Bob and Alice are managing a team of warriors, Alice ’s team consist of warriors with strength 3, 4, 6, 7 and Bob has 4 warriors of strength 9,8,4,2 respectively.
If a warrior with strength x fights another warrior with strength y, warrior with strength x wins with probability $x/(x+y)$ and there is no draw (i.e. the other one dies), and the victorious gains confidence and after the match his strength becomes x+y.
The tournament consists of Bob and Alice picking up warriors to fight against one another, one at a time. Alice is going to pick first for each fight (she picks randomly) and winner of the tournament is the person who has at least one warrior left at the end.
What is the optimal strategy for Bob To maximise his probability of winning the game given Alice chooses warriors randomly?
I have a gut feeling that there is no optimal strategy , and all strategies lead to the same probability , which is $(sum(Bob_{Warriors})/(sum(Bob_{Warriors})+sum(Alice_{Warriors}))$
But i cant seem to find a way to prove is formally