Jump to content
BrainDen.com - Brain Teasers


  • Posts

  • Joined

  • Last visited

  • Days Won


Posts posted by Yoruichi-san

  1. Still not sure about "power,"but if it is correct, then the answer has to be 10. If you look at the original post, the colors of the words indicate groupings, so the 10 applies to the e, not the natural log.

    Very clever, Yoruichi-san. B))

    "support proposal" is not 10. Good job on noticing the color groupings, though!

  2. I get the whole math terms

    vibe that is going on, but I must bow out as I am so mathematically handicapped. Good luck ya'll

    Wow, being on the right track made you give up...that's ironic. :/

  3. [spoiler='

    Well, this changes my previous answer']Apparently I was wrong in assuming that since no desire to see others die is mentioned, we can assume the last pirate will be happy with 50. I guess I didn't realize "seeing someone else die" is human nature...I thought it was in"human"e...haha, jk. Anyways, in that case my new answer:

    Situation: Paid (taxed)

    Order: [...pirate 5, pirate 6, pirate 7]

    2 pirates(6&7): 50(100)

    3 pirates(5-7): 50(100), 49(49)

    4 pirates(4-7): 33(100), 49(49), 33(100)

    5 pirates(3-7): 33(100), 33(100), 48(48), 32(32)

    6 pirates: 25(100), 25(100), 25(100), 47(47), 31(31)

    7 pirates: 25(100), 25(100), 25(100), 24(24), 46(46), 30(30)

    Captain gets: 100 (own) + 175 = 275 coins...

    4 pirate case is funny...if you tax pirate 7 48 coins, the captain gets 33+33+48. But if you tax pirate 6 49 instead of 100 and pirate 7 100, then the captain gets 33+49+33. One more coin...woo hoo...

    Well, actually, you can apply what you said about 4 pirate case to 6 pirate case as well...

    And, well, it *is* human nature...hence the coliseums, public hangings/burnings/beheadings, reality shows... ;P

  4. in the 4 step, pirate 6 (3) knows that the captain and pirate 7 (4) are going to vote yes, so to save the most coins pirate 6 (3) will always vote no. so if the captain only taxes pirate 6 (3) 49 coins, pirate 6 (3) would still vote no and only owe 49/(2+1)=16 coins.

    Actually, I meant that pirates 4 and 6 are going to vote yes, and pirates 5 and 7 are going to vote no. The tax scheme would be: 5 (100), 6 (49), and 7 (100). Pirates 5 and 7 vote no, and each have to pay 33. Pirate 6 votes yes, since he knows that if he votes no they would go down to the the 3 pirate case he would have to pay 50. So pirate 4 (the captain in this case) would get 33+49+33=115 instead of 33+33+48=114 coins.

  5. do all pirates vote simultaneously?....or do they vote by seniority?

    I don't think it really matters, since they are all game theory geniuses who can figure out exactly how everyone else is going to vote based on the proposed tax scheme...

    The way the vote is going to should already be known before the vote takes place. The voting itself is just a formality...

    Actually I was thinking about that, too, but the way I worked it out, whether the vote is simultaneous or by seniority may affect who gets taxed what amount but it won't affect the amounts that are taxed and the total amount the captain gets, which is what the question is asking for.

  6. keeping in mind that the pirates are greedy, there are 2 things each pirate will want. one is to keep/get the most coins. the second is to be captain or closest to being captain (so they can get more coins the next time).

    work backwards (in each scenario, the higher number pirate is the lower ranking pirate):

    2 pirates: captain will tax pirate 2 100 coins and will vote yes, pirate 2 will vote no. captain wins and gets 50 coins (100/(1+1)).

    3 pirates: captain taxes both 100, both vote know captain dies. no good.

    captain taxes pirate 2 100 coins and pirate 3 49 coins. pirate 3 is happy (keeps more coins than if there were 2 pirates) and votes yes along with captain. captain gets 50+49=99 coins.

    4 pirates: captain taxes pirate 2 100 coins, pirate 3 100 coins and pirate 4 48 coins. captain and pirate 4 vote yes. captain can't tax pirate 4 49 coins because then pirate 4 has no reason to vote yes this time (they could vote no and now be 3rd in line for captaincy instead of 4th and keep the same number of coins). captain gets 33+33+48=114 coins

    5 pirates: captain taxes pirate 2 100 coins, pirate 3 100 coins, pirate 4 32 coins and pirate 5 47 coins. captain, pirate 4 and 5 vote yes. captain gets 33 + 33 + 32 + 47 = 135 coins.

    6 pirates: captain taxes pirates 2,3 & 4 100 coins, pirate 5 31 coins and pirate 6 46 coins. still gets 3 yes votes and gets 25+25+25+31+46=152 coins

    7 pirates: captain taxes pirates 2,3 & 4 100 coins, pirate 5 24 coins, pirate 6 30 coins and pirate 7 45 coins. gets 4 yes votes and 25+25+25+24+30+45=174 coins.

    That's close to what I got, except in the 4 step, the captain would be better off taxing pirate 6(or what you call pirate 3 in that step) 49 coins (since pirate 6 would have to pay 50 coins if the vote failed and it went down to the 3 pirate case) then taxing pirate 7 (your pirate 4) 48. And then the rest of my reasoning goes from there.

  7. Captain gets 285 coins.

    Here's my reasoning: the captain wants to avoid getting hanged, and if any other pirate were captain he would as well. Every pirate that's not the captain wants to do better than they would if that captain were hanged, and the next pirate becomes captain. This means that you can reason up from a two pirate situation. Assign them numbers 1-7 with 1 being the current captain, and 7 being the lowest rank:

    2 pirates: 7 will always vote no and lose the tie to 6, so he will be taxed 100 and pay 100/(1+1)=50 to maximize 6's revenue.

    3 pirates: 5 will always vote yes, 7 will vote yes as long as he does better than he would if 6 were captain, so he votes yes on 49. To max the payment from 6 (as he is the only no vote) he will pay 50.

    4 pirates: 4 votes yes, 7 votes yes if less than 49, and 6 votes yes if less than 50. The most 4 can get as captain is then 50 + 49 + 48 = 147.

    Continuing this reason, the current captain maxes out at 50 + 49 + ... + 45 = 285, where everyone votes "yes" except 2.

    Does that make sense? I feel like this would be odd because every pirate can improve their situation by voting no until there is enough for mutiny.

    Well...right method of thinking about it (good), but...

    For the 4 pirate case, if pirate 7 knows that pirates 4 and 6 are going to say yes, there is no reason for him to vote yes as well, since it only needs two votes to pass. Hence he would vote no, and in that case both his and pirate 5 only have to pay 1/3 of what the tax was.

  8. Trying to examine things from bottom up, starting with 2 pirates, but what is a pirate's preference if there is a tie?

    Pirates labeled 1 to 5, 1 being most senior(captain). In the case with only two pirates, only 4 and 5 remain. Four doesn't need 5 to agree for the vote to pass, so he will tax 100 coins from 5. Five will oppose the vote and keep 50 coins.

    Case with 3 pirates:

    Now pirate 3 is in charge and he needs the support of one other pirate and this is where I have a question. Pirate 5 knows that the most gold he can get is 50, so if pirate 3 offers to tax pirate 4 100 gold coins and pirate 5 only 50 gold coins, will pirate 5 accept? Does he care if pirate 3 lives? Either way, pirate 5 will only get 50 pieces of gold.

    As I see it, whether ties matter or not changes the answer.

    Same order of preference as pirates 2:


    Getting money (or in this case, losing less money)

    Seeing someone else die (human nature...)

  9. Also, the wording "The captain must obtain approval (yes votes) from at least half his crew (including himself)" implies that a tie goes to the captain if there are an even number of pirates. Is that the case?

    Yep, ties go to the captain. Same voting rules as the previous pirates (if you haven't looked at them, you may want to since it will probably help you figure out the solution).

  10. Question - if a captain gets outvoted and walks the plank, what happens to his 100 coins?

    Uhh...they probably just go into the sea with him. Got to let a pirate die with his treasure... ;P

  11. Making correction

    1. Share lease of pastry over front

    2. Boring way to go of had lunch

    3. Organic diary of squeal to the support proposal

    4. Important from pessimistic pizza to loser of omen of former over also latin gods hug

    If they ask "can you put that in plain engish?", show them this post...;P

    Look at my signature.

    Whoops, just realized that 2 should be "had lunch", not "ate lunch"...kind of gives away that part ;P

  12. Good job. Even simpler than my solution.

    Any ideas on the other part?

    So if we take hmmm...'s algorithm for drawing the squares and put it into mathematical terms:

    For even n (number of squares), we get (n-1) of squares that are 1/(n/2) in length around the outside and 1 square of (n/2-1)/(n/2) in the remaining space.

    If we apply this formula to n=2, we get (2-1)=1 square of length 1/(2/2)=1 and 1 square of length (2/2-1)/(2/2)=0. This is a contradiction (we get one square that takes up the whole square, not 2 squares!). So we can't do this for n=2.

    For odd n, hmmm...'s algorithm becomes (n-4) squares of length 2/(n-3) and 4 squares of length (n-5)/(n-3).

    Applying this to n=3 gives us lengths of 2/(n-3)=2/0 and (3-5)/(n-3)=-2/0 not to mention (3-4)=-1 squares, which are all contractions (negative one squares of length infinity?!!!) , so we can't do it for n=3.

    For n=5, we get (5-4)=1 squares of length 2/(5-3)=1 and 4 squares of (5-5)/(5-3)=0. So again we have 1 square that takes up the entire square, not 5 squares, so again a contradiction, and hence we can't do it for n=5.

    This proves by contradiction that we can't divide a square into 2,3,or 5 using the algorithm proposed by hmmm.... However, I think there are other ways of dividing the square for higher numbers, and we still need to show those methods don't work for 2,3, or 5 either...

  13. Seeing the Pirate's game posted (thanks magicalcheese!) reminded me of my classes in polysci/game theory, so I thought I'd take a shot at making Pirates 3. So here's my spin:

    So after some more looting and plundering and general pirating, the pirates have picked up an additional member (making for 7 pirates total) and now they each have 100 gold coins.

    Somehow (I highly suspect there was rum involved ;) ), they agreed to allow the captain to levy a "pirate tax". The captain can choose tax each crew member a different amount, but the proposed tax scheme (like all proposed tax schemes) must be voted upon in the usual pirate format.

    (For those of you who missed the previous pirates strings created by magicalcheese, here are the rules: The captain must obtain approval (yes votes) from at least half his crew (including himself) or else there will be a mutiny, the current captain will be forced to walk the plank, and the next highest ranked member of the crew will become the new captain, and the process will repeat until a proposal passes.

    However there is one extra catch in this case: for the pirates who vote "no" on a proposal that passes, the amount of tax they have to pay is the proposed amount divided by (the number of pirates who vote no+1), rounded to the nearest whole coin. For example, if out of 7 pirates, if three of them are taxed 100 gold coins and vote no, then they each have to pay 100/(3+1)=25 gold coins to the captain.

    As before, assuming they are all really smart (and relatively sober during the voting...;)), and really greedy, what is the maximum amount the captain can get out of taxes without risking his life?

    Remember, the captain is not only trying to get enough yes votes to pass, he is trying to get the maximum number of gold coins as well.

    I just tried working this out myself, but maybe someone will find a better answer than I did...

  14. Umm...

    There is only one color. So the probability of getting a pair is always one. Otherwise, I don't know if it is possible...

    The probability of getting the a pair is (A/N)^2+(B/N)^2+(C/N)^2...=(A^2+B^2+C^2...)/(N^2)

    where A,B,C...represent the numbers of the different colors, and N is the total number of balls.

    Adding 20 balls of color A makes the probability: (A+20)^2/(N+20)^2+B^2/(N+20)^2+C^2/(N+20)^2...=(A^2+40A+400+B^2+C^2...)/(N^2+40N+400)

    which only equals the original probability (for real positive integers)if A=N and B=C=D=E=...=0.

  15. Hey - I liked it, the red herring was the cartoon capers and the smell of breakfast ( they love their herring in Netherlands). Reminded me of my little uns eating them - so just a little side tracked. Actually I had no focus on this one - it just would not lead me anywhere!

    Nothing wrong with a little misleading, or originality - better a post that is difficult than a repeat!

    Thanks, and sorry, the last paragraph of that post wasn't really meant for you...I was just to lazy/pressed for time to post two separate replies. I like the originality of your scoresheet posts as well! :)

  16. Well, I think I've figured out the first part of 1.

    1. currant ______

    "Shared lease" would be co-rent, or as relating to a pastry --> currant. As to what "pastry over front" means, or any of the other questions, I have no idea.

    Good! You are starting to think about it the right way (although I'll let you know that "co-rent" was not the term I was going for).

  17. Well, that's what differential equations are for (and why I say that they are the only useful form of math). ;)

    You can always write a differential equation describing what we observe. Solving that differential equation is another matter though...

    Thanks a lot for the words of praise! :)

    Oh, and thanks to engspangussian for the clever pattern...and the people who figured out the multiplying thing! This was totally a team effort, which is why I like this forum!

  • Create New...