Alice and Bob each have 9 soldiers. There are 4 battle fields. Each player sends a non-negative integer number of soldiers to each battle field, and whoever has strictly more wins that battle field(if they send the same number, neither wins). The player that wins 3 or more battle fields gets a payoff of 1, and in all other cases gets a payoff of 0. What strategy should be played?
I remember this game from a math tournament event, but nobody explained the solution afterwards. I am completely lost with this problem. Here is what I know so far:
Notation: $(a,b,c,d)$, $a+b+c+d=9$ means put $a$ soldiers in battle field $1$, $b$ in field $2$, etc.
- Never allocate all $9$ soldiers onto two fields, since you will never win.
- Never allocate $(0, 1, 1, 7)$, since you can win only if the other plays $(x, 0, 0, <7)$, which by 1 is never optimal.
- Never allocate $(1, 1, 1, 6)$ by the same reasoning as above.
- Let $(a, b, c, d), 0 \leq a \leq b \leq c \leq d, a+b+c+d=9$ be an allocation of troops. Then $a+b+c \leq 6$, and so $(a+1, b+1, c+1, 6-a-b-c)$ beats $(a,b,c,d)$. So there's no allocation that beats all others.
- $(2, 2, 2, 3)$ is only beaten by $(3, 3, 3, 0)$
- $(1, 2, 3, 3)$ is only beaten by $(2, 3, 4, 0), (2, 3, 0, 4)$
- $(3, 3, 3, 0)$ is only beaten by $(4, 4, 0, 1), (4, 0, 4, 1),(0, 4, 4, 1)$
- $(4, 4, 1, 0)$ is only beaten by $(5, 0, 2, 2), (5, 0, 3, 1), (5, 1, 2, 1)$ [and swapping 0 and 5]
All possibilities (not including permutations) are: $$(0, 1, 2, 6), (0, 1, 3, 5), (0, 1, 4, 4), (0, 2, 2, 5), (0, 2, 3, 4), (0, 3, 3, 3)\\ (1, 1, 2, 5), (1, 1, 3, 4), (1, 2, 2, 4), (1, 2, 3, 3)\\ (2, 2, 2, 3)$$