Two generals are fighting a five front battle. Each general has 1 unit of army, which he divides into five separate armies that he sends to the five fronts. If one general sends more army to a front than the other, then the other army instantly gives up and the front is taken over by that general. If they send exactly equal amounts of army to a front, then the leaders of the armies play a game of Backgammon to determine the owner of the front. Whichever general takes over at least three of the fronts wins the battle (the other general gives up).

Neither general has any spies, and they split their armies simultaneously, so each general would like you to find him a probabilistic strategy that guarantees that he will not lose the battle more than half of the time.

Background: I found the three front version of this problem on the xkcd forums. I don't know if generalizations of the problem to larger numbers of fronts have been considered before.

As a starting point, here is a class of strategies that can never work:

Suppose that you have a distribution such that the expected number of fronts you win is at least $\frac52$ regardless of the other player's strategy. Then you will lose against a random permutation of the splitting $(0, 0, \frac3{10}, \frac3{10}, \frac25)$ more than half of the time.

Proof: Assume your strategy has probability $f(x)$ of sending at least $x$ units of army to a front. Then expecting to win $\frac52$ fronts implies that for any $a + b + c+d+e = 1$, we have $f(a)+f(b)+f(c)+f(d)+f(e) \ge \frac52$, and combining this with the fact that the average amount of army sent to a front is $\frac15$ it isn't hard to show that the number of armies sent to any particular front must be uniformly distributed between $0$ and $\frac25$. Thus, the number of fronts that you expect to win against a random permutation of $(0, 0, \frac3{10}, \frac3{10}, \frac25)$ is exactly $\frac52$.

It isn't hard to calculate that the probability of winning the battle is given by (E(fronts won) - Pr(win even number of fronts) - 1)/2. So, we just need to check that the chance of winning an even number of fronts is more than half. To start off with, we will always lose the front he sends $\frac25$ armies to, and win the fronts he send $0$ armies to. Thus, the probability of winning two fronts must be at least the combined probability of winning three or four fronts (since you never win one just one front), so you will lose more than half of the time as long as you have a nonzero chance of winning four fronts, and you will have a nonzero chance of winning four fronts as long as you send at least $\frac3{10}$ units of army to two of the fronts. But the chance of sending at least $\frac3{10}$ to a front is $f(\frac3{10}) = \frac14$, and if your strategy never sends $\frac3{10}$ to two fronts at a time then the chance of sending at least $\frac3{10}$ to a particular front would be at most $\frac15$, which is impossible.