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  1. I'm not a pirate

    Right on, Wilson  
  2. I'm not a pirate

    Not a hockey goalie, fisherman, or cartographer. I like how goalie fits with these clues very well even though the clues aren't interpreted in the way I had in mind. But the second stanza is difficult to fit with that answer. I like the reasoning behind the fisherman answer as well, but once the answer is revealed there will be a much clearer interpretation of the clue about Atlas and the last line's use of the word dearth implies that two would be an unacceptably low number (and for the sake of hinting, the dearth in the last line refers to Atlas' feet from the previous line). A cartographer is actually closer to my answer than you might think.
  3. I'm not a pirate

    Indeed not an asteroid impact. I'm thinking that once the answer is seen, all of the clues will become visible and make sense. (There might be hinting in here.)
  4. I'm not a pirate

    Not a blacksmith; while it could certainly fit parts of the riddle, the last stanza would seem to go unexplained. Also not a fencer, as I would consider that to be too similar to the decoy answer of not being a pirate. Not a lobster trap. I'm afraid I don't know enough about them to be able to judge how well it fits the clues, but that also of course means it's not what I was thinking of. Hint: Spoiler Thalia and Tyler were on the right track in going after a profession  
  5. I'm not a pirate

    Hi DD. Not a jigsaw; while in some ways it's on the right track, I'm looking for something with a more specific interpretation for the opposing sword and masking of thieving hand.
  6. Fill the vacancy

    I don't have a full solution, but this might help get things started. Spoiler   Describe the Praesidium as a vector (n1, n2, n3, …) where nm is the number of people in the Praesidium of priority m. So a Praesidium of (2, 5, 0, 1) would have eight people: 2 of priority one, 5 of priority two, and 1 of priority four. First note that Arcsin can win if there are at least 2m people of priority m for any value of m. Just put 2m-1 people in S so there are also at least 2m-1 people in not(S), and Arctan will have to leave at least 2m-1 of them who will be priority m-1 in the next round. Keep repeating that, and you'll get 21 people of priority 1 and one of them will get through the next round to reach priority zero. Conversely, if the Praesidium is filled only with people of one priority level (call it m) and there are fewer than 2m people, then Arctan can just keep eating whichever of S or not(S) has the most people and will eat them all before any reach priority zero. The interesting stuff happens when there are people at multiple different priority levels, but there are fewer than 2m people at any given priority level. As a simple example of that, consider the Praesidium (0, 2, 5). If Arcsin just split people into S and not(S) evenly [like making (0, 1, 2) and (0, 1, 3)] then Arctan would just eat the larger one and the next round would be (1, 2), then Arctan would eat whichever of S or not(S) holds the priority one person to make it (1) in the next round, and then Arctan would win. But instead if Arcsin started off by making S and not(S) be (0, 2, 0) and (0, 0, 5), then the next round would either be (2) or (0, 5), and either of those situations would be an easy win for Arcsin because they have nm > 2m. Let's try to generalize that. Suppose there is some priority level m with at least 2m-1 people. Then Arcsin could make one group S with 2m-1 people of priority m and the other group not(S) containing the rest of the Praesidium; Arctan would have to eat S or else he would be in a losing situation, and proceeding to the next round would have converted everyone in not(S) to one lower priority level. So if there are two priority levels with nm at least 2m-1 (call them ma and mb) then Arcsin can win: put 2ma-1 priority ma people in S and pub 2mb-1 priority mb people in not(S). Or if there's a priority level m with at least 2m-1 people and two different priority levels mc and md with 2mc-2 and 2md-2 people, then Arcsin could make S be 2m-1 people of priority m so Arctan has to eat them with everyone else in not(S) and then could with because mc-1 and md-1 now both hold 2m-1 people. In general terms, you can calculate the vector where each term is given by (m - log2 nm), and if there is any series of terms with at least one value less than or equal to 1, at least one value less than or equal to 2, at least one value less than or equal to 3, etc eventually reaching some point where there are two or more values less than or equal to that number, then Arcsin could win. That vector can come in handy, so I'll define it as a term for the next paragraph. Since using "l" or "o" from "log" could get confusing: gm = m – log2 nm, and call gm undefined if nm = 0 and omit such terms from the vector. Now consider cases where there's not a series of gm terms going 1, 2, 3, …, x, x to give Arcsin a win with that approach. Suppose the gm vector looks like (3, 2, 3, 2, 3, 3). Arcsin could put everyone from both priority levels with gm = 2 into S and put everyone from the four priority levels with gm = 3 into not(S). If Arctan doesn't eat S, then there would be two priority levels with gm = 1 in the next round so Arctan would lose, so Arctan must eat S. Then the next round would have four priority levels with gm = 2. Arcsin could put everyone from two priority levels in S and everyone from two other priority levels in not(S), so no matter what Arctan does there will be two priority levels with gm = 1 in the next round and Arcsin could win. Stated in more general terms, you can take the gm vector and calculate a new vector ngm which is a count of all gm terms over the range (m-1 to m] (excluding m-1 but including m), and if ngm is at least m, or if there is a series of terms ngm – m incrementing from 1 upward until there are two terms of the same value, then Arcsin could win.   After that point, it gets too abstract for me to continue comprehending.
  7. Engineers vs Managers

    Question #2 Spoiler   In general, consider a group with an equal number of engineers and managers. The engineers will, of course, always act as if the engineers are engineers and the managers are managers. If the managers always act as if the managers are engineers and the engineers are managers, I suspect you would have a system with two groups that behave symmetrically with respect to each other and thereby with no way to differentiate the two groups. For the specific case of applying the method for question #1 to case #2: I think the queues will always end up emptied if the managers always play correctly. Any time a manager is preceded by an engineer, he'll incriminate the engineer to get them both thrown out. If there are a string of consecutive managers (not at the beginning of the queue), the first one will incriminate the engineer immediately ahead of him, so once those two are out of the queue the next manager will be asked about the person who was ahead of the engineer that just got removed. So a string of multiple managers can remove a string of multiple engineers and vice-versa.    
  8. Three-birthday paradox

    This is a way to calculate probabilities exactly (if you're willing to ignore leap years), although it uses an iterative approach rather than providing a closed form solution. Spoiler Define S (singleton) as the number of birthdays that are held by at least one person in the room, D (duo) as the number of birthdays that are held by at least two people in the room, and T (triplet) as the number of birthdays that are held by at least three people in the room. Note that if a birthday is held by two people, that birthday will be counted in both the S and D terms. A room with N people and no more than three sharing a birthday can have its state described as a vector (S, D, T) where S + D + T = N. If you start with an empty room and add one person, you go from state (0, 0, 0) to state (1, 0, 0) with 100% probability. If you have one person in the room already and you add a second person, you go from state (1, 0, 0) to state (2, 0, 0) with probability 364/365, or you go from state (1, 0, 0) to state (1, 1, 0) with probability 1/365. In general, if you add one person to any state (S, D, 0) then you have a D/365 chance of forming a triplet, an (S-D)/365 chance of adding a new duo from a singleton, and a (365-S)/365 chance of adding a new singleton. Using this approach, you can make a spreadsheet where the rows represent S and columns represent D. In Excel or Open Office, use column A to label the rows (S) from 0 upward and row 1 to label the columns (D) from 0 upward. In cell B2 put 1 to represent 100% chance of having state (0, 0, 0) if no one is present. In each cell in column B below B2, put the formula representing getting an new birthday added from the state immediately above, which would be (365-D-S)/365 times the probability that you were in the preceding cell to begin with, so in spreadsheet format would be cell B3 with B2*(365-A2)/365 (remember that in column B the value of D is zero so it doesn't need to be included in the formula) and copy/paste that into all of the lower cells in column B. Since we know the answer to the famous two birthdays question is 23, the state (23, 0, 0) ought to have probability less than 0.5 if you set the spreadsheet up right. For the other cells in row 2, put zeros because you can never reach a state where D>S. In all other cells, the probability that they will be occupied is the sum of (the probability of having a singleton added from the state immediately above) + (the probability of having a duo added from the state to the left), which for cell C3 would be B3*(($A3-B$1)/365)+C2*((365-$A2)/365) which can be copy/pasted to all other cells. Note that the cells only represent states with probability (S, D, 0). What happens if you have a triplet so T is greater than 0? If that happens, the probability is no longer accounted for in the spreadsheet. And that means you can take the sum of probabilities along a diagonal from lower left to upper right to find the probability that out of N people there are no triplets that share a birthday. Alternatively, if you're a real geek, you can instead use the following perl code so you don't have to sum along a diagonal in a spreadsheet. The probability of having no triplets drops to 48.89% when you have 88 people. #!/usr/bin/perl use strict; use warnings; # The @vector is an array of arrays such that # $vector[S][D] will hold the probability of getting # to state (S,D). # This initializes @vector and makes $vector[0][0] be 1. my @vector = (); $vector[0] = [1]; # $N is the total number of people in the room my $N = 0; # The variable $sum is the sum of all probabilities # where N people are in the room and the state # can be desribed as (S, D, 0), which is the # probability that there are no triplets. The loop # will stop when the $sum for a given value of $N is # less than half. I would have initialized it within # the do-while loop but then it wouldn't work as an # until criterion :( my $sum = 0; do { $N++; $sum = 0; # Because of the way the array-of-arrays is set up, # this will have to push zeros to the end of previous # arrays when you add new potential values for D. # I'm overkilling with this by including impossibly # high values of D because I'm lazy. # Just ignore this unless you're a hardcore perl geek. for (my $i=0; $i<$N+1; $i++) { $vector[$i][$N] = 0; } # Make $vector[N][0] be $vector[N-1][0] times # (365-(N-1)) $vector[$N][0] = $vector[$N-1][0] * (365-($N-1))/365; $sum += $vector[$N][0]; # Next consider the $vector[N-D][D] terms for (my $D=1; $D < 1+$N/2; $D++) { my $S = $N-$D; $vector[$S][$D] = $vector[$S][$D-1] * ($S-($D-1))/365 + $vector[$S-1][$D] * (365-($S-1))/365; $sum += $vector[$S][$D]; } } while ($sum > 0.5); print "Stopped at N=$N with probability of no triplets $sum\n";    
  9. Diamond Tiling

    Spoiler Consider the top row of triangles in the hexagon. It has four triangles in a V and five triangles in an A orientation. Consider the four triangles in a V. They obviously can't be part of the blue shape since the other triangle that would contribute to the blue shape isn't present in the hexagon, so they must be part of a red or green shape. Those four red/green shapes that occupy the Vs must also occupy four of the A triangles in the top row, so only one of the A triangles could be involved in forming a blue shape. Then consider the second row of triangles, with five triangles in a V and six triangles in an A orientation. Since there could be no more than one blue shape in the preceding row of triangles, only one of the five V triangles in the second row could be involved in a blue shape and the other four must be involved in red or green shapes. And therefore at most two of the six triangles in an A orientation can be part of a blue shape. This process continues for each subsequent row. So the first row has no V triangles in a blue shape and (at most) one A triangle in a blue shape, the second row has at most one V triangle and two A triangles in a blue shape, etc. Also note that a symmetric argument works on the bottom row. Putting everything together, the maximum total number of triangles involved in forming blue shapes would be Row 1: 1 triangle (an A) Row 2: 3 triangles (a V and two As) Row 3: 5 triangles (two Vs and three As) Row 4: 7 triangles Rows 5-8: mirror image of rows 1-4 And the total is 32 small triangles involved in forming blue shapes, which is 1/3 of the total hexagon. Symmetric arguments apply to the red shape if you consider the upper left and lower right borders like we considered the top and bottom borders for the blue shape, and if you consider the upper right and lower left borders for the green shape. So each of the three colors can occupy no more than 1/3 of the hexagon.    
  10. Bus Schedule

    After further consideration Spoiler Consider a person waiting at the bus stop, and define the time that they arrive at the bus stop and start waiting for a bus as time 0. The odds that any given bus will arrive at the bus stop at any given time x between 0 and 60 minutes is uniform. The odds that you will arrive at the bus stop before time x (and therefore be able to ask the person how long they've been waiting) is x/60, and the odds that all three of the other buses will come after time x (therefore meaning the person will get on a bus at time x and not get on an earlier bus) is [(60-x)/60]3. So the odds of all of those happening is proportional to x(60-x)3 / 604. To turn that into a probability density function, scale it so the integral from 0 to 60 minutes is 1. So to scale it to make the integral be 1 you should multiply it by 1/3 and you have the probability density function P(x) = x(60-x)3 / 3*604 To find the average wait time, integrate x*P(x) dx    
  11. Bus Schedule

    planglazed is right. Spoiler Assume you show up at time t=0 and there will be four buses arriving at random times between t=0 and t=60 minutes. What are the odds that you will be waiting at least x minutes for any given value of x? Arbitrarily number the buses #1, #2, #3, and #4 (the numbering does not necessarily correspond to the order in which they arrive). The probability that bus #1 doesn't arrive within x minutes is (60-x)/60. And the probability that all four buses arrive later than x minutes is [(60-x)/60]4. The probability that any one of the four buses will arrive before x minutes is one minus that, so 1 – [(60-x)/60]4. "The probability that a bus will arrive within x minutes" is equivalent to "the integral of the probability density function of the next bus arriving at a specific time, integrated over the range from 0 to x minutes". So the probability density function is that thing's derivative, which is d/dx 1 - [(60-x)/60]4 define u = 60-x, du/dx = -1 du/dx * d/du 1 - u4/604 4 u3 / 604 4 (60-x)3 / 604 To find your average wait time (ignoring all the other people waiting at the bus stop and their wait times for now), take the integral of (the wait time) times (the probability that you will be waiting for that period of time) over the range from x=0 to x=60. Your average wait time is twelve minutes, but now how about all of those people that you met at the bus stop and asked how long they've been waiting? Since you will arrive at the bus stop at a time that can practically be considered as random, you can consider yourself to be a fifth bus, and ask how long people will be waiting until one bus out of five would arrive. Without going through all that math again (I'm, uh, intentionally leaving it as an exercise for the reader (and not eager to deal with the math involved in accounting for the fact that the wait time of everyone else waiting for a bus when you arrive will affect the probability that a bus will arrive for you within a following period of time)) it's probably somewhere in the ballpark of 10 minutes, so the average total wait time of everyone you interview would probably end up being around 20-24 minutes.  
  12. Engineers vs Managers

    Answer to question 1   Spoiler Starting at person number m=2, ask person m whether person (m-1) is an engineer. a) If m says (m-1) is an engineer, consider m to be vouching for the engineerhood of (m-1). Increment m by one and continue the process, so for example the second question would be to ask person #3 whether person #2 is an engineer. b) If m says (m-1) is not an engineer, then you know for sure that m and (m-1) can't both be engineers, so throw out both m and (m-1). By throwing them both out, you will have reduced the number of managers by at least one, and you would have reduced the number of engineers by no more than one, so the engineers will still be in the majority. After you've thrown those two out, when you increment m by one and ask the next person whether the preceding person is an engineer, since person (m-1) has been thrown out, ask them whether the person with the largest number less than m that's still remaining is an engineer. If everyone with a number smaller than m has been thrown out (such as if you start off by asking person #2 whether person #1 is an engineer and they say no so you throw them both out and proceed to m=3) then skip that person and increment m by one again and ask them whether the preceding person is an engineer. At the end of the process, you will end up with the remaining people forming a chain where each remaining person vouched for the person with the number just smaller than theirs. You know that there cannot be a manager that is immediately followed by an engineer, or else the engineer would have ratted out the manager and you wouldn't have a chain of people vouching for each other. A manager could be within the group, but only if they're followed by another manager so they wouldn't be called out. So if there are any managers, they must occupy consecutive spots at the end of the queue. Since there are still a majority of engineers, you know that the entire first half of the remaining people must be engineers.  
  13. Biggest splash

    How did you get that Vsw decreases with R in the Iceberg 2 regime? It's true for the case of theta = 15, but I don't think it's true in general -- in the extreme case of a goblet with theta = 89.999 degrees (where the goblet is very close to a flat plane) I'd imagine a sphere with a large radius would get more underwater than a sphere with its center below the surface.
  14. I'm not a pirate

    Although I'm not that, it's probably worth noting that that answer seems to have picked up on the "thieving hand now masked" clue.
  15. Biggest splash

    This is a solution if the sides are 30 degrees away from vertical. There is another way to go about solving it that makes it more tractable if you have an arbitrary angle away from vertical, but I found this result particularly interesting. Unfortunately spoilers are still not working at the moment and figures can't be hidden by making them white, but at least the text will be hidden as white so you can click here and press Ctrl-A (or whatever the equivalent on a Mac would be) if you want to make it visible.