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 Topic: Math & logic problems :-)

 (Read 12753 times)
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  • Re: Math & logic problems :-)
     Reply #60 - August 16, 2010, 06:11 PM
    Quote from: Prince Spinoza on August 16, 2010, 06:09 PM
    Cut the rope into 2 pieces. One 25m, one 50m. Tie the 25m to the top hook. Make a loop in the other end. Loop the other piece through this. Tie the other end(s). Now you have a 50m rope. Climb down. Cut the knot. Tie oone end to the hook.  Pull the other out of the loop. You have another 5om of rope to climb down to the ground.


    You got this without searching on the Internet?! Bravo! Afro
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    Relativism | Cognitive dissonance | Cognitive biases | Memeplex | Consequentialism
  • Re: Math & logic problems :-)
     Reply #61 - August 16, 2010, 06:14 PM
    I tried searching the internet, couldn't find it anywhere lol, thought u just made it up. Where did you get it from?
  • Re: Math & logic problems :-)
     Reply #62 - August 16, 2010, 06:36 PM
    Quote from: zoomi on August 16, 2010, 05:40 PM
    here's a classic, I'm sure many of you have seen it before, so if you have, let someone who hasn't figure it out!

    1
    11
    21
    1211
    111221
    312211
    13112221
    1113213211

    whats the next line...Huh?


    Oh!
    I never saw it.

    But I think I solved it.
    The next should be:

    31131211131221

    Each line after the first "describes" the previous line in terms of consecutive characters.
    Do not look directly at the operational end of the device.
  • Re: Math & logic problems :-)
     Reply #63 - August 16, 2010, 06:37 PM
    You can solve a question like that, but then you can't do a simple google. I am skeptical, Mr Prince Spinoza, very skeptical you figured it out! Tongue Smiley

    http://www.lmgtfy.com/?q=rope+hook+knife+riddle
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  • Re: Math & logic problems :-)
     Reply #64 - August 16, 2010, 06:44 PM
    Quote from: Prince Spinoza on August 16, 2010, 06:03 PM
    Another famous one -

    There are 3 doors. Behind 1 is a thousand dollars. Behind the other two is an empty bucket. You have to select one door. Once you do this, one of the other two doors will be opened to reveal one of the empty buckets. You now have the chance to stick with your original choice, or to switch to the other unopened door.

    What should you do, and why.

    Switch.

    Because the first one you picked has a 1/3 chance of being the winning one, while the last one left has 2/3 chance.
    Do not look directly at the operational end of the device.
  • Re: Math & logic problems :-)
     Reply #65 - August 16, 2010, 06:53 PM
    Quote from: zoomi on August 16, 2010, 05:40 PM
    here's a classic, I'm sure many of you have seen it before, so if you have, let someone who hasn't figure it out!

    1
    11
    21
    1211
    111221
    312211
    13112221
    1113213211

    whats the next line...Huh?



    Googled after giving up (after 15min!), good one! Afro
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  • Re: Math & logic problems :-)
     Reply #66 - August 16, 2010, 07:27 PM
    what do you do for a living tlaloc? you've been through the drill.
  • Re: Math & logic problems :-)
     Reply #67 - August 16, 2010, 07:32 PM
    Quote from: s12345 on August 16, 2010, 07:27 PM
    what do you do for a living tlaloc? you've been through the drill.

    I like to pretend to be a programmer ^_^
    Do not look directly at the operational end of the device.
  • Re: Math & logic problems :-)
     Reply #68 - August 16, 2010, 07:37 PM
    This is a bit hard to solve and demonstrate, but it's fun:

    On a single 6-sided die, what is the average number of consecutive throws you need to finally obtain either a 1 or a 2?
    (Or, the general version of the problem: on a single N-sided die, what is the average number of consecutive throws you need to finally obtain a number from 1 to M?)
    ((Or, the even more general version: on average, how many consecutive times you need to "run" some action so that the event with probability p happens?))
    Do not look directly at the operational end of the device.
  • Re: Math & logic problems :-)
     Reply #69 - August 16, 2010, 08:05 PM
    Quote from: Tlaloc on August 16, 2010, 07:37 PM
    This is a bit hard to solve and demonstrate, but it's fun:

    On a single 6-sided die, what is the average number of consecutive throws you need to finally obtain either a 1 or a 2?
    (Or, the general version of the problem: on a single N-sided die, what is the average number of consecutive throws you need to finally obtain a number from 1 to M?)
    ((Or, the even more general version: on average, how many consecutive times you need to "run" some action so that the event with probability p happens?))


    i think its 3 throws.

    for a probability of any M on a N sided die would be N/M.

    1/p runs isn't it.

    or have i screwed up again Smiley
  • Re: Math & logic problems :-)
     Reply #70 - August 16, 2010, 08:08 PM
    Quote from: s12345 on August 16, 2010, 08:05 PM
    i think its 3 throws.

    for a probability of any M on a N sided die would be N/M.

    1/p runs isn't it.

    or have i screwed up again Smiley

    Correct.

    Now demonstrate/calculate why Grin
    Do not look directly at the operational end of the device.
  • Re: Math & logic problems :-)
     Reply #71 - August 16, 2010, 08:10 PM
    i did the whole agp series summation for expected length , but i think there should be a simpler and more elgant argument than the shit i did.
  • Re: Math & logic problems :-)
     Reply #72 - August 16, 2010, 08:17 PM
    I think the series of np/(1-p)^(n-1) is the only way to solve it.

    I cannot see any other way.
    Do not look directly at the operational end of the device.
  • Re: Math & logic problems :-)
     Reply #73 - August 16, 2010, 08:20 PM
    just seems that way. the result is so beautifully simple that there must be an elegant argument to justify it without going into a raw expectation computation.
  • Re: Math & logic problems :-)
     Reply #74 - August 16, 2010, 08:23 PM
    There is one similar but more complex problem I always wanted to solve but I never bothered yet.

    It's like this:

    Take a sequence of N "bits", all set to 0.
    At each "run", EACH bit has a probability p of being switched to the opposite value (1 if 0, 0 if 1... which, in information technology terms, means XORing it with 1)
    What is the average number of runs needed for the whole sequence of N bits to become 1?

    Solving this is very interesting because it can be at the core of calculating how many "average generations" some genetic code needs in order to change into something specific by simple mutation and duplication.
    Do not look directly at the operational end of the device.
  • Re: Math & logic problems :-)
     Reply #75 - August 16, 2010, 08:36 PM
    Quote from: Tlaloc on August 16, 2010, 07:37 PM
    On a single 6-sided die, what is the average number of consecutive throws you need to finally obtain either a 1 or a 2?


    So am I wrong to think it is 3 because the theoretical chance is 1/3?
    Quote from: Tlaloc on August 16, 2010, 07:37 PM
    ((Or, the even more general version: on average, how many consecutive times you need to "run" some action so that the event with probability p happens?))


    Again, I'd say the theoretical value and the number of times it must be repeated before, in theory, it would occur. Is this not the correct way to think statistically?
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  • Re: Math & logic problems :-)
     Reply #76 - August 16, 2010, 08:41 PM
    Quote from: Tlaloc on August 16, 2010, 08:23 PM
    There is one similar but more complex problem I always wanted to solve but I never bothered yet.

    It's like this:

    Take a sequence of N "bits", all set to 0.
    At each "run", EACH bit has a probability p of being switched to the opposite value (1 if 0, 0 if 1... which, in information technology terms, means XORing it with 1)
    What is the average number of runs needed for the whole sequence of N bits to become 1?


    That is one complicated but interesting question indeed!
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  • Re: Math & logic problems :-)
     Reply #77 - August 16, 2010, 08:45 PM
    Quote from: HighOctane on August 16, 2010, 08:36 PM
    So am I wrong to think it is 3 because the theoretical chance is 1/3?
    Again, I'd say the theoretical value and the number of times it must be repeated before, in theory, it would occur. Is this not the correct way to think statistically?

    The final formula IS simple like that... but the reason is not simple.

    To compute the average throws, you need to calculate the chance of it happening in exactly 1 time, multiply it by one... then sum it to the chance of it happening in 2 times multiplied by 2, summed to the chance of it happening in 3 times multiplied by 3... and so on to infinity.

    So, basically, you have to find the solution to a series.

    Such solution ends up being an elegant 1/p
    Do not look directly at the operational end of the device.
  • Re: Math & logic problems :-)
     Reply #78 - August 16, 2010, 08:50 PM
    Interesting, good to know, thanks. Afro
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  • Re: Math & logic problems :-)
     Reply #79 - August 16, 2010, 08:51 PM
    Quote from: HighOctane on August 16, 2010, 08:41 PM
    That is one complicated but interesting question indeed!

    Yes.
    It's also solved by a series. But a more complex one than the consecutive dice throw.

    Interestingly, the dice throw problem is a subset of this one.
    Because it's this same one but with just 1 bit ;P
    Do not look directly at the operational end of the device.
  • Re: Math & logic problems :-)
     Reply #80 - August 16, 2010, 08:57 PM
    I see! I've never given series that much interest. I've been more at awe of the awesome powers of matrices in solving unknowns! (Well, recently anyway).
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  • Re: Math & logic problems :-)
     Reply #81 - August 16, 2010, 09:04 PM
    NEW PROBLEM  grin12

    This is rated 16 & Astaghfirullah! So look away kids!

    Only a hedge funds trader would ask a question like this  bunny

    A woman would like to have sex with three men. There are only two condoms. How can the condoms be used in such a way that all three men can safely have sex with the lady using only the two condoms?

    PS: Don't try the solution at home!
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  • Re: Math & logic problems :-)
     Reply #82 - August 16, 2010, 10:42 PM
    I'll call them person A, B and C and condoms 1 and 2.

    Person A has sex with the woman using both condoms, with 1 inside 2.
    Person B has sex with the woman using only the outer condom 2.

    Now the insides of condoms 1 and 2 are filthy.
    The outside of condom 1 is still clean.
    The outside of condom 2 is "contaminated" by the vagina. Smiley

    Person C has sex with the woman with condom 1 turned inside out inside condom 2.

    If I were person C, I would run down and buy a new condom.  Afro
    Bukhari 62:142 - Narrated Anas bin Malik:
       The Prophet used to pass by (have sexual relation with) all his wives in one night, and at that time he had nine wives.
  • Re: Math & logic problems :-)
     Reply #83 - August 16, 2010, 10:49 PM
    Here is a little trick that shouldn't be too difficult to figure out.

    I can instantly see that
    41*39 = 1599

    and
    52*48 = 2496

    and
    63*57 = 3591

    Can someone give me a brief mathematical explanation on how I do it?
    Bukhari 62:142 - Narrated Anas bin Malik:
       The Prophet used to pass by (have sexual relation with) all his wives in one night, and at that time he had nine wives.
  • Re: Math & logic problems :-)
     Reply #84 - August 17, 2010, 12:20 AM
    Quote from: TheLastKnight on August 16, 2010, 10:49 PM
    Here is a little trick that shouldn't be too difficult to figure out.

    I can instantly see that
    41*39 = 1599

    and
    52*48 = 2496

    and
    63*57 = 3591

    Can someone give me a brief mathematical explanation on how I do it?

    (a+b)*(a-b) = a^2 - b^2

    So:
    (40+1)*(40-1) = 1600-1
    (50+2)*(50-2) = 2500-4
    (60+3)*(60-6) = 3600-9
    Do not look directly at the operational end of the device.
  • Re: Math & logic problems :-)
     Reply #85 - August 17, 2010, 12:25 AM
    I'm guessing but I see a pattern:

    ab X cd = (a*c+c),d*10+d*b

    example:

    41*39 = 4x3+3 > 15, 9x10+9 >99 > 15,99 > 1599

    interesting...
  • Re: Math & logic problems :-)
     Reply #86 - August 17, 2010, 08:12 AM
    Quote from: Tlaloc on August 17, 2010, 12:20 AM
    (a+b)*(a-b) = a^2 - b^2

    So:
    (40+1)*(40-1) = 1600-1
    (50+2)*(50-2) = 2500-4
    (60+3)*(60-63) = 3600-9

    Tlaloc, youre a frikkin genius
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  • Re: Math & logic problems :-)
     Reply #87 - August 17, 2010, 10:43 AM
    Quote from: IsLame on August 17, 2010, 08:12 AM
    Tlaloc, youre a frikkin genius

    True :S
    Too bad I am the laziest person I know, or I'd conquer the world or something ^_^
    Do not look directly at the operational end of the device.
  • Re: Math & logic problems :-)
     Reply #88 - August 17, 2010, 10:45 AM
    i am not so sure about the genius bit. the line between genius and near genius is very thin Tongue i am on the wrong side of it
  • Re: Math & logic problems :-)
     Reply #89 - August 17, 2010, 11:23 AM
    Well done, Tlaloc, that was the explanation I wanted.

    Here, have a cookie!  Smiley
    Bukhari 62:142 - Narrated Anas bin Malik:
       The Prophet used to pass by (have sexual relation with) all his wives in one night, and at that time he had nine wives.
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