[NT] How do you think through this?

[Little_Sticks]

I brute forced it with a program, using variable finite constraints, and using whole numbers only in combinational sequences. From this, if we consider a Tie to be as good as losing then I am 100% certain that you have the best chances if you place your troops accordingly: 1/3:1/3:1/3.

 

[Little_Sticks]

I figured it out. It's a cube. The biggest cube covers the most space, covering the most possibilities.

1/3 * 1/3 * 1/3 > anything else

 

[Chunes]

send 1/2 to both fronts and let the third be forfeit.

Anyone suspecting this strategy would send 1/2 + 1 to a battlefield, 1/2 - 2 to the other, and 1 to the third. And hope for the favorable matchups. I'm pretty sure that spread gives better odds of winning than 1/2 | 1/2 | 0.

And ironically 1/2 | 1/2 | 0 beats the suggestion in the post above 100% of the time. For that I put forward that 1/3 | 1/3 | 1/3 might cover the most possibilities versus random selections, but strategists are rarely random. Going for all three battlefields is a poor strategy that spreads you too thin.

Edit: damn, just realized we're fighting against a random, mindless opponent. Where's the fun in that? :\

 

[forzen]

JustHer,

(1) Did it ever occur to you that one might draw more utility from the debates surrounding infinity than the post itself?

(2) In order to be able to back the currency of your comment up you must have intellectual reserves. Having the proper reserves entails that you must be aware of the difference between the literal as opposed to the figurative definition of infinity, which I propose you lay down. upon laying down this definition/conception, you will then need to demonstrate how the figurative conception can still be infinity and yet allow for fractions of it to exist--i.e. 1/4 of infinity. And so on.

Alright, we know what infinity is, a number that goes on forever. Because you cannot prove what number is the highest number. Anyone can see that by using simple logic; by adding one proves that a number is not the biggest. Like I keep telling people, math is how we describe the working foundation of our reality based on our observation, any abstract concepts are used to fill holes or are not observable. Infinity happens to be one of this concept because numbers can go on and on and on.

Infinity is used to describe a number so large or so small that it no longer matter. You can see this by one of the limit concept:

lim x --> infinity 1/x = 0

It states, that as x starts to become really really big, the fraction 1/x starts to become really really small that it no longer matter so it might as well be zero.

So how do we divide infinity. We'll we can't because it's a number that goes on forever. But, we can take a snapshot of a number going on forever to get a result. Look at this for a second:

n --> infinity: n as it goes to infinity

1/3 (n--> infinity) + 1/3 (n--> infinity) + 1/3 (n--> infinity) = n --> infinity

So lets say we were trying to prove the highest number possible which we will call n. Now, as we know n keeps going and going and going since we simply have to add one to make it bigger. But, the statement above would still be true since if we add a number to one side (infinity), that number would get divided by 1/3 and get added to the three individual pieces on the other side which would still make it true. Of course, given that infinity goes on forever we would need to take a snapshot of a number as it gets infinitely massive to get a result.

If you see any logical fallancy, point them out since this is my own interpretation to try to explain where I'm coming from.

 

[Little_Sticks]

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.

 

[forzen]

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.

I don't know if I understood how you set up the graph, but these are the points I got from doing a (x, y, z) graph.

(3,0,0) (3,0,1) (2,0,1) (2,0,2) (1,0,2) (1,0,3)
(0,3,0) (0,3,1) (0,2,1) (0,2,2) (0,1,2) (0,1,3)
(0,0,0) (0,0,1) (0,0,2) (0,0,3) (3,3,0) (3,3,1)
(2,2,1) (2,2,2) (1,1,2) (1,1,3)

This is the pyramid which states that the army is spread out to ratio of 1/3 | 1/2 | 1/6. 1/3 being at the bottom, 1/2 being at the middle, and 1/6 being on top. I used A = lwh on every cube and added them together to get the total area:

A = (3X3X1) + (2X2X1) + (1X1X1) = 14

But, i'm not going to calculate every deviations in the squares as it change size...that's too much work lol. I'm just trying to see if I understood where you were coming from which I kind of see how you worked out the solution .

 

[goodgrief]

Hey was mine wrong? Because there are a lot of complex systems running here and I cbf understanding them right now but I thought the question was simple.

 

[BlueGray]

I came up with a 2/3 chance of winning with (1/3,1/3,1/3)

This situation seems the easiest to solve for. Chance of winning any given two battles is (1/3 * 1/3)/(1/2) so 2/9 for any given two. Since there are 3 ways to have two battles you get 3 * 2/9 = 6/9 = 2/3.

I'm going to agree that (1/3,1/3,1/3) is the best odds.

There are two variables. Any three dimensional representation can be rotated to provide simply a face. Orders are (x,y,1-x-y) If the opposing force didn't necessarily use all their forces three dimensions would be useful.

 

[Little_Sticks]

Crap well my probability is definitely wrong. And my pyramid idea might actually be a parabolic cone idea. I think I know how to figure out the possibilities with shapes if that is the case.

BlueGray, out of curiousity, did you ask this because you don't know how to think about it or because you want to see how other people think about this problem? Or both?

 

[Daedalus]

Hey folks, I donno if this has already been posted. But in general, i would assume that if the commander splits his forces into 2 huge groups+ one token force he will have a good chance of winning.

(eg...if there are 1000 men, there will be 500 in one army, 499 in another, and 1 in the last)

this would give a high probability of win against an enemy that randomly splits its forces, because we need to win only two battles!

the first army is ALMOST guaranteed a win (in case the opponent has more than 500 in one army, then then its a loss BUT in that event, the second army is a guaranteed win, as the enemy has >= 501 in his first army, resulting in <= 408 in his second one , as he needs at least one for his third army)

so we get 1 guaranteed win with this method + 1 VERY high probability of win for the second army. We don't have to worry about the third, cos 2 out of 3 battles wins the war

what do you guys think?

 

[forzen]

Ok, I found a solution. First, I used 9 sticks to represent the armies.
I got 6 combinations and we'll label them from 1-6 from top to bottom.

1 (1)
111
11111

111 (2)
111
111

11111 (3)
1111


11 (4)
11111
11

1 (5)
1111
1111

111111 (6)
111

Each combination may or may not win because of the rule that if it's a tie, from a combination's perspective, it wins. This created conflict because each may or may not win depending on which point of view you're looking from. For example:

1
111
11111

111
111
111

The two configurations above may or may not win depending on which side you're looking from. But I found that the best configuration that wins majority of the time is

1
1111
1111

which beats 2, 4, and 6 two out of three and gets a win from 1 if looking from this configuration's point of view. Only lost to 3.

 

[Daedalus]

But I found that the best configuration that wins majority of the time is

1
1111
1111

which beats 2, 4, and 6 two out of three and gets a win from 1 if looking from this configuration's point of view. Only lost to 3.

yep!

This is what I found too, in the post right above yours. In my hypothetical example of 1000 men per army,
splitting them into armies having 500, 499 and 1 men each respectively, results in the most number of wins. with at least one guaranteed win, with a high possibility for a secondary win.

 

[forzen]

yep!

This is what I found too, in the post right above yours. In my hypothetical example of 1000 men per army,
splitting them into armies having 500, 499 and 1 men each respectively, results in the most number of wins. with at least one guaranteed win, with a high possibility for a secondary win.

Yes, I read your post and it was the same result that I got .

...........
...................__
............./´¯/'...'/´¯¯`·¸
........../'/.../..../......./¨¯\
........('(...´...´.... ¯~/'...')
.........\.................'...../
..........''...\.......... _.·´
............\..............(

 

[Ming]

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?

 

[Daedalus]

Yes, I read your post and it was the same result that I got .

...........
...................__
............./´¯/'...'/´¯¯`·¸
........../'/.../..../......./¨¯\
........('(...´...´.... ¯~/'...')
.........\.................'...../
..........''...\.......... _.·´
............\..............(

yep, now we have to decide which brave man amongst the soldiers will volunteer for the one-man army

Like Horatio on the bridge.lol



Ps: ive been thinking a bit about our solution, and i think that the larger the number of men in each side, the higher the probability of win, for this solution.

lets look at the example of 1000 men.

split into 500, 499 and 1 men

the Only way the enemy can win..not even a win...a stalemate is if he can have the same setup as ours.

that is, he has to 500, 499,and 1 men in each army respectively. Any other combination he has, will result in a loss or a stalemate for him


say he has 501 in one army, 498 in the other and 1
he wins the first battle, loses the second, and is stalemated in the 3rd

he has 502,497,1
wins first, loses second, stalemate on 3

he has 499,500,1
same, 1 win, 1 loss, 1 stalemate

he has 499, 499,2
loses 2 wins one (in this event we win)


I think its suffice to say that with this setup there is NO WAY for the enemy to defeat us. the best he can hope for, is a stalemate, and even that on very special cases. The larger the forces, the probability of our win/stalemate increases!

therefore this setup is LOSS-PROOF

 

[Little_Sticks]

First consider in 2d. You get the following line of possibilities in 2d.

And we are using ratios so I take the lengths from 0 to 1.

Then consider in 3d. You are essentially collecting a bunch of those two-dimensional lines along 0 to 1 to contain all possible ratios.

Now normalize it.

And the probability for (1/3, 1/3, 1/3) becomes 3/9 => 33.3% chance of losing, so 1-.333 = 66% chance of winning.

The shaded regions are the points where the enemy will win. The two shaded triangles above (1/3, 1/3, 1/3) along the Z-axis are where the enemy will win one battle, but since it also won the battle on the Z-axis, it is winning possibilities. The shaded triangle below (1/3, 1/3, 1/3) along the Z-axis is where the enemy will win both of the other two battles, making it winning possibilities.

I'm pretty sure this is right...someone think it over and respond.

Edit:

Consider (1/2, 1/2, 0),


(1/2 + 1/2 + 1/4 + 1/2 + 1/2 + 1/4) / (Numerator + 1 + 1/2 + 1/2 + 1/4)

2.5/(2.5 + 2.25)

2.5/4.75

1 - 2.5/4.75 = (4.75-2.5) / 4.75 = 2.25 / 4.75 ~= 47.4% chance to win. Is this correct?

 

[durentu]

Anyone suspecting this strategy would send 1/2 + 1 to a battlefield, 1/2 - 2 to the other, and 1 to the third. And hope for the favorable matchups. I'm pretty sure that spread gives better odds of winning than 1/2 | 1/2 | 0.

And ironically 1/2 | 1/2 | 0 beats the suggestion in the post above 100% of the time. For that I put forward that 1/3 | 1/3 | 1/3 might cover the most possibilities versus random selections, but strategists are rarely random. Going for all three battlefields is a poor strategy that spreads you too thin.

Edit: damn, just realized we're fighting against a random, mindless opponent. Where's the fun in that? :\

The thing about war is that there are way too many variables to make a detailed analysis. True, there's fun in figuring it out, but in practical terms, the solution would require too narrow conditions for it to be really applicable.

But really, the "Art of War" covers a lot of the variables one would generally be confronted with. Even a really good pep talk could swing the odds.

Perhaps the army/battle analogy should be removed and replaced with positive/negative pixels or atoms. This would be much more applicable to nano-warfare. specifically nano-defense.

edit:

now that I think of it, I think there were probability studies done with viruses and bacteria in the body. If one were interested enough, I'm sure there's a paper to draw inspiration from.

(The English plural of "virus" is "viruses"[1]. It is NOT "viri" or "virii". - learned something new)

 

[fragrance]

The opposing general uses no strategy and distributes his troops randomly. What is the probability that you win the war?

Formally, it'd be 0.25-0.25-0.50 against his army which is probably 0.33-0.33-0.33, so I'd lose two battlefields. In that case, I'd override your command and split my team into 0.5-0.5 and send them to two battlefields, preferably those at the sides, then I'll close in on his 0.33 center force. He can't win a two-front-war.