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Fun fun riddles/problems

[Deleted User][Deleted User] Posts: 0
edited August 2013 in Off-Topic
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Post edited by [Deleted User] on
SCARY_WIZARDTroodon80CrevsDaak
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  • elminsterelminster Member, Developer Posts: 16,151
    edited August 2013
    #3 I assume is...

    the North Pole. By that I mean the Geographic North Pole and not the Magnetic North Pole.

    [Deleted User]artificial_sunlightMoomintroll
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    elminster
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  • elminsterelminster Member, Developer Posts: 16,151
    edited August 2013
    Do what you feel like concerning spoiler tags :)

    From what I can see from the first riddle...

    In terms of the overall equation n=1. Past that I'm not sure if you are asking what the sum of n+n is or the sum of the first "1" integer. If the later is the case I'm not sure how a person goes about getting the sum of a number from just one number. In any case I'm going to go with the real answer is either 2 or 42 (the answer to life, the universe, and everything)

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    elminster
  • Montresor_SPMontresor_SP Member Posts: 2,164
    #1:

    This is proved by induction: If the sentence holds true for any integer, then it is also true for the next integer.

    It is easy to show this is true for n=1:
    1*2/2 = 1

    Assume it is true for n=k; that is:
    1+2+3+...+k = k*(k+1)/2.

    Then it must also be true for n=(k+1):

    1+2+3+...+k+(k+1)
    = k*(k+1)/2 + (k+1)
    = k*(k+1)/2 + 2*(k+1)/2
    = (k^2 + 3*k + 2)/2
    = (k+1)*(k+2)/2 which is the similar formula for (k+1) QED

    So if it is true for k, then it is also true for (k+1) and therefore for (k+2), etc. But since it is true for 1, it must be true for 2, and for 3, and for 4, and ... up to infinite.

    [Deleted User]
  • KaltzorKaltzor Member Posts: 1,050
    For #2...

    Well, if we ignore variables placed onto it all by the throw... We are really left with 3 possibilities of which one is quite rare... It can land on Heads, Tails or on it's side, though the side will require it to be leaning against something... If we assume it's a flat surface that can't really happen...

    For chances of getting heads, it's 1 out of 2, so you for one heads you would need 2 flips...

    I don't think there's really a reliable method of calculating how many it would take to get x amounts of heads in a row.

  • TsyrithTsyrith Member Posts: 180
    edited August 2013


    What is the expected number of coin flips required to obtain your first 2 consecutive heads? for 3? for 4? Is there a patern?
    May I ask the inevitable follow-up?

    Is my name Booker DeWitt?

    /scoffchortle I use wit to disguise a mental shortcoming.

  • ajwzajwz Member Posts: 4,122
    1. In English, name a four letter word with 3 syllables

    2. What is the only number whose letters are written in alphabetical order (for example 2 = two: t is before w in the alphabet, but w is after o, so two doen't work)

    3.
    0 to the power n is always 0 for any n
    n to the power 0 is always 1 for any n
    0 to the power 0 has no solution, but if it did have a solution, would it be closer to 0 or 1?

    [Deleted User]
  • Montresor_SPMontresor_SP Member Posts: 2,164
    edited August 2013
    ajwz said:

    1. In English, name a four letter word with 3 syllables

    2. What is the only number whose letters are written in alphabetical order (for example 2 = two: t is before w in the alphabet, but w is after o, so two doen't work)

    3.
    0 to the power n is always 0 for any n
    n to the power 0 is always 1 for any n
    0 to the power 0 has no solution, but if it did have a solution, would it be closer to 0 or 1?

    1.

    iota


    3.

    I'd say closer to 1, because the value of lim(n^0) for n->0 is 1.


    I pass on 2 for now. :o

  • JalilyJalily Member Posts: 4,681
    edited August 2013
    ajwz said:

    1. In English, name a four letter word with 3 syllables

    Aria.
    ajwz said:

    2. What is the only number whose letters are written in alphabetical order (for example 2 = two: t is before w in the alphabet, but w is after o, so two doen't work)

    Forty.
    ajwz said:

    3.
    0 to the power n is always 0 for any n
    n to the power 0 is always 1 for any n
    0 to the power 0 has no solution, but if it did have a solution, would it be closer to 0 or 1?

    1.


    [Deleted User]CrevsDaak
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  • Awong124Awong124 Member Posts: 2,643
    edited August 2013

    Fourth riddle:


    4 men are wearing a hat and buried neck-deep into sand, unable to move even their head. They are buried in such a way that they are looking in the same direction (let's say left). Also, a wall separates the first and second man, as displayed here:

    snip
    To what extent that they can't move their heads? If they can move their mouths, then one of them can just tell another what color their hat is.

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  • Awong124Awong124 Member Posts: 2,643
    So they can't talk amongst themselves before the 5 minutes are up.

  • zurathanzurathan Member Posts: 53
    i found the answer. but how should i tell it? how does the spoiler thing works? or should i just message it?

  • zurathanzurathan Member Posts: 53
    testing spoiler, excuse my ignorance

    4th guy is able to see 2nd and 3rd and if their hats same he would know his own hat because of there is only 2 black and 2 white hat. and if he is not able to talk, 2nd and 3rd guy should have different hats. so 3 guy could talk after 4 minute and 59 seconds as a last resort.
    the thing is 3rd guy can see 2nd guys hat and if he can trust that 4th guy is not a moron then he can talk!
    bu there is the possibility to hearing 4rd guy saying "hodor, hodor hodor" and they are all dead

    [Deleted User]
  • CorianderCoriander Member Posts: 1,667
    @JonSnowIsAlive


    3 will know his hat is different from 2's if 4 doesn't answer.


    Montresor_SP[Deleted User]
  • FinneousPJFinneousPJ Member Posts: 6,456
    @Coriander, @zurathan already got that. Also, it's not strictly true, as demonstrated in the answer of @zurathan.

  • KaltzorKaltzor Member Posts: 1,050
    edited August 2013
    If it was to know all 4... if the 4 people know their own hat color, 4 since he should be able to see his own, 3 and 2... leaving the last one easily guessable.

    And I wonder... What kind of hats are they? As even though they can't move their heads, they can still move their eyes and look up to see their hat unless it's a tiny thing on the top of their heads they can't see...

  • FinneousPJFinneousPJ Member Posts: 6,456
    If they knew their own hat the question would be trivial...

  • CorianderCoriander Member Posts: 1,667
    @FinneousPJ I win for being succinct.

    Also, it's got to be up to the person asking the question to say whether an answer is correct. I'm not really interested in seeing the other answers until I know if I'm right or not. So instead of waiting to see if @zurathan is correct, I posted my own answer.

  • Morte50Morte50 Member Posts: 161
    First riddle from the OP is I'd say most elegantly solved with proof by induction, as provided by @Montresor_SP

    As for the second:

    Assuming a fair coin, the number of flips needed for a single heads is a random variable with a Geometric Distribution (with p = 0.5), so the expected number of flips needed is 2 (EV of a Geometric Distr. is 1/p).

    For two consecutive heads, we can start with the same Geometric random variable and flip an additional coin. If it's heads, we're done. If it's tails, we need to repeat the process. Interestingly, the number of times we need to repeat the process is itself a Geometric random variable (with p = 0.5). The expected number of repetitions we need is therefore, again, 2. The expected number of flips 'inside' each repetition is 3: 2 from the 'inner' Geometric variable, plus the additional flip. Those 'inner' Geometric variables are independent of each other and of the 'outer' Geometric variable, so the expected number of flips = 3*2 = 6.

    For three consecutive heads, we can simply repeat the same principle: get to two consecutive heads, then flip an additional coin: again, the number of repetitions is a Geometric random variable, EV = 2; the 'inner' variables have EV = 6+1, so the expected number of flips needed to get 3 consecutive heads is 14.

    This progresses quite predictably:
    4 heads: 30 flips
    5 heads: 62 flips
    6 heads: 126 flips

    This, perhaps predictably, is a (partial) geometric series, so in general we have
    N heads: 2^(N+1) - 2

    There's probably a more mathy way to prove this, but it's been a while since I've worked with Geometric distributions (Normal distribution FTW!), so I'm a bit rusty.



    [Deleted User]
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  • zurathanzurathan Member Posts: 53
    ok now is the time for a puzzle that puzzled me soo bad. i just hope that its not well known or someone solves it too easy.

    we have four numbers and any calculator operation like +, -, squareroot, power,..... the goal is founding 100
    and there were two example too; (5+5)x(5+5) and (99+9/9)
    oops forgot to say the four numbers we given in this puzle was 7

    this puzzle has an other version too; this time we have five numbers 1,7,7,7,7 and limited operations to +,-,/,x and ofcourse ( ) parenthesis.

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