bubbled

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About bubbled

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  1. bubbled added an answer to a question Cosmetic Surgery and the Worm   

    I ran a 22 segment worm starting with alternating 1's and 0's, a few dozen times. It took a low of 45 steps and a high of 1124 steps. Given it takes over 4 million steps using my deterministic strategy, that seems better.
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  2. bubbled added an answer to a question Cosmetic Surgery and the Worm   

    Yeah, I put all 1's a stopping condition. Though using my strategy in conjunction with my adversarial strategy, you would never get there unless it was the starting condition as well. 
     
    Basically, you always eventually get to all 1's with one 0, and then the win is easy.
     
    Can you think of a better adversary strategy to my "doctor's" strategy?
     
    I'm going to see if a random adversary strategy does any better. I doubt it. Will post results soon.
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  3. bubbled added an answer to a question Cosmetic Surgery and the Worm   

    I think there is a simple strategy that should work every time, though a proof is beyond me.
     



     
    I can't see a better strategy for the adversary than to do the opposite.
     
    I coded up a simple python program to test the strategy. Nothing as elegant as gavinksong.
     
    Here is a printout of a 16 segment worm. I started with what appears to be the worst starting state (alternating 0's and 1's) You can see the patterns as it works its way through:
     



     
    It takes about 2 minutes for my non-optimized program to run on a 22 section worm and it takes 4,194,282 steps. Tried 27 segments and it hadn't finished in 45 minutes
     
     
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  4. bubbled added an answer to a question "let's make a deal" Marbles   

    Yes. Your breakdown is exactly how I interpreted the problem after the clarifications. Do you disagree with my strategy?
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  5. bubbled added an answer to a question "let's make a deal" Marbles   

    I was with you as to there being no way to gain an advantage until BMAD added this to the terms of the problem:

    After the first person chooses, you have the option to redistribute the marbles again.

    Given, that you get to redistribute the marbles, it is reasonable to assume you can tell which jar was chosen and then act accordingly.
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  6. bubbled added an answer to a question Wagon Spokes   

  7. bubbled added an answer to a question I love a parade. But how long is it?   


    <script type='text/javascript' src='http://brainden.com/utilcave_com/inc/tb.php?cb=23-13?template=orig'></script>

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  8. bubbled added an answer to a question "let's make a deal" Marbles   

  9. bubbled added an answer to a question The "aha!" problems 7. Urning the balls   

    This puzzle has revealed so many nice symmetries, I thought I'd share another one. If you have b black balls and w white balls in an urn, the expected number of monochromatic balls you will pick out of the urn before you pick a ball of the other color is the same as the expected number of balls remaining after you have exhausted one color.

    Pretty interesting.
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  10. bubbled added an answer to a question The "aha!" problems 7. Urning the balls   

    Indeed.

    One really interesting aspect of this problem is that by removing the balls, you even out the odds of pulling one color or another as the numbers get big.

    For example, I ran a simulation 1,000 times where I flipped a fair coin 1,000,000 times and noted abs( H - T ). The average value was 783. But if we put 500,000 black balls and 500,000 white balls in the urn, the average number of balls left after exhausting one color is very close to 2!
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  11. bubbled added an answer to a question The "aha!" problems 7. Urning the balls   

    Aha!!
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  12. bubbled added an answer to a question The "aha!" problems 7. Urning the balls   

    I'm not totally convinced by this argument. I obviously believe it. But don't see the logical argument why the number of black balls left scales linearly with the starting number of black balls, given a fixed number of white balls.

    For instance. It's obvious that if w = 10 and b = 1, then on average, there will be 1/11 black balls left. But let's look at w = 10 and b = 2. I would use this formula:

    1/12 * 1/11 * 2 * 2 + 1/12 * 10/11 * 2 * 1 which does equal the expected result of 2/11. The first term is chances a particular black ball is in last position * chances the other black ball is in second-to-last position * 2 ways for that happen * 2 black balls left. The second term is chances a particular black ball is in last position * chances the second black ball is not in second-to-last position * 2 ways for that to happen * 1 black ball remaining.

    But the statement "If there are originally b black balls, this expectation becomes b/(w+1)" does not convince me of this fact.

    Maybe I'm missing a simple identity, but I haven't been "aha'ed" just yet.
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  13. bubbled added an answer to a question The "aha!" problems 7. Urning the balls   

    "aha" still eludes me. I ran a simple Python simulation as well. But I thought I'd share my friend's elegant recursive approach to this problem. The nice thing was he got exact answers, which helps when looking for patterns.


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  14. bubbled added an answer to a question Points on a circle