grrrr :steam::steam::steam: .......oooooooooooooooooooooooohhhhhmmmmm....guys

I swear that's the best answer to the problem if everything is random. I'll try to explain what I'm thinking.

It's like this

Those are your possibilities. As the first battle goes up in number the choices available for battles two and three go down in number, forming a square. The lower the number of the first battle, the larger the square of the next two battles will be.

Now put all these possibilities together and what shape do you get? A pyramid formed by stacking the squares on top of each other and organized along the z-axis by the number of the first battle in increasing order of the first battle. THIS BECOMES ALL YOUR POSSIBILITIES, which is important to understand.

So from this pyramid, if we want to look up how many possibilities a given sequence accounts for, then all we need to do is locate the square stack for the given sequence along the z-axis and send its height right down to zero on the z-axis - this forms a cube and gives you all the possibilities that the sequence covers. And what you find is that if you take the stacks to infinity, and converge on the sequences that give you the biggest cube, 1/3:1/3:1/3 will have the biggest cube and cover the most possibilities. I really wish I could model in 3d because this is really neat and it gives you a way to determine probability of winning. So given any finite pyramid we see that using 1/3:1/3:1/3 always yields the probability of

Volume = 1/3*AreaB*H / 2 //Divide by 2 because we have half the area to consider from the pictures.

Ex1: AreaB=100, H=10, volume = 333 1/3 / 2 = 166 2/3, volume of (1/3:1/3:1/3) = 1/3 * 1/3AreaB*1/3H = 37.03

Probability of Winning = 37.03/166 2/3 = 22.21% chance of winning

Someone check the Math.