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  1. #61
    Senior Member Little_Sticks's Avatar
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    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.

  2. #62
    Senior Member Little_Sticks's Avatar
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    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

  3. #63
    Senior Member Chunes's Avatar
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    Quote Originally Posted by durentu View Post
    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? :\
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  4. #64
    Senior Member forzen's Avatar
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    Quote Originally Posted by Provoker View Post
    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.
    This post grammatical errors had been intentionally left uncorrected.

  5. #65
    Senior Member Little_Sticks's Avatar
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    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.

  6. #66
    Senior Member forzen's Avatar
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    Quote Originally Posted by Little_Sticks View Post
    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 .
    This post grammatical errors had been intentionally left uncorrected.

  7. #67
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    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.

  8. #68
    Senior Member BlueGray's Avatar
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    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.
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  9. #69
    Senior Member Little_Sticks's Avatar
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    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?

  10. #70
    Senior Member Daedalus's Avatar
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    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?
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