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  1. #71
    Senior Member forzen's Avatar
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    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.
    This post grammatical errors had been intentionally left uncorrected.

  2. #72
    Senior Member Daedalus's Avatar
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    Quote Originally Posted by forzen View Post
    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.
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  3. #73
    Senior Member forzen's Avatar
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    Quote Originally Posted by Starsiege View Post
    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 .

    ...........
    ...................__
    ............./´¯/'...'/´¯¯`·¸
    ........../'/.../..../......./¨¯\
    ........('(...´...´.... ¯~/'...')
    .........\.................'...../
    ..........''...\.......... _.·´
    ............\..............(
    This post grammatical errors had been intentionally left uncorrected.

  4. #74
    Senior Member Ming's Avatar
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    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?

  5. #75
    Senior Member Daedalus's Avatar
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    Quote Originally Posted by forzen View Post
    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
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  6. #76
    Senior Member Little_Sticks's Avatar
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    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?
    Last edited by Little_Sticks; 05-17-2010 at 03:49 PM. Reason: Oopps. Simple math error.

  7. #77
    Senior Member durentu's Avatar
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    Quote Originally Posted by Chunes View Post
    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)
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  8. #78
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    Quote Originally Posted by BlueGray View Post
    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.

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