bonanova

In the Land of Knights and Knaves [LKK], Knights told the truth, and Knaves always lied. When the rare averitas flu virus infected some of the inhabitants, their behavior was reversed: sick Knights began to lie, and infected Knaves had to tell the truth! Your job is to find out the ones who have been infected so they can be given the costly AFV vaccine. You've just dealt with six of them, and now a group of ten LKK citizens is brought to your office, and they venture the following information: Who are the sick ones? and who are the Knights? 01. Alan: Jack is lying, or he's a Knight; but not both. 02. Bervin: Jack is lying, or I am a Knave; but not both. 03. Clarence: Harry might be a Knave, but only if he is well. Harry is well. 04. David: Harry might be a Knave, but only if he is sick. 05. Evan: Alan might be a Knave, but only if he is well. 06. Bervin: Alan might be a Knave, but only if he is sick. Alan is sick. 07. Fred: If Harry is sick, then Jack is lying. 08. Fred: Alan is lying. 09. George: Fred is lying. 10. George: No fewer than 4 of us are lying. 11. Evan: No more than 4 of us are Knights. 12. Alan: I am a Knight. 13. Harry: I am a Knave. 14. Ingolf: None of our statements is a paradox. 15. Jack: If 2+2=5, then New York is a small city. Edit #1: Numbering the statements, to make discussion easier. Edit #2: Bervin and Clarence change their stories, slightly. [thx to s54 and wb!] ------------------------ If you like this type of puzzle, here are some others, recently posted to this forum. Knights and Knaves get the flu [bononova] Honestants and Swindlecants: The Delegation [octopuppy] Knights, Commoners & Knaves [roolstar] Four Knight and Knave problems [bonanova] knights and liars [2] [bonanova] knights, knaves and liars [bonanova] 3 Princess Puzzle [O'Beckon] The Barber of Honestants and Swindlecants [spoxjox] The liar, the truth teller....and the random answerer [Martini]

Zed: Hi, Ned! LTNS! Ned: For sure. How are things? Zed: Well for one thing, me and the missus have a family now. Ned: No! Zed: Yep, three rug rats. Ned: Is one of them a girl? Zed: I could tell you, Ned, but then I'd have to shoot you. That question is part of a different puzzle, OK? Ned: Sorry. So, how old are they then? Zed: Well, the product of their ages is 72. Ned: You know that doesn't tell me their ages, Zed. Zed: Well, their ages add up to the day of the month you were born. Ned: C'mon Zed, you're leaving me in the dark, man! Zed: Well since you asked, the youngest is in fact a girl. Ned: Thanks. Now I know their ages. Was Ned telling the truth?

Length.

Length.

Nope. When you choose a door, you know one of the doors you didn't choose has a goat. Your winning chances are, nevertheless, 1/3. You are saying that somehow, magically, if you see that goat, your odds increase to 1/2. Final try: You pick a door. Nothing else happens. You are given the opportunity to pick the other two doors instead. Do you swap? Of course. No one would argue that choosing two doors has the same winning odds as choosing one door. Ah.... but be careful; one of those other doors has a goat! Here ... I'll even show you which one. OK, you're right. No sense swapping now that I see that both of the other two doors aren't winners. If you believe that two outcomes are always equally likely, you should buy more lottery tickets.

Great discussion ... my rationale was on a much lower level. The dangers are: ravenous wolf roaring lion angry bear sure death - note: it isn't said to be imminent death, just sure death. As is commonly known, nothing is sure but death and taxes. So sure death lies on all the paths. Why add a beast to that? Go East.

The cylinder is what is removed from the sphere. Think of beads on a string. The string is the cylinder.

You're on the right track - you should struggle with the idea of a 1/3 chance changing to a 1/2 chance. Why should seeing a goat that you already knew was there affect your 1/3 chance? It shouldn't. And it doesn't. The odds are 100% that the car is behind one of the doors: Door 1: The door you picked has a 1/3 chance. The other two doors, combined, have the remaining 2/3 chance. Door 2: The opened door that revealed a goat has zero. Now you know where the 2/3 chance lies - the third door. Door 3: The third door has the remaining odds: 1 minus 1/3. Roughly 2/3. Should you make the switch? Hint: The number of choices open to you is not what is important here, because the success probabilities of the choices are not equal. Lottery tickets have only two outcomes - you win or you lose. But not with equal probability.

Very nice puzzle ...

I'm certain the question was meant to be a bit demanding. It was recalled by a friend from an MIT entrance exam. It requires visualizing the drilled sphere. Once you see it, drilled, the height of end caps is a non-issue. The hole has an unambiguous length.

1, 11, 21, 1211, 111221, 312211, _____, _______ ?

OK. We have answers for 1, 2, 4, 5, 7, 8, 15, 16. Have at it ...

Visualize the material that's removed from the sphere by the drill. It comprises two spherical caps sandwiching a right circular cylinder. Define the length of the hole to be the height of the cylinder. That's the same as if you climbed inside the hole and measured its length from end to end with a measuring tape. Note that because the end caps have height of their own, the length of the hole is less than the diameter of the sphere. Also note that as the diameter of the drill approaches the diameter of the sphere, the length of the cylinder [length of the hole] approaches zero. That is, all the sphere is removed: comprising two hemispherical end caps and no cylinder. This means that any sphere of diameter greater than 6 inches can have a 6 inch hole drilled completely through it. To understand the puzzle is to understand that it could have been worded worded more helpfully. See next. But that's part of the puzzle - to understand the condition of "6-inch hole drilled through a sphere", even tho the puzzle is worded [accurately, but] not helpfully: Drill a hole through a sphere. Make the diameter of the drill large enough that the inside cylindrical surface of the hole measures exactly 6 inches in length. Now calculate the volume of the portion of the sphere that remains = volume of: (sphere - cylinder - 2x [end caps]).

Kudos to schmod54 and octopuppy <- don't break a leg .... !. Perhaps I was too easy with this one ... There will be others.

OP does not say whose steps [officer's or thief's] measured the 27-step lead. It's reasonable to assume it was the thief's. If so,

OK, now give mine a try Tip of the hat to octopuppy for a nice puzzle.

By using a drill with a larger diameter.