bonanova

Hint: Solve these problems first: Break the stick at two random points. Break the stick at a random point. Break one of the pieces at a random point. These can be solved in closed form; are they the same? Verify the answers by simulation if you like.

I would agree.

Does the game end the moment the cards in one hand say so? Are the groups of three cards dealt simultaneously? Else how does seat position come into play?

There is a standard problem that asks, if you randomly break a stick into three pieces, the probability that the pieces can form a triangle. The answer is not unique because there are different ways to randomly create three pieces. The method that is the most challenging to analyze breaks the stick at a random point, randomly chooses one of the pieces, and breaks it at a random point. If you can solve that puzzle you may have insight to solve the following variant: Break a stick at a random point. Break both pieces at random points. Randomly discard one of the pieces. What is the probability the remaining pieces can form a triangle?

Well we've finished the discussion of the puzzle so I won't spoiler this. You made several points and raised new questions. I'll try to reply to the extent that I know. Inarguably the cube contains points that its edge does not. For finite sets that gives a Yes answer. For infinite sets, and these sets are uncountable, we express size by cardinality. The edge has the cardinality of the reals, so the question really asks the cardinality of the cube. Bushindo's bijection expresses Cantor's finding that geometric figures of all dimensions, arbitrarily large ones, have the cardinality of the reals. So the cube has other points, but not more points. The power set is the set of all subsets. Is there a bijection between the elements of a set and the elements of its power set? The answer is No, and so the cardinality of the power set is greater than the cardinality of the set. For finite sets, it can be seen by inspection. The set {1,2,3} has 23 = 8 subsets, including the empty set. No bijection exists. For infinite sets, the proof is almost as simple. It's a proof by contradiction. Suppose the elements of an infinite set have been paired with the elements of its power set. We then color the set's elements blue that are paired with a subset that contains the element, and red if they are not a member of their paired subset. Now consider the element of the set that is paired with the subset of all the red colored elements. Can it be red? No, because red elements are paired with subsets that do not contain them. Can it be blue? No, because blue elements are paired with subsets that do contain them. So the subset of all red elements cannot be paired, and this contradict the premise of a bijection. So power sets have increasing cardinalities. We know the integers are countably infinite, and the reals are not; and we have an understanding of both these sets. Do we have an understanding of a set that has cardinality Aleph2? One such set is the set of all functions of real numbers. Or of all curves in a plane. No one has suggested a set of cardinality Aleph3 other than the power set of the functions of a real number. What is AlephnAlephn? I would say it lacks any recognizable similarity with a power set or a real number raised to the power Alphan because the infinite cardinals are not algebraic. So it's a symbol raised to its own power. What value or significance we could thus assign to it I do not know.

OP says "Without bending, breaking or overlapping dowels, only connecting them at the ends ..."

And has the title become germane?

Meaning that one player will eventually run out of his H or V moves and lose. Edit: there is a fixed number of moves, but how they distribute between H and V moves depends on how the game progresses. At some point all the remaining moves will be H moves and the V player loses, or all the remaining moves will be V moves and the H layer loses.

So this brunette walks into the Doctor's office and says, Doc, something is terribly wrong -- I hurt everywhere! What do you mean? the Doctor asks. Well I hurt here, she said as she touched her head, and here, touching her knee, and here, touching her shoulder, and here, touching her stomach, and here, touching her elbow, and .... OK I understand, said the Doctor. I'm scheduling you for a comprehensive set of tests immediately. Two hour pass, the results are in, and the woman is back in the Doctor's office. The Doctor approaches her and asks, You're not really brunette, are you? No, she admits, I'm blonde. How did you know? You have a broken finger.

OP says: One of us will break a piece of the bar along a vertical line. The other will break a piece along a horizontal line.

Tres kewl!

Nice puzzle. I'm in the dark. What is the back story?

Yeah, true. For instance, if my (first) move was to break on the third vertical line, as described in OP, what (horizontal) second move would win for you?

mmiguel, would you say there are different cardinal numbers for one edge, 12 edges, faces, solids? BTW just to clarify, I mean by cube to denote a solid, as per EH's reasoning. Any other views?

I unwrap a Hershey bar and notice it is 7 squares wide and 4 squares high. I propose a game. One of us will break a piece of the bar along a vertical line. The other will break a piece along a horizontal line. We alternate moves until one of us has no remaining moves. That player loses the game. As an example, the first to move might break the bar along the third vertical line, leaving a 3x4 rectangle and a 4x4 square. The second player to move might break the square along the topmost horizontal line, changing it into 4x1and 4x3 rectangles. And so forth. Someone will ask: can you rotate the pieces? The answer is No. Although it is permissible to eat the Hershey Bar and play the game with pencil and paper. Here is the puzzle. You may choose the orientation of your breaks or which of us moves first. Choose wisely and you can force a win.

Pi vs e Pi goes on and on and on . . . And e is just as cursed. I wonder: Which is larger When their digits are reversed? - Martin Gardner

Consider the unit cube x, y, z = [0,1]. Does it contain more points than its edge x = [0,1], y=z=0? Edit. Let's state it simply: Does a cube contain more points than are contained in one of its edges?

Ever ask two questions then realize they answer each other? I'm not good at this type of puzzle -- math and logic is more to my liking. But thought I'd look into them. Part of the solution is to discover what to look for.

Yup! Nice.

The title of this puzzle was a bumper sticker's humorous way, a while back, of expressing the famous 2nd law of thermodynamics. Entropy is a measure of disorder, and the Law states that it always increases, when properly accounted for. Examples are shuffling a deck of cards, shaking salt and pepper in the same shaker after carefully placing all the salt on top, and the ongoing need to straighten up the house. Even if I used tweezers to replace each grain of salt, or restored the cards to suit order, or cleaned the house, the Law still holds. The system would then have to include me. And, sad but certain to say, my entropy would have increased more than the decrease in entropy I would have produced in the cards, salt or house. The gloomy fact is that the Universe is running down. In the struggle between order and disorder, disorder wins out. In my early years that seemed unfair to me. I thought it should at least be a draw. You should only have to clean the house once. Unless there are small children about ... But it's not so. So here is the puzzle. Does the Second Law give evidence to a universal evil force that wanders about, looking for ordered states to destroy? Or is there a view that permits neutrality in that regard, while still asserting the Second Law? Well of course there must be a view like that, because that's how things actually are. The puzzle is, how would you simply and intuitively state such a view? How would you explain to my younger self, or to your child or grandchild, that fairness [neutrality] and the Second Law can and do coexist? Coveted bonanova Gold Star goes to the most concise answer. Enjoy this one!

Good puzzle. I didn't reply very quickly.

If I understand their posts correctly, and I may not have, sp's expression gives 5, and PT's algorithm gives 6, when, e.g., m=5 and n=7. Neither seems possible without stacking, or, equivalently, holding two or more pieces together next to each other, so as to keep the configuration planar. OP asks for the minimum straight-line breaks, simultaneous or not. So as an aid to counting, it disallowed stacking. Break a piece. Repeat. Count the steps.