Jump to content
BrainDen.com - Brain Teasers

jasen

Members
  • Posts

    204
  • Joined

  • Last visited

  • Days Won

    3

Posts posted by jasen

  1. after a week long journey abroad, now I'm active again.

    @phil1882: The solution does not require a computer. I do not see well encrypting decrypting by hand.

     All you need: scissors, paper, pencil.

    After thinking again, and using the clue harrey give us (scisors, pen, and paper),

    Yes it is possible to prove it without computer, but it takes many steps.

    1. You write your solution on a answering paper.
    2. Put it with back side up on the floor, and ask your friend (who need prove) to numbering it. Better numbering it like (horz, vert) format
    3. Cut it to smaller peaces, every squere (with a number) 1 peace.
    4. Rearrange it with back side up, then turn over/show the numbers in question in the right place. (means you are not cheating)
    5. now take all the squares in first row, suffle it, turn then over secreatly, and show them to your friend, that all number from 1 to 9 are there.
    back to step 4 and do the same with other rows (vert and horz). and other 3x3 squares.

  2. Suppose the leading bullet has 1/k chance of being the fastest.  Then we will not get complete annihilation.
    So far, so good. ....

     

    After rethinking this, I understand the doubter opinion.

    We have to avoid using "first bullet fastest"
    Although first bullet is not the fastest, it it still possible that the first bullet escape

    1 4 2 3  -> in this arrangement  3 (first bullet) is not the fastest, but it escape.

    -----

    we better explain it in another way.

    with n bullets, if the fastest bullet is not the first, there is (n-1)/n cases that we are sure there will be a collision. n-2 bullets will remains.  an so on...

  3. The interesting fact is : Whatever number of bullets fired, the highest probability is 2 bullets left.


    initial  2 bullets  1000000 tries
     P of 0 bullets anhilated =  .500683
     P of 2 bullets anhilated =  .499317

    initial  4 bullets  1000000 tries
     P of 0 bullets anhilated =  .041601
     P of 2 bullets anhilated =  .582647
     P of 4 bullets anhilated =  .375752

    initial  6 bullets  1000000 tries
     P of 0 bullets anhilated =  .001354
     P of 2 bullets anhilated =  .076568
     P of 4 bullets anhilated =  .609111
     P of 6 bullets anhilated =  .312967

    initial  8 bullets  1000000 tries
     P of 0 bullets anhilated =  .000029
     P of 2 bullets anhilated =  .00346
     P of 4 bullets anhilated =  .104317
     P of 6 bullets anhilated =  .619309
     P of 8 bullets anhilated =  .272885

    initial  10 bullets  1000000 tries
     P of 0 bullets anhilated =  0
     P of 2 bullets anhilated =  .000075
     P of 4 bullets anhilated =  .006025
     P of 6 bullets anhilated =  .127299
     P of 8 bullets anhilated =  .620703
     P of 10 bullets anhilated =  .245898

    initial  12 bullets  1000000 tries
     P of 0 bullets anhilated =  0
     P of 2 bullets anhilated =  .000002
     P of 4 bullets anhilated =  .000186
     P of 6 bullets anhilated =  .008569
     P of 8 bullets anhilated =  .147078
     P of 10 bullets anhilated =  .618606
     P of 12 bullets anhilated =  .225559

    initial  14 bullets  1000000 tries
     P of 0 bullets anhilated =  0
     P of 2 bullets anhilated =  0
     P of 4 bullets anhilated =  .000002
     P of 6 bullets anhilated =  .000258
     P of 8 bullets anhilated =  .011055
     P of 10 bullets anhilated =  .164173
     P of 12 bullets anhilated =  .614903
     P of 14 bullets anhilated =  .209609

    initial  16 bullets  1000000 tries
     P of 0 bullets anhilated =  0
     P of 2 bullets anhilated =  0
     P of 4 bullets anhilated =  0
     P of 6 bullets anhilated =  .000008
     P of 8 bullets anhilated =  .000366
     P of 10 bullets anhilated =  .013871
     P of 12 bullets anhilated =  .178564
     P of 14 bullets anhilated =  .610811
     P of 16 bullets anhilated =  .19638

    initial  18 bullets  1000000 tries
     P of 0 bullets anhilated =  0
     P of 2 bullets anhilated =  0
     P of 4 bullets anhilated =  0
     P of 6 bullets anhilated =  0
     P of 8 bullets anhilated =  .000005
     P of 10 bullets anhilated =  .000545
     P of 12 bullets anhilated =  .016474
     P of 14 bullets anhilated =  .191395
     P of 16 bullets anhilated =  .606078
     P of 18 bullets anhilated =  .185503

    initial  20 bullets  1000000 tries
     P of 0 bullets anhilated =  0
     P of 2 bullets anhilated =  0
     P of 4 bullets anhilated =  0
     P of 6 bullets anhilated =  0
     P of 8 bullets anhilated =  .000001
     P of 10 bullets anhilated =  .000012
     P of 12 bullets anhilated =  .000665
     P of 14 bullets anhilated =  .018998
     P of 16 bullets anhilated =  .203064
     P of 18 bullets anhilated =  .601149
     P of 20 bullets anhilated =  .176111

     

  4.  

    Hidden Content

    Hidden Content

    Right !

    You can create a compass from a battery, copper coil, a nail, and thread.

    1. take a battery, and coils from the clock.
    2. take a nail from wall. (to hang a clock, we need a nail).
    3. get thread from your cloth.
    4. create magnet. http://www.wikihow.com/Create-a-Magnet-With-a-Wire-and-a-Nail 
    5. using right hand rule to find south and east. https://en.wikipedia.org/wiki/Right-hand_rule
    6. Find center ballance, and hang the magnet using thread.

    now you know direction.

     

  5.  

    Hidden Content

    Right

    You can create a compass from a battery, copper coil, a nail, and thread.

    1. take a battery, and coils from the clock.
    2. take a nail from wall. (to hang a clock, we need a nail).
    3. get thread from your cloth.
    4. create magnet. http://www.wikihow.com/Create-a-Magnet-With-a-Wire-and-a-Nail 
    5. using right hand rule to find south and north. https://en.wikipedia.org/wiki/Right-hand_rule
    6. Find center ballance, and hang the magnet using thread.

    now you know direction.

  6. Doesnt Sun consider as  a star?

    Well, it's nighttime, so you shouldn't see the Sun.
    The Sun is a star indeed, and it didn't occur to me the moon is not a star. Supposedly you could deduce the cardinals from the hour and the moon position?

    In the night you can't use sun as guidance.

    Unfortunately no astronomer between the prisoners,
    nobody could deduce the cardinals from the hour and the moon position.
    You see arround, wheather there are tools you can use.
    But you only can find electric lamp, hanging photo, and the working clock.

    1st clue : use component from the clock.

  7. you are a prisoner in a huge jail. All prisoner will be executed in the morning next day. Fortunately No guard there. Suddenly in midnight a prisoner can break his cell. and take all keys. But he will not free and let others escape with him, except there is someone can tell  directions (south, east, north or west), because wrong directions means recaptured. Nobody can read star for directions. You see a clock on the wall ticking, and your execution time get closer.

    What should you do to know the right direction?

  8. Should be:

     

    Hidden Content

    This is a subject in cryptography

    A don't want to send his key to B.
    B also don't want to send his key to A.
    But they want to communicate in insecure Internet connection.
    So they have to use an encryption method which work like this.
    Only commutative encryption method can answer this problem.

  9. P(1 Black) = 0.99 100 
    P(2 Black) = 0.98 100
    P(3 Black) = 0.97 100 
    P(4 Black) = 0.96 100 
    P(5 Black) = 0.95 100 
    .
    .
    P(99 Black) = 0.01 100 

    ----------------------------------------------------- +
    P(with 1 or some black) =  0.572....


    P (all white) = 1
    P (all white) : P(some black) = 1 : 0.572

    Conclusion : do not take the bet !!

  10. There are 2 cases

    case 1. Someone want to fool you, he just put one black
    case 2. Nobody try to fool, maybe there are no black,1 or more than one black.

    in case 1 :
    you should not take the bet.
    there is 35% probability that in 100 tries you cannot find the black one.

    in case 2 :
    I still confuse how to solve this case.

  11. Yes, I can agree with that logic. Thanks.

    In creating a question in An IQ test, the creator must make sure that there is no double interpreatation to the question (the question can be answered differently)

    example if you ask :
    1,2,....

    a. 3
    b. 4
    c. 5

    this is a bad question, 
    because there is a double interpretation there 

    you usually answer a.3 
    but b.4 is also correct, because : 20 = 1,   21 = 2,   22 = 4

    so to avoid this, one of the correct answer must be removed from the question.

    -------

    in the question above you may think the logic that I over is questionable.
    But it also acceptable, so if the question maker do not allow my logic, so he must
    prevent this kind of logic appear, so he must create another option that is also not a duplicate.

    but since my logic, and pavan logic come to one option, so this is still acceptable.

    So, if we have found a logic in an IQ test question, just use it.

  12. I think:

     

    Hidden Content

    Right answer, but I'm sure some of us still wonder why.

    Let's cek the combinations
    note (x,y) means : bullet at  xth hole in gun1 and bullet at yth hole in gun2
    since gun1 is shot 1st so if x<= y then means 1st player win.
    so (1,1) or (2,2) means 1st player win.

    for 2 holes and 1 hole guns
    (1,1) 1st win
    (2,1) 2nd win

    for 3 holes and 2 holes guns
    (1,1) 1st win (1,2) 1st win
    (2,1) 2nd win (2,2) 1st win
    (3,1) 2nd win (3,2) 2nd win

    for 4 holes and 3 holes guns
    (1,1) 1st win (1,2) 1st win (1,3) 1st win
    (2,1) 2nd win (2,2) 1st win (2,3) 1st win
    (3,1) 2nd win (3,2) 2nd win (3,3) 1st win
    (4,1) 2nd win (4,2) 2nd win (4,3) 2nd win
     
    for 5 holes and 4 holes guns
    (1,1) 1st win (1,2) 1st win (1,3) 1st win (1,4) 1st win
    (2,1) 2nd win (2,2) 1st win (2,3) 1st win (2,4) 1st win
    (3,1) 2nd win (3,2) 2nd win (3,3) 1st win (3,4) 1st win
    (4,1) 2nd win (4,2) 2nd win (4,3) 2nd win (4,4) 1st win
    (5,1) 2nd win (5,2) 2nd win (5,3) 2nd win (5,4) 2nd win


     and so on....

    if we pay attention at the combinations above, they create pattern of 2 triangle 

    1st player win at upper triangle and 2nd player win in lower triangle
    both triangle are same size, so both player have same chance to win.

     

×
×
  • Create New...