bonanova

Kudos to TSLF and the solvers. Great puzzle.

Hi dtdt, welcome to the Den. Share the numbers with us and we'll do our best with them. The title has XM in it. Are they the [roman] numbers?

Inspector Craig of Scotland Yard was called to Transylvania to solve some cases of vampirism. Arriving there, he found the country inhabited both by vampires and humans. Vampires always lie and humans always tell the truth. However, half the inhabitants, both human and vampire, are insane and totally deluded in their beliefs: all true propositions they believe false, and all false propositions they believe true. The other half of the inhabitants are completely sane: all true statements they know to be true, and all false statements they know to be false. Thus sane humans and insane vampires make only true statements; insane humans and sane vampires make only false statements. Inspector Craig met two sisters, Lucy and Minna. He knew that one was a vampire and one was a human, but knew nothing about the sanity of either. Here is the investigation: Craig (to Lucy): Tell me about yourselves. Lucy: We are both insane. Craig (to Minna): Is that true? Minna: Of course not! Did Craig learn the vampire's identity?

Can we conclude that a request for confirmation of a premise by a type A questioner is equivalent to the assertion of that premise by a truth teller? In answering that question assume that I am type A.

Somewhere in the vast reaches of the ocean, there is a very strange island known as the Island of Questioners. It derives its name from the fact that its inabitants never make statements, they only ask questions. The inhabitants ask only questions answerable by Yes or No. Each inhabitant is one of two types, A and B. Those of type A ask only questions whose correct answer is Yes; those of type B ask only questions whose correct answer is No. For example, an inhabitant of type A could ask, "Does two plus two equal four?" But he could not ask whether two plus two equals five. An inhabitant of type B could not ask whether two plus two equals four, but he could ask whether two plus two equals five. I once visited this island and met a couple named Ethan and Violet Russell. One morning Ethan asked me, "Are Violet and I both of type B?" Since I am not an inhabitant of the island, I may now ask you, and you now may answer, What type is Violet?

Two friends, whom we will call Arthur and Robert, were curators at the Museum of American History. Both were born in the month of May, one in 1932 and the other a year later. Each was in charge of a beautiful antique clock. Both of the clocks worked pretty well, considering their ages, but one of them lost ten seconds an hour and the other gained ten seconds an hour. On one bright day in January, the two friends set both clocks right at exactly 12 noon. "You realize," said Arthur, "that the clocks will start drifting apart, and they won't be together again until, let's see, why, on the very day you will be 47 years old. Am I right?" Robert then made a short calculation. "That's right!" he said. Who is older, Arthur or Robert?

.. Answer from a truth teller no doubt ..

You're driving from your home town to visit a friend in Elmdale. You reach a complex intersection where roads lead out in nine directions, evenly spaced, 40 degrees apart. Unfortunately, Hurricane Sandy has blown down the sign, with its nine arrows now pointing out at a random orientation. You have no other indication of which road is the one that leads to Elmdale. Fortunately, perhaps, you meet five travelers at the intersection. One of them tells you that two of the group are truth tellers, two are liars and one answers truthfully or falsely, at random. Two others of the group confirm that statement, while the other two seem disinterested and offer no comment. After making it clear to them you are looking for the road that leads to Elmdale, one of the silent travelers tells you that the group is not from the area and in fact none of the five can tell you which road that is. Suspecting that he is a liar, you start to ask the one that spoke first which road is the desired one, but the Original Poster interrupts you and says that you have used your allotment of questions. Without guessing, how do you make the right choice, and meet your friend?

René Descartes was flying home from a conference when the flight attendant asked, "Monseur Descartes, would you like a cocktail?" To which the philosopher replied, "I think not," and promptly disappeared.

An old favorite. Imagine an infinitely stretchy elastic band is tied to a tree on one end and to the bumper of a pickup truck on the other end. The truck, with an infinite tank of gas, drives away from the tree on an infinitely long straight road at a constant speed of 10 mph. Andy Ant walks with a constant speed of 0.1 mph relative to whatever surface he is standing on. When the truck has reached a point 0.1 miles from the tree, you place Andy on the elastic band at the tree and start him walking toward the truck. Does Andy ever reach the truck?

Does that put Y in the 28th position?

Another hokey approach has come to my attention Inscribe a random right triangle. Take the diameter as the chord.

Hokey, at best. I agree.

Inscribe a regular polygon of n sides where n > 1. Take the chord to be one side of the polygon; (n = 2 gives a diameter.) Let n be chosen at random. What is that average chord length?

I'm not sure you can define random in any sense other than an unbiased selection from all available instances. The challenge that arises is that several exhaustive, random [in some sense] and uncountable subsets each yield a distinct value of average length.

This is an interesting approach. The radius is a middle measure of the extremes of zero and the diameter, but It is the median, not the mean. One must represent (find an expression for) all the uncountably infinite chords and then examine the distribution of their lengths. If the distribution were found to be uniform from 0 to 2r then r would be correct. All the examples given previously in this thread, however, show the lengths not to be uniformly distributed.

You got it. Kudos for hanging in with it. The missing factor 2 comes when you recognize the integration range is 1/2. You have to divide by that to get the average over that range. I didn't notice till now that grifri238's result is a little high.. I got: After a first break at point a <.5. the second break must be in the central region of width a within the larger [1-a] piece. If you average a/(1-a) over all values of a from 0 to .5 you get -1 + 2log2 = -1 + 1.386294361... = 0.386294361... Half that is 0.1931471806... I have not analyzed the variant mentioned in OP. I'm waiting for the 'aha!' Moment that reduces it to this case.. http://brainden.com/forum/public/style_emoticons/#EMO_DIR#/smile.png

Well since I neglected to use the word "average" in the OP, another answer is anything between 0 and 2. That was the "fourth" answer.

All of the above, probably. Let's make it a competition to show the greatest and least among any reasonable definition of randomness.

What is the length of a random chord drawn through a unit circle? There are at least four answers.

Are you sure that yours and curr3ent's images are identical? Does your second question which row is mostly black, or at which row there are more cumulative blacks?

So now I have to figure out the OP--randomly break the stick, then randomly break each piece. Discard one piece. Next time... You are in the right ball park, but a little low. Ask where the longer stick must be broken given the first break was at f < .5. Then average the probability of that second break for 0 < f < .5.

Noah preceded Abraham; he was not Jewish. This topic will be moved to Others.

Good one. Yes and yes. Your approach should serve well for the second case as well.

That's close, Cap'n. 10x more cases should get you to 3-decimal-place accuracy.