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Posts posted by plainglazed

  1. answer


    Charlie.  Thought I was close this morning in the shower but it wasn't until my commute into work that I realized that possibility was indeed unique.  Will leave the details to MM :P.Very fun puzzle my friend.


    • Upvote 1
  2. 20 hours ago, BMAD said:

    This contradicts the solution below.  As your a > b > c > d > e > f > g solution does not follow the condition that 

    a + b must equal the sum of the rest


    yes, but...


    ...was referring to the specific case of only five weights to prove a solution cannot exist


  3. Spoiler

    Flip coins to make the bazillion coin the center of the longest string of the same value (heads or tails).  Would expect about nine consecutive heads or tails would be enough to be both easily spotted and not take very many flips.  At some point it could make more sense to break up other existing long strings vs adding to the indicating string.  The downfall of this strategy is if the valuable coin is near either end, but you could strategize to flip the first or last coin to heads if the money coin is in the say first eight coins and the next three coins then tell at what position taking at most five flips, otherwise flip (if necessary) the first and last coins to tails.


  4. Spoiler

    Flipping a coin eleven times results in 11C8+11C9+11C10+11C11 or 232 trials that have 8 or more heads.  There are 11C7 or 330 trials that would have exactly seven heads, half of which would have eight heads with one more flip:  



  5. better than 50%?


    Half of the time there will be a point where there are more black cards than red cards flipped, in which case the odds the next flip is red is greater than 50%.  If that does not occur, the long term odds that the last card is red is still fifty/fifty?  May be faulty logic - conditional probability is a bear.

    For example:  Odds of black, black are .5(25/51) = ~.2451 then the odds the next card flipped is red is (26/50) or .52.  the other 1-.2451 of the time, the odds the last card is red is .5? So with this scheme the combined odds of picking a red are (.2451)(.52)+(1-.2451)(.5)=.5049.

    But if this logic stands, how to optimize?


  6. I think I see the pattern here and am encouraged by the beauty of binary symmetry so here goes:


    For integers i and the number of prisoners n equal to 2^i-1, the probability of success p is (2^n-m)/2^n or n/2^i where m is the number of strings needed to memorize and is 2^(n-i).

    i          n=2^i-1          m=2^(n-i)               p=(2^n-m)/2^n or n/2^i
    1             1                      1                                         1/2
    2             3                      2                                         3/4
    3             7                     16                                        7/8
    4           15                   2,048                                   15/16
    5           31              67,108,864                              31/32


  7. hey bn - Thanks for getting this one started.  You are correct in all you say above including your initial summary of the problem.  Have edited the OP to include that explanation.  I assure you there is a scheme in which one can guarantee finding seven heavy coins.  Perhaps start off by finding one guaranteed heavy coin in two uses of a balance scale.  Or not.  I have no doubt you or others here will discover the method for finding seven heavy dimes.

  8. You are shown a pile of dimes all of which have one of two distinct weights differing by a small amount not detectable by feel.  Forty eight dimes are separated from this pile and you are told of these forty eight, light ones are a dime a dozen (literally - i.e. 44 heavy dimes and 4 light dimes).  Using a balance scale twice, find seven heavy dimes.


    EDIT:  for clarification

    • Upvote 1
  9. okay, not too proud to start this off with some probably way simplistic maths:


    six or seven clusters.  if you can have a fraction of a cluster would guess 6 5/9.

    Think the time and speed intervals are red herrings.  Once the slowest car enters the highway all other vehicles will line up behind/cluster with it, eventually.  The slowest cars expected position would be 50.  Then the same scenario repeats so the second slowest expected at 25; then 12 1/2, 6 1/4, 3 1/8, 1 5/9.  But can you have a fraction of a car thus six or seven or on average 6 5/9?


  10. had thought perhaps


    kilometer - since also measure of distance but not fitting the clues

    then perhaps


    dyne - assume from some unknown force but now one too many letters.  or the bug's obvious lack of energy could be the result of a foot pound.

    rats, still thinking

  11. ooh, nice one Cyget.  Spoiler for some additional thoughts:


    Perhaps the oddly adulating could be thumbs up and dispensing slight could be thumbs down (or the bird if not referring to the same finger).  Woven invoking serendipity = fingers crossed?  But am also having trouble with the last couplet - maybe the ok sign?  Still think fingers could very well be it.


  12. 1.


    Replace the three fifty/fifty counters with one black and one white then add another white counter to keep the odds of drawing the original sole white counter at 25%. So now starting with 5 white and 3 black counters I think gives the same odds as the original scenario.  And after picking two white, there are three of each left so 50%?

    EDIT:  yeah, as I continue to wake up here, am less convinced of my logic above.

  13. Not a generalized formula but better than strategy for three?


    Label prisoners A-G.  Label Yellow hats 0 and Green hats 1. 

    Each prisoner must memorize these eight strings and their inverses as well as their relative position within the group:

            A B C D E F G
    S1 - 0 0 0 0 0 0 0
    S2 - 0 0 0 1 1 1 1
    S3 - 0 0 1 0 1 1 0
    S4 - 0 0 1 1 0 0 1
    S5 - 0 1 0 0 1 0 1
    S6 - 0 1 0 1 0 1 0
    S7 - 0 1 1 0 0 1 1
    S8 - 0 1 1 1 1 0 0

    These sixteen strings and all strings created by changing just one bit of each of the sixteen comprise all 128 possible two

    color hat combinations with seven prisoners.

    So now the strategy:  Each prisoner sees six bits (hats) and can create two seven bit strings by inserting a 0 and a 1 at

    their respective position.  If neither of those two strings match any of the sixteen memorized strings, abstain.  If one of

    those two strings matches one of the memorized strings, choose the other bit (hat color) that did not create a match.

    This effectively packs seven wrong answers in each of the sixteen memorized strings and gives one correct response in each

    of the other 112 cases.  So with 7 prisoners (or more) freedom is possible 7/8ths of the time.


  14. Spoiler

    Given the constraint of the OP that all pairwise distances are distinct, one such distance must be the greatest.  And since there are an odd number of soldiers, that soldier (n) with the greatest distinct pairwise distance is watching another soldier (m) but that soldier(m) is watching another leaving no one watching soldier (n).


  15. Silently be in doubt and subtle...


    the number in the OP should be read with a silent B

    Note:  Although the word "number" as it pertains to pain (killer) exists as an adjective as in "I can still feel that Doc, can you make it number?" it does not formally exist as a noun.  Although in the spirit of a riddle I feel its use here is perfectly acceptable and quite clever.  Certainly I think all would understand what is meant if one says, "I need a number Doc, the pain is killing me."  Nice riddle Biotop!


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