There are 3 battles, right? Thus to win, you'll have to have 2 wins.
Is it needed to participate in all 3 battles? If not, then just split 1/2, 1/2; because then it's a definite both win. Because if as many or more = win.
If needed, then we'll have to go with that. Let my 1st, 2nd and 3rd battle be x,y,z respectively. Let the enemies 1st, 2nd and 3rd battle be a,b,c respectively.
x must be > or = to a
y must be > or = to b
z and c can be of any value, because once you've got two wins, you've won.
x+y+z = a+b+c (since they have equal numbers right?)
since z and c can be of ANY value to win; then all it needs x and y > a and b respectively.
For the highest chance of x and y > a and b respectively, c has to be the smallest value possible. Because it can not be zero (because you HAVE to send someone!) then you have to choose 1 to be z.
Let the sum of x+y+z = n. If we take z out (which the best value is 1) then we are left with x+y = n - 1.
For x and y to win against a, b and c (remember that for z to beat c, all they have to do is to contribute more than 1) we make the following formula..
(x+y)/2 (because of 2 matches) > or = (a+b+c)/3 (because of 3 matches a,b,c)
(n-1)/2 > or = n/3
Since n can't be negative (you can't have negative people!), that means the inequality sign can't change. Thus you get...
The Probability of (n-1)/2 > or = n/3 when n > or = to 3 (remember that x,y and z each have to be at least 1, so thus added together they have to be at least 3).
If you apply it there, then as long as n > or = to 3; then n can be of any infinite value; because it apply to all values of n.
Thus as long as z is equal to 1, and either x or y is one half because for n/2 (because
-1 represents z, remember? So if you take z away, you're left with n/2) has the highest chance of win over n/3, as long as n > or = 3.
Thus the highest chance of winning this stupid battle is by having z battle as 1, and either x or y battle having 1/2.
Stupid question. Did they really need infinite numbers?