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Posts posted by bonanova

  1. My interpretations

    1. Country(Pitch2) = France
      Fav(USA) = wherigos.

    2. Pitch(Marie) + 1 = Pitch(Chile)
      Fav(Marie) = multi
      Finds(Marie) = 2 Finds(Spain)
      Fav(Spain) /= letterbox.

    3. Fav(Heidi) = mystery
      Finds(Heidi) > Finds(Hugo) > Finds(Frank)

    4. Contry(Bridgette) = Germany
      Pitch(Bridgette) = Pitch(Frank) - 2
      Pitch(Frank) /= 7
      Pitch(UK) - Pitch(Bridgette) > 1

    5. | Pitch(Chile) - Pitch(Belgium) | /= 1 (*)   EDIT: should read /= 2
      Pitch(Chile) > Pitch(Stanley)
      Pitch(earth) = 1

      (*) Interpret "next door but one" to mean exactly one intervening pitch.
      Clue 5 then says this does
      not apply to Chile and Belgium.
    6. Pitch(Juan) = 3
      ManX = { Hugo Juan Stanley }
      Finds(Manx) = 16
      Pitch(Manx) < Pitch(Frank)

    7. Pitch(Heidi) = Pitch(USA) + 1
      Finds(USA) > Finds(UK)
      Finds(Fav = traditional) = 40
      Country(40 = Finds) = { Belgium France Germany Spain UK }


    I believe gives this solution

    1. Stanley UK earth 2
    2. Marie France multi 10
    3. Juan Chile letterbox 16
    4. Bridgette Germany traditional 40
    5. Hugo Spain virtual 5
    6. Frank USA whereigo 4
    7. Heidi Belgium mystery 8



  2. On 5/27/2018 at 2:38 AM, CSIQ said:

    ... the girl/boy, boy/girl options are, in fact, only one option.

    They are equally likely but they are not the same.
    But it's worse than that. The OP is deficient, because it does not tell us how we came to know what we know.


    Suppose the person that is telling us about Teanchi and Beanchi has first decided this: If a family has at least one son, I will say the family "has at least one boy." If not, then I will say the family "has at least one girl." But let's say we don't know that is his algorithm for describing families to us. Which is true, because the OP does not disclose it. But what would happen in that case?

    We know all two-child families belong, in equal numbers, to these four gender groups: { GG GB BG BB }. If T&B are GB BG or BB, our informant will say T&B have at least one boy. Only in the GG case, where there is not "at least one son," will he say T&B have at least one girl. The probability that the other kid is a girl is then 100%!

    Or suppose our informant has first decided this: If the elder of two children is a son, I will say the family "has at least one boy." If not, then I will say the family "has at least one girl." What happens in that case? If T&B are BG or BB, he will say T&B have at least one boy. Otherwise, when T&B are GG or GB, he will say T&B have at least one girl. The probability that the other kid is a girl is then 50%.

    Or suppose our informant has decided this: If there is at least one daughter, I will say "the family has at least one girl." Otherwise, I will say "the family has at least one boy." Now, if T&B are GG GB BG he will say T&B have at least one girl, and the probability that the other kid is a girl is 1/3.

    Ouch. That is the case that seems most likely, but we can't assume our informant's algorithm with certainty. The OP as stated is deficient.

    Instead, let's create a situation where we know how we know what we know, and therefore will let us find the probability that "the other kid is a girl," unambiguously.


    If births are gender-neutral, we know that families with exactly two children divide equally into four categories for the sex of their older/younger child. Namely: { GG GB BG BB }. The idea now is to eliminate the last category, but do it in a way that does not inject any bias. If we can do that, then we can say GG is one of three equally likely cases { GG GB BG } so the answer is 1/3.

    Here's one way.

    Pick a number from a hat containing all number from 1-100, and say it turns out to be 13. Now assemble thousands and thousands of two-children families from the general population. This means equal numbers of GG GB BG and BB families. Now, in some random order, we ask them this question: "Is at least one of your children a girl?" and let's say that the 13th family to answer "Yes" is the Smith family.

    We can be certain that the Smith family is type GG GB or BG with equal probability.

    Doing that, we know the answer is 1/3.

  3. Once letters start getting removed it quickly gets easier. At the start it's very hard. How much of a clue are you willing to share? I don't want to disclose too much, but would you be willing to confirm the politician is male and contemporaneous? Initials or country of affiliation might be too revealing.

  4. 7 hours ago, Donald Cartmill said:

    Capt Eds' answer is not quite as straight forward as could be.

    Both answers state correctly that (a) winning distances give speed ratios and (b) combined speed ratio gives combined winning distance. What part of that can be more (or less) straightforward?

  5. At the annual Brain Denizen picnic, there were the inevitable games, among them the ever-popular three-legged race. Three teams were formed by tying one contestant's right leg to another' s left leg. Fortunately all six contestants made it to the finish line without any broken bones! For purposes of this puzzle we assume all three teams ran the 100-meter course at constant speed.  Team 2, comprising BMAD and Thalia, were able to beat Team 3, comprising rocdocmac and DejMar by 20 meters, but lost to the winner, Team 1, comprising plasmid and plainglazed, by 20 meters as well. By how many meters did Team 1 beat Team 3?

  6. On 5/31/2018 at 9:10 AM, DejMar said:
      Hide contents

    One of the solutions (due to symmetry), with the original empty peg-hole at #1 is with the jumps {7, 29, 24, 18, 15, 2}. This pattern of jumps will leave 8 pegs -- 3 in the third and 5 in the 5th (bottom) row of the triangle.


    Very nice!

    There is another solution that ends two moves earlier, leaving pegs only on the edges.

  7. You've probably seen this puzzle. There are 15 holes in triangular array. (See sketch below.) The game begins with pegs in 14 of the holes. The play is to jump pegs over adjacent pegs, removing the "jumped" pegs afterward, as in checkers. The jump is made in a straight line. To make a jump, you need a contiguous group consisting of { peg1, peg2, hole } in a straight line. Peg1 ends up in the hole, and peg2 is removed. The object is to make 13 legal jumps and end up with a single peg.

    This happens about 6% of the time. That is, about 94% of the time you get a configuration, with more than one peg remaining, that permits no further legal jumps. In some games the peg must end up in the original empty hole, and that happens only about 3% of the time. So, it's not a trivial puzzle.

    This puzzle asks for something different, and easier: Lose as badly as possible. That is, select a location for the empty hole, and then find a sequence of moves that leaves the greatest number of pegs on the board where there are no more legal jumps.

    It's simple enough to play, even without the game, by marking hole locations on a sheet of paper and using pennies.


    Hint: you might first want to sketch patterns of pegs that have no further legal jumps. Then decide where to place the initial empty hole and how to make the jumps to get to the desired configuration.

    As already stated, there are 15 holes. There are also 36 possible jumps. For convenience in writing sequences of jumps, they can be numbered, as follows:

    Number the jumps like this:         and the holes
    --------------------------->  o     -------------     1
    So Jump #1 means the         / \      like this:
    peg in hole #1 jumps        1   2     ---------->   2   3
    over the peg in hole                                        
    #2 into the empty                                 4   5   6
    hole #4.                 o         o  
                            / \       / \           7   8   9  10
    Jump #18 is peg 7      3   4     5   6
    over peg 8, into      7              13       11  12  13  14  15
    hole 9.              /                 \
                        o-8       o      14-o
    Holes 4, 6, 13     / \       / \       / \
    begin 4 jumps;    9  10    11  12    15  16  
    the others      17        19    21        23
    begin two.      /         /       \         \
                   o-18      o-20   22-o      24-o
    There are
    36 jumps.
                25       27    29  30   33        35
               /         /       \ /      \         \
              o-26      o-28   31-o-32  34-o      36-o

    With symmetries taken into account, the holes have four equivalence classes:

    • Corners (1, 11, 15)
    • Adjacent to corners (2, 3, 7, 10, 12, 14)
    • Edge centers (4, 6, 13)
    • Centers (5, 8, 9)

    This means that there are just four distinct places for the empty hole to start a game: { 1 2 4 5 }. All other holes are symmetrically equivalent to one of these.

    Just to be sure the numbering above is understood, here is a winning game of the normal type. Start with pegs in every hole except #1. (The top hole is empty.) Then make these jumps: { 7 14 2 17 23 27 34 26 30 6 35 14 7 }. If done correctly, the original hole #1 contains the final peg.


  8. On 5/9/2018 at 7:10 AM, plainglazed said:

    You and I each have a coin, select its orientation (heads or tails), and reveal to each other simultaneously.  If both are heads, you win a dollar.  If both are tails, you win a quarter.  But if both are different, I win fifty cents.  Fair enough?



    A selection strategy exists that gives us equal expected winnings, namely that we select H with probabilities p and q respectively, where

    100pq + 25(1-p)(1-q) {my winnings} = 50[p(1-q) + q(1-p)] {your winnings}

    This happens when p = q = 1/3 (Wolfram solution / contour plot); both sides of the equation evaluate to 200/9.

    In fact, if either of us shows H with probability 1/3, the equation holds, regardless of opponent's strategy:
    p = 1/3 both sides evaluate to (50/3)(1 + q), and vv, due to symmetry.

    Knowing this, both of us can avoid a net expected loss by presenting H 1/3 of the time.

    Additionally, no advantage can be gained by either of us by slightly increasing or decreasing that value. If we both increase or both decrease, I come out slightly ahead. If one increases while the other decreases, you come out ahead by the same amount: in the case of 1/3 +/- 0.1, that amount is 2.25.

    And as already stated, if only one of us increases or decreases, no advantage or disadvantage is gained if the other stays at 1/3.




  9. Doctoring the figure a bit (while I think about solving it.)

    o  -  o  -  o  -  o  -  o
    |                    /
    o     o     o     o
    |              /
    o  -  o  -  o

    Question: You mention the number of nails (12) and also the number of nails in the perimeter (here that's 10.) So for this example, would we be given N=12 or N=10? I'm thinking the latter, but would like to confirm.

  10. Looks like


    The angles now are

    [ - 90o/1 - 90o/2 + 180o/3 ] + [ - 90o/3 - 90o/4 + 180o/5 ] + ... + [ - 90o/(2n-1) - 90o/2n  + 180o/(2n+1) ] + ....

    = (90o) Sum(n=1, inf) { - 1/(2n-1) - 1/2n  + 2/(2n+1) }

    = 90o { ln(2) - 2 } (Wolfram)

    = -117.6167...o

    Twice as much as the first answer. Showing that rearrangement of terms in an infinite series that is not absolutely convergent (both answers were forms of alternating harmonic series) can change the result. If the terms of a convergent series are all positive, the order does not affect the result.


  11. With that interpretation,


    The successive angles are 180/3 - 90/1 + 180/5 - 90/2 + ... so that

    Final angle = 180o {  1/3 - 1/2  +  1/5 - 1/4  +  1/7 - 1/6  + ... }

                    = 180o { (1/3 - 1/2) + (1/5 - 1/4) + (1/7 - 1/6) + ... }
                    = 180o { -1/6 - 1/20 - 1/42 - ... - (1/(2
    n+1) - 1/2n) - .... }
                    = -90o { 1/3 + 1/10 + 1/21 + ... + 1/(
    n(2n+1)) + .... }
                    = -90o Sum(
    n=1, inf) 1/(n(2n+1))
                    = -90o { 2 - 2 ln(2) } (Wolfram)
                    = -55.2335 ...o (CCW from


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