown to survive the duel with Smith is 2/5.
Then Jones shoots Brown. In the first round, Brown has a probability of 1/2 * 4/5 = 4/10 to win. In the second round, he has a probability of 1/2 * 1/5 * 1/2 * 4/5 = 4/100.
Thus, Brown has a chance to survive Jones: 4/10 + 4/100 + 4/1000 + 4/10000 = 0.4444(4) = 4/9.
The probability of Brown to survive both of his opponents is equal to 2/5 (over Smith) * 4/9 (over Jones) = 8/45. The probability of Jones to survive = 1 - 3/10 - 8/45 = 47/90.
In bold are the calculations I'm confused about and don't know how to arrive at. Can someone please explain it? Thank you.
UPDATE: They shoot one by one in the same order until one person survives.