Alice and Bob have $N_A$ and $N_B$ warriors under their command, numbered $1$~$N_A$ and $1$~$N_B$ respectively. Alice has $1$ fighting power at her disposal, and Bob has $b$ ($b\gt 0$). Before the game, they privately distribute their power between their warriors. When the game begins, both send their warrior #1 to a 1-to-1 fight. If a warrior with power $x$ fights one with power $y$, the former wins with probability $\frac{x}{x+y}$, and the latter with $\frac{y}{x+y}$. If #1 is defeated, #2 is sent to continue the next round of fight, so on and so forth until one side has all of their warriors defeated and loses the game. There's a small twist though: after a warrior defeats an $x$ power opponent, his power will increase by $cx$ ($c\geq 0$).

Clearly, a player's strategy is their method of distributing fighting power.

Question 1: given $N_A=N_B$ and $b=1$, does Alice have a strategy that can guarantee her no less than 50% winning probability, no matter how Bob plays?

Question 2: is there always a dominant strategy for Alice?

Note: question 2 is a stronger claim, and implies question 1. For $c=1$, I know the answers for both questions are yes, because in that case the game is essentially the gambler's ruin problem, so every strategy gives the same winning probability for Alice. I suspect but can't prove an equal distribution is dominant for $c=0$ and give-it-all-to-one is dominant for $c\gt1$. A simulation for $N_A=N_B=2$, $b=1$ and $c=1/2$ suggests that giving 0.420341... to the first warrior guarantees 50% winning probability for Alice.

Alice and Bob have $N_A$ and $N_B$ warriors under their command, numbered $1$~$N_A$ and $1$~$N_B$ respectively. Alice has $1$ fighting power at her disposal, and Bob has $b$ ($b\gt 0$). Before the game, they privately distribute their power between their warriors. When the game begins, both send their warrior #1 to a 1-to-1 fight. If a warrior with power $x$ fights one with power $y$, the former wins with probability $\frac{x}{x+y}$, and the latter with $\frac{y}{x+y}$. If #1 is defeated, #2 is sent to continue the next round of fight, so on and so forth until one side has all of their warriors defeated and loses the game. There's a small twist though: after a warrior defeats an $x$ power opponent, his power will increase by $cx$ ($c\geq 0$).

Clearly, a player's strategy is their method of distributing fighting power.

Question 1: if Alice and Bob are equally matched ($N_A=N_B$ and $b=1$), does Alice have a strategy that can guarantee her no less than 50% winning probability, no matter how Bob plays?

Question 2: is there a dominant strategy for Alice?

Note: question 2 is a stronger claim, and implies question 1. For $c=1$, I know the answer is yes for both questions, because in that case the game is essentially the gambler's ruin problem, so every strategy gives the same winning probability for Alice. I suspect but can't prove an equal distribution is dominant for $c=0$ and give-it-all-to-one is dominant for $c\gt1$. A simulation for $N_A=N_B=2$, $b=1$ and $c=1/2$ suggests that giving 0.420341... to the first warrior guarantees 50% winning probability for Alice.

Alice and Bob have $N_A$ and $N_B$ warriors under their command, numbered $1$~$N_A$ and $1$~$N_B$ respectively. Alice has $1$ fighting power at her disposal, and Bob has $b$ ($b\gt 0$). Before the game, they privately distribute their power between their warriors. When the game begins, both send their warrior #1 to a 1-to-1 fight. If a warrior with power $x$ fights one with power $y$, the former wins with probability $\frac{x}{x+y}$, and the latter with $\frac{y}{x+y}$. If #1 is defeated, #2 is sent to continue the next round of fight, so on and so forth until one side has all of their warriors defeated and loses the game. There's a small twist though: after a warrior defeats an $x$ power opponent, his power will increase by $cx$ ($c\geq 0$).

Clearly, a player's strategy is their method of distributing fighting power.

Question: is there always a dominant strategy for Alice?

Note: for $c=1$, I know the answer is yes, because in that case the game is essentially the gambler's ruin problem, so every strategy gives the same winning probability for Alice. I suspect but can't prove an equal distribution is dominant for $c=0$ and give-it-all-to-one is dominant for $c\gt1$. A simulation for $N_A=N_B=2$, $b=1$ and $c=1/2$ suggests that giving 0.420341... to the first warrior dominates all other strategies.

Alice and Bob have $N_A$ and $N_B$ warriors under their command, numbered $1$~$N_A$ and $1$~$N_B$ respectively. Alice has $1$ fighting power at her disposal, and Bob has $b$ ($b\gt 0$). Before the game, they privately distribute their power between their warriors. When the game begins, both send their warrior #1 to a 1-to-1 fight. If a warrior with power $x$ fights one with power $y$, the former wins with probability $\frac{x}{x+y}$, and the latter with $\frac{y}{x+y}$. If #1 is defeated, #2 is sent to continue the next round of fight, so on and so forth until one side has all of their warriors defeated and loses the game. There's a small twist though: after a warrior defeats an $x$ power opponent, his power will increase by $cx$ ($c\geq 0$).

Clearly, a player's strategy is their method of distributing fighting power.

Question 1: given $N_A=N_B$ and $b=1$, does Alice have a strategy that can guarantee her no less than 50% winning probability, no matter how Bob plays?

Question 2: is there always a dominant strategy for Alice?

Note: question 2 is a stronger claim, and implies question 1. For $c=1$, I know the answers for both questions are yes, because in that case the game is essentially the gambler's ruin problem, so every strategy gives the same winning probability for Alice. I suspect but can't prove an equal distribution is dominant for $c=0$ and give-it-all-to-one is dominant for $c\gt1$. A simulation for $N_A=N_B=2$, $b=1$ and $c=1/2$ suggests that giving 0.420341... to the first warrior guarantees 50% winning probability for Alice.