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Posts posted by jasen

  1. I can prove this hipothesis is wrong !

    let say the tipewritter only have 5 hammers and it was set like this :

    letter 2 placed at hammer 1
    letter 3 placed at hammer 2
    letter 5 placed at hammer 3
    letter 1 placed at hammer 4
    letter 4 placed at hammer 5

    our expected word is "12345"

    1st iteration 2 3 5 1 4
    2nd iteration 3 5 4 2 1
    3rd iteration 4 1 2 5 3
    4th iteration 5 4 1 3 2
    5th iteration 2 3 5 1 4  (it is back to 1st iteration)

    so it will never reveal the the expected word.



  2. bonanova, this your puzzle drives me mad

    Only communication can make them survive !
    So I think they must create unseen communication between them,
    I've find a trick for them to communicate.
    They must communicate with time !!

    let say a,b,c,d ..... are the prisoners names in sequence, no need to sort.

    every prisoner must know who will enter the room after him, 
    so 'a' know 'b' will enter the room after 'a'.

    'a' open 50 first boxes (1 to 50), if he find 'b' name there, so he must leave the room before 10 minutes.
    'b' notice that 'a' leave the room before 10 minutes, so he know that his name is at 50 first boxes (1 to 50).
    but if 'a', leave the room after 10 minutes, means that his name is not there so he must open boxes 51 to 100.

    with same way, 'b' give information to 'c'.

    and the probability they all save are 50% !!

  3. I think this strategy is better

    everybody have to memorize each person position

    say :
    n = his position
    x = (name in the opened box) position

    every person open n-th box 
    then exchange the box (with the x-th box)
    repeat this to the next box 50 times, but he have to skip the box if it have been exchanged.
    if it reach the last box, go to first box

  4. can they sort the boxes (by names) they open ?

    if they can :
    first person open 1st box then exchange the box (the 1st box with the n-th name position of the box)
            repeat this to the next box 50 times, but he have to skip the box if it have been exchanged

    2nd person open 2nd box then exchange the box (the 1st box with the n-th name position of the box)
            repeat this to the next box 50 times, but he have to skip the box if it have been exchanged

    all prisoner do the same

  5. put 1 dice at the right side and 1 dice at the left of a ballance scale.
    after several times changing different side of the dices, 
    we can find the normal weight.

    then let the left dice as standard.
    then weight other three dices, each side once.
    if a dice show anomally (get heavier/lighter) when the side changed,it must be cheat dice.
    if we find 1 dice without anomally then it is the true dice.

    if all 3 dices is not true dice, so the left dice which we use as standard is the true dice.

  6. Now I am confused.

    I think I followed the strategy you described yesterday, when you postulated that it would deliver 100% chance of winning.

    But, now your simulation it gives you a 50% chance of winning.


    Did I miss something??

    I was quoting bonanova solution/answer, not my proposed strategy.

    I was working on the strategy that you describe:

    • Bet half my money [rounded to an integer]; or
    • bet the "ceiling" [twice my original money - Current money].

    Testing this in a spread sheet [with only about 100 trials] I am winning about 62% of the time.

    Your strategy is not applicable, you can't bet more money than you have,  if your current money is 50$, your original money is 100$, so 2*100-50 = 150$, which you don't have.

    Now I explain, my proposed strategy.

    If I have 100$, then I bet 50$, if I lose, my money remains 50$
    Now I have 50$, then I bet 25$, if I lose, my money remains 25$
    Now I have 25$, then I bet 12.5$, if I lose, my money remains 12.5$

    this keep happens, the game will never over, to infinitely small amount of money.

    but :

    if I have 100$, then I bet 50$, if I win, my money is 150$
    Now I have 150$, then I bet 50$, if I win, my money is 200$

    game over, and I win.

    so. it seem that I will win 100%.

    But, My proposed strategy although theoretically true, it is not applicable.
    If I'm lucky I will win in a few step, but if not, it will take an infinite time to win.
    then I will give up, an bet rest of my money.

    so I conclude, there is no strategy to win, we only have 50% chance to win.

  7. If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

    Find the sum of all the multiples of 3 or 5 below 1000.

    It is easy to find the answer using computer, but it's more interesting if we can solve it without coding.

  8. In a gambling game, the rules are :

    you bring an amount of money,
    you can bet any amount of money you like,

    the game stop if :
     1. your money have doubled (exactly 2x your first amount of money)
     2. or you lose all of your money. 

    you can't stop in the middle of the game.

    example : if you bring 10 dollars:

    1. If you bet all your money in the first bet, so if you win you get 20 dollars and game over, or if you lose, you lose all your money and game over.
    2  If you have already win 19 dollars, so in the next game you maximum bet is only 1 dollar. 

    what is your strategy to win the game ?

  9. Hidden Content

    Right !

    This case is one of the famous misleading sequence

    For n=1, you get 1 region.
    For n=2, you get 2 regions.
    For n=3, you get 4 regions.
    For n=4, you get 8 regions.
    For n=5, you get 16 regions.

    See a pattern?

    Yes, but if you conjecture that n points produces 2n-1 regions, you would be wrong!

    The correct answer is a little more subtle: 
    it is the sum of the first 5 binomial coefficients for power (n-1):

                  1                        = 1
                1   1                      = 2
              1   2    1                   = 4
            1   3   3     1                = 8
          1   4   6    4     1             = 16 
        1   5   10  10    5     1          = 31
      1   6  15   20   15    6     1       = 57
    1   7  21   35   35   21    7     1    = 99

  10. hint

    Hidden Content

    modified Stern-Brocot Tree to third sequence.
    3rd sequence  34/55   157/254  123/199 212/343  89/144  233/377  144/233  199/322 55/89

    since it use fibbonaci number, which is relatively prime, so the fraction is already in simple form.

    so there is no fraction with 55<b<89 

  11. An observation without an explanation

    Hidden Content

    maybe this have something to do with math golden ratio, 
    which it close to 1.6180339887...

    2/3 = 1.5
    3/5 = 1.6666....
    5/8 = 1,6
    8/13 = 1.623
    144/233 = 1.618055556....
    233/377 = 1.618025751....

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