voider

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About voider

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  1. voider added an answer to a question Which Chair you will choose ?   


    Referring to the website, not this particular thread: (Maybe in the wrong section of the forum, but it is a reply.)

    Firstly I agree on the value of collaboration and openness to creativity. But I believe in the context of logic puzzles there is usually a distinction between "something I haven't thought about" and "something that makes no sense, and is based on assumptions that are difficult to intelligently comprehend". Obviously I must have some pride in what I believe to be logical or at least reasonable thinking (even if it is inaccurate) because it affects me when others can't see this distinction, or get it the wrong way round. In fact, this is the first thing I discovered on brainden, other than that it had puzzles and problems on it. Of 200+ comments on the Epimenides Paradox, I dare say 20% or less showed accurate comprehension of its meaning and solution. I discussed this with rookie and proposed a way to improve people's thinking, not just define an answer, a rare opportunity. It didn't come together. I know that generally you can't change people nor the way they think, and that's why my prior post is just an expression of my observations. If it is assumed that nothing can be done about it, and no one has tried, then what does it say about an appropriate environment on brainden? Is it appropriate that most people don't "get" one of the most famous paradoxes, doesn't it virtually encourage the notion that "it doesn't matter what's true, as long as others agree with you", or even "it doesn't matter what's true, as long as you agree with yourself [, and have fun in the process]" (deliberate hint of Hedonism, it's hard to initiate a discussion on that level) ? I would imagine a problem-solving context to do a bit better than that.

    Can you express whatever you think about a logic puzzle on brainden? Yes
    Should you be able to express it without second thought, and expect it to be accepted for what it is? I don't know.
    Do all expressions have value to the "system"? I don't believe so. Spam, redundancy, etc.

    Although not directly relevant, I will point out that the ideal logic problem solver would probably read problems, solve alone (never give up) and never come to a forum. But if I'm communicating any sense to you, you'll see that I think the worth of the community with respect to its social and objective values (if any) is a conundrum. You can't have everything, and I'm observing what brainden doesn't have.
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  2. voider added an answer to a question Which Chair you will choose ?   


    Personally I find it bizarre. I've noted that the "signal to noise ratio" (if signal means correct/true/good reasoning/answer/process, noise means bad/incorrect/false) on brainden is worse than most ad hoc forums or messages, even for the trivially easy puzzles. One of these forms is where people give answers that don't mean a thing to anyone else; the words and logic is practically gibberish in English. How does this happen? Even for most people to acknowledge a correct logical answer seems to be impossible here. If this place lacks common sense, you have to wonder what "communal" value there is given the objectives of contributing here.

    In this case, the fact that we have completely different solutions means most of us are completely wrong, and yet believe we are all on the right track.
    I have no problem accepting I could be wrong, but I'm the only one who has given a proof, and I have implicitly disproved all other answers. But in my experience, proofs and disproofs mean nothing to most people.
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  3. voider added an answer to a question Which Chair you will choose ?   

    Got to say, it's __________ that five of us have completely different solutions, and no working in common.
    Someone find the right word for me
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  4. voider added an answer to a question Which Chair you will choose ?   

    The probability of four same color (thus first chair winning):
    8/8 * 3/7 * 2/6 * 1/5 = 1/35. Same answer results from using combinations.
    There are 70 (ordered) permutations of the 8 stamps, each with equal chance.
    There are 16 (ordered) permutations of 4 of the 8 stamps, not with equal chance.

    If this doesn't occur, then first chair cannot know the answer.

    Second chair now knows that first chair does not see: YYXXXX
    Specifically it leaves 68 permutations with equal chance.
    Then, if he sees XXYYXX he wins. As in the first case, there are two permutations that satisfy this.
    So he wins this way with 1/34 chance.
    I believe he learns nothing else.

    Third chair now knows that second chair does not see: XXYYXX
    This eliminates 2 possibilities, leaving 66 with equal chance.
    If he sees XXXXYY he wins, 2/66 chance.
    Also if he sees XXYY?? then he wins knowing he has XY (since the others haven't won yet).
    This could happen as BBWW**** or WWBB**** where **** is permutations of BBWW. 2 * 6 = 12 ways
    However XXYY?? intersects with XX??XX at XXYYXX so there are 8 ways left.
    So if he didn't win by XXXXYY he wins this with 8/64.
    Equivalent overall: 2/66 + 64/66 (8/64) === 10/66 = 5/33
    I believe he learns nothing else.

    First chair:
    He's survived this far, so there are 56 possibilities left. YYXX?? is not possible.
    He applies same thing as third chair, he wins if he sees ??XXYY where he must have XY. Still 12 possibilities, minus 4 intersections.
    Wins with chance 8/54 = 4/27.
    I can't see him learning anything else.

    Second chair:
    48 possibilities left, YYXX?? and ??XXYY are excluded.
    If he sees XX??YY he must have XY and he wins. 8 ways => 8/48 = 1/6 chance of winning like this.
    If he doesn't win like that, he knows the XX, YY, XY pairs aren't there (no one sees them). XXYYXY and XXYYXX and XXYYYY aren't there, it must mean XX YY isn't there. What about XX XY? (XX XY YY, XX XY XX can't be). One mixed pair doesn't exist, this means there are two or three mixed pairs.
    Thus if he sees XX??MM or MM??XX where MM = mixed pair, then he knows he has XY.
    XXMMMM has 8 forms: 2x for swapping X with Y, 4x for two MM pairs as XY or YX. No more multipliers because the rest must be YY.
    MMMMXX has 8 ways also.
    He wins with 16/40 chance now = 2/5
    Overall: 8/48 + 40/48 (16/40) = 24/48 = 1/2 chance.

    Third chair: 24 left.
    So I think there are 2 or 3 mixed pairs. Second chair would have won if one of 1st or 3rd pair was not mixed, therefore they must both be mixed.
    I believe possibilities left are MMMMMM or MMXXMM. This would mewan third must have XY no matter what.
    Can check this by seeing how many possibilities MMMMMM and MMXXMM form:
    MMXXMM: 2x for XY or YX on third pair, 2x for XY/YX on 1st pair, 2x for identify of X, 1x for the rest must be YY. 8 ways.
    MMMMMM: 8x for three pairs XY/YX. Remaining is also a pair, 2x for that. 16 ways.
    Totals 24 so correct.

    Patterns:
    1 XXXX??, 2 ways
    2 XX??XX, 2 ways
    3 ??XXXX, 2 ways
    3 XXYY??, 8 new ways
    1 ??XXYY, 8 new ways
    2 XX??YY, 8 new ways
    2 XXMMMM or MMMMXX, 16 ways
    3 MMXXMM or MMMMMM, 24 ways

    By intuition or otherwise, the order of events is independent of the ways of winning (must be better way to explain). Anyhow, you can just add up the number of ways.
    First chair: 10 ways
    Second chair: 26 ways
    Third chair: 34 ways

    So the logical choice is third chair, with 34/70 chance of winning, assuming the other two students are not (color)blind and are equally logical.
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  5. voider added an answer to a question Which Chair you will choose ?   

    I assume she will randomly choose the stamps because e.g. if you chose the first chair + the four stamps you see are the same color. So you want to maximise your chances.
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  6. voider added an answer to a question A Game of Probabilities   

    You can't say STOP after they've already won.
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  7. voider added an answer to a question A Game of Probabilities   

    Consider the decision tree.
    You can calculate the probability that you will win, when you predetermine on which round you will say STOP (if you survive until that round).
    The first round probability is 4/12=0.33
    Second is 0.397979...
    This increases, until some round to say STOP where the probability of winning will begin to drop.
    Obviously the "general solution" is to say STOP on the round that produces the overall highest probability of winning relative to the beginning of the decision tree.
    My solution would be a computation, rather than a calculation. There are multiple ways of presenting it, but none of them would look nice.
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  8. voider added an answer to a question Impossible Paper   

    It looks strange but it's too easy. Might be harder if you have to construct it in your mind with your eyes closed.
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  9. voider added an answer to a question   

    133 is correct, = 7C2 * 6 + 7


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  10. voider added a post in a topic One Up Me   

    Sounds like a way to earn reputation... if you're into that...
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  11. voider added an answer to a question   

    abortoperation
    d?????????????

    day in monday
    izj

    sday in tuesday
    oizq

    day in days
    iaq

    Does the alphabet/number map change after using a letter?
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  12. voider added an answer to a question   

    Oops correction:

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  13. voider added an answer to a question   

  14. voider added an answer to a question   

    Boolean algebra: 1 + 1 = 1
    Galois field 2: 1 + 1 = 0
    Subspaces: 1 subspace + 1 subspace = 1 subspace
    Sand: 1 pile + 1 pile = 1 pile
    Speed of light: 1 c + 1 c = 1 c

    Networks: power of 1 component + power of 1 component < power of (1 component + 1 component)
    Problems: 1 small problem + 1 small problem = RAGE
    Memorisation: time taken to learn something twice as long
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  15. voider added an answer to a question   

    I've looked at a few patterns, none of which seem to fit perfectly, but my intuition definitely favours D and rejects A and B overall.

    The more obvious observations are the near-symmetry along one or both diagonals, that (3, 2) is symmetrical, (3, 1) is a rotation of (3, 2), that each block has 3 of each colour, that the columns of the first row blocks have all colours.


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