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Posts posted by nana77

  1. Is there time to get the new actions in?

    I am suspicious of Flame for pushing the tie d1. Even though the results were nice. And I am wary of Gavin. Plas scores some goodie points for his plan. idk about Bona.

  2. Your analysis left out the scenario in which you are baddie.

    It is also not clear how my being nk would clear Plas or implicate Flame. It would clear Bona though if he survived. And would be 3vs1. Every scenario that does not end tonight is 3 vs 1. 2 vs 1 requires 2 goodie deaths and that can not happen with the plan as given. I also do not see how anything blows Plas cover.

    Bona can be cleared by surviving. I can be cleared by being spied. Flame can be cleared by successful save of nk. Gavin and Plas can not be cleared (other than by baddie being killed and game ending).

    Odds are Bona or me die tonight and the three others will be the suspects.

  3. Since Aura 2.0 knows Aura 1.0's role, should 1.0's role be revealed to all for balance? Especially if 2.0 is replacing a baddie or indy who just got a free claim.

    3 hours ago, flamebirde said:

    Well, this is interesting at least. Nana's claimed one of two roles, keeping herself from the RID but still somewhat clearing herself. We won't know if the Indy acted, but I wasn't blocked, so we know the Indy didn't act on me (or at least I know that; you guys have to trust me here). 

    My bet is that the baddie is either Nana, Bona or Plas. Whoever it was had to be active and had to know that Aura is experienced/otherwise a good target. It's unlikely that it would be random, in my opinion.

    I'll vote Bona for now, just to get some conversation going.

    By by the way, is there any reason you said "Never vote for the person who's voting for you" and then immediately did just that? :P

    Of course, we could have gotten ridiculously lucky and Aura was the Indy. But that's obviously pretty unlikely.


    Lesson 5: Goodies vote our suspicions and don't care how it looks.

  4. Lesson 1: give a little info about yourself but not too much.

    I acted last night.

    Lesson 2: sharing info

    that would narrow me too much. nothing interesting to share, at least.

    Lesson 3: try to solve the game

    Aura was nk.  Maybe it was random, but if not, Plas is most likely culprit.

    Lesson 4: Never vote for the person who voted you. It is called omgus (oh my god you suck) and looks scummy as heck.


    I vote Plasmid.

  5. To avoid messing with spy and save, it is a good idea for goodie blocks to not act unless the indy or baddie is known or at least strongly suspected.


    And weeee, another BD mafia! Or is that wiiiii?

    *quietly sits back as Aura logics the above into a role claim*


    My answer:



    Let's call the spies a, b, c, d, e, and the phone numbers 1, 2, 3. Have a, b, c call 1, 2, 3.

    If all three spies get the same answer, that is the correct answer.

    If 2 spies (let's say a and b) get the same answer, then we know the traitor must either be the person they named, or one of the two accusers (with the other accuser having called the bad number). If one of the two are bad, the third spy should be accusing them. If the third spy accuses someone else, then we know whoever a,b accused is the traitor and the third spy must have bad results. So we either have the answer (whoever a,b accused) or have it narrowed down to 2 (the one accused by a,b and the one of a,b accused by c. The other 3 spies are all clear and may proceed with a 2nd round of calls to figure it out. Whoever is accused by 2 or more in the 2nd round is guilty.

    If all three spies get different results, that is only possible if one of them is a traitor and one of the other two called the bad number. D and E are thus clear. If D and E were both accused, that only leaves on of ABC to have been accused and that one must be the traitor (since at least one result must be correct, and 2 incorrect results means a traitor must have made the accusation thus the traitor must be in ABC). If only one of D or E was accused, then whichever of ABC not accused is also clear. So that person along with D and E can make 3 calls to get the traitor. Whoever is accused by 2 or more in the 2nd round is guilty.

    The only hope for the traitor is to create a round robin in which neither D nor E are accused. Let's say A accuses B who accuses C who accuses A.

    In this case, D and E choose one of the 3, say A, to join in a 2nd round of calls. And A will be calling the number that B had called, 2.

    If all three get the same results, whoever that is must be guilty. Assuming they do not all answer the same:

    If A is traitor, 2 is the bad number and D and E will both be told A is the traitor.

    If B is traitor, 3 is the bad number. A will be told B, D will be told B, and E will be told something else. Since D and E are told different things, they know one of them has the bad number, and therefor 2 is a good number and therefor A is clear. A can only be bad if 2 was a bad number. So they know that anytime D and E get different results, they can trust what A says.

    If C is the traitor, 1 is the bad number, A will be told C, E will be told C, and B will be told something else. Just as above, they know they can trust what A says as true.


  7. Answers in spoilers, please.

    Two people wish to cross a river. There is only one boat. The boat can only carry one person at a time. A person can not cross the river unless in the boat. The boat can not cross the river without someone in it. Both people cross the river. Can you explain how it happened?

  8. Three castles, each occupied. One of them is king, other 2 are not. One always tells the truth, the other 2 always lie. Red and yellow castles are by the lake, green castle is not.

    The three each make a statement:

    Green: I am the king.

    Yellow: I am not the king.

    Red: The king lives by the lake.

    The riddle's question: in which castle is the king?




    Green must be lying since green's statement means yellow would true as well.

    Since green is not the king, red statement must be true, since king must be by the lake. Since red is true, yellow must be false. Which means yellow castle has the king. And the king is a liar.


  9. for 11


    per Erdos, at least 4 are already in order, so at most 7 dvd's must be moved to their correct locations.

    for 22 on one shelf


    per Erdos, at least 6 are already in order, so at most 16 dvd's must be moved to their correct locations.

    for 22 on 2 shelves


    Since shelf order does not matter, worst case scenario is each shelf has 6 dvds which will remain on the shelf and 5 it will trade with the other shelf. Any dvd leaving one shelf will have a counterpart on the other shelf that needs to change shelves as well. Of the 6 that remain on the shelf, worst case is 2 will be in order. So 4 dvd's total are in order and on the correct shelf, meaning the other 18 dvd's must move.


  10. This was answered on the original forum.

    To see:


    The answer was:

    A. I. Breveleri  parses the question as :


    "::" : ":" :: ":" : ?

    which should be read "as is to is to as is to is to what"

    which seems to indicate the answer "" i.e. nothing. 

    The parsing is correct.  But the solution to the question isn't nothing, it's "."

    or, "four dots"  is to "two dots"  as "two dots" is to dot.


  11. P is proven as first letter by the folllowing:

    Plant-1 means one of those letters is right. -LANT disproven by:

    -L disproven by Blimp-0

    --A disproven by Again-0

    ---N disproven by Wrung-0

    ----T disproven by Reset-0

    only the P is left for Plant

    S is proven as final letter.

    Coins-1 means one of those letters must be right.

    C is disproven since P is 1st letter.

    -OIN is disproven since the 1 point in Point-1 is the P.

    That only leaves the S for Coins.

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