bonanova

Right answer, marked solved. It's similar to DeGe's result, which is the probability that a particular point is a terminus of the arc. A very simple words-only description also gives the result. In this case it shoul explain why simulating a line segment works also when the ends are joined (circle.)

continuing... That's the right formula, and it fits plasmid's simulation numbers and k-man's insights. I'm marking it solved. If the simplicity of the expression suggests a words-only derivation, I'll hold off saying anything more.

Cool. Best so far. There is a slightly longer path. What's the fraction of diagonal moves if the dimensions are 8x2?

It seems you understand the OP. Now about the maximal path length ...

What if you picked a different starting point?

Julia, I think we're all with you on that point. It's hard to find a defining aspect that C uniquely has in common with the first three.

If you like, warm up with "What's 64 times 1?" before getting to the actual question.

Just a personal preference - arguments can be made for each.

By modifying the OP slightly I believe the four points may be completely arbitrary: Take any four points A, B, C, D in the plane, no three of which are collinear. They describe a unique quadrilateral, if we take the points as being its vertices. But they also describe a square, if we require only that the points lie on its sides or their extensions. The OP did not state, but it implied, that A B C D are taken in sequence. That is, the sides assiciated with A and C are opposite sides, (and so also with B and D) for the construction to be unique. That is, any four points, no three collinear, define two unique pairs of parallel lines, one point on each line, that intersect to form a unique square. Find a compass - ruler construction for those four lines.

The Black King sets out one day to tour his kindom, a standard 8x8 chessboard. He's not feeling well, though, and he wants to return by the shortest path to his starting square. We'll assume all squares are one unit on a side and ask, what is the length of such a trip? Wait. This is Brainden. You all are geniuses. Let's add a wrinkle. The King is actually feeling fine, and he wants his walk to provide him the maximum possiblle workout. Diagonal moves now come into play. To avoid radical complications , we'll count their length as two N-S or E-W moves. So here's the actual question: what is the maximal length of a complete King's tour of an 8x8 chessboard? Bonus points if a proof is given.

Some of the considerations discussed in that paper bear on the problem at hand. But in the spirit of the Aha! theme, let's rule out exhaustive computer constructions and go with compass and straightedge.

Yes, clearly not every set of four points (no three of which are collinear) lie on a square. Let's limit the puzzle to a set of points similar to the ones in the OP. Using standard constructions (compass, ruler) can you draw a square that has these four points (one each) on its sides?

Heroic work. Can you (aha) enumerate the equally likely cases more simply?

Consider the line segment [0, 1] and the entire real axis. Both have the cardinality C of the reals. f(x) = 1 / [ x (x-1) ] is a bijection. This is a paradox only if real numbers themselves are paradoxical.

Johnny was asked to write the sum of the two fractions p/n and q/n where p, q and n are single-digit numbers. He remembered that the numerators were to be added, but he thought maybe the denominators should be added as well. Also, he decided that putting digits next to each other could suggest addition. So his first try was to say that p/n + q/n = pq/nn. (where pq and nn are two-digit numbers.) Then he remembered the result should not depend on the order of addition, but he saw that p/n + q/n = qp/nn was just as bad. So he decided to use them both. His final representation of the sum was p/n + q/n = (pq + qp)/nn. The puzzle asks: Let p, q and n be chosen at random from {1, 2, 3, 4, 5, 6, 7, 8, 9}. What is the expected error in Johnny's formula?

From the set of integers {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} I randomly choose an element, say 3. I subtact 3 from 10, getting 7. Now I have the set {1, 2, 3, 4, 5, 6, 7}. I choose another element at random, say 5. I subtract 5 from 7, getting 2. Now I have the set {1, 2}. I randomly choose one of these elements, say 1. I subtract 1 from 2, getting 1. Now I have the set {1}. I randomly choose one of these elements. It turns out to be 1. I subtract 1 from 1, getting 0. Now I have the empty set. Each step took away a nibble, leaving a smaller set. This example nibbled a set of 10 elements down to zero in four steps. Starting with a set of p elements, what is the expected number n of nibbles required to empty the set?

What is the probability that n points placed at random on the circumference of a circle all lie on a semicircular arc? That is, for at least one of the points, all the others lie [0, pi] radians from it.

Agree with R_G.

m00li's post #3 answers the OP. Post #4 stated two objections. Let's look at the objections. [1] Differences between 5 and 6. OP asks for proof that in a group of 6 people condition A or condition B exists. Post#3 proves that if A or B does not exist the group can have no more than 5 people. "Not (A or B) implies less than 6" is logically the same as "6 implies (A or B)." [2] "Mutually unacquainted" is different from "not mutually acquainted." Post #3 (and post #7, but post #3 was first) assume they are the same. If there is a third possibility then no proof exists. To show this, let's construct a third, asymmetrical relationship: Let's say that I am acquainted with someone who is not acquainted with me. Say a groupie (not the fish) is "acquainted" with a Rock Star who is not acquainted with the groupie. Since the groupie knows the star, they cannot be "mutually unacquainted." Since the star does not know the groupie, neither can they be "mutually acquainted." All that can be said of them is that they are "not mutually acquainted." Using "mutually unacquainted" and "not mutually acquainted" as different relationships prevents the requested proof. Here's how that might play out. Among three rock stars, Elvis knows Mick and Bob, but Mick and Bob are strangers. Among three groupies, Sam knows Pete and Jane, but Pete and Jane are strangers. Sam, Pete and Jane, being groupies, all know Elvis, Mick and Bob None of Elvis, Mick and Bob know any of Sam, Pete and Jane. No three of six are "mutually acquainted" No three of six are "mutually unacquainted" Ergo, with that interpretation there is no proof. Mick, Bob, Pete and Jane, however, are "not mutually acquainted." With that interpretation the proof exists.

That is amazing.