bonanova

You must start with a deck that contains a triangular number of cards. From a standard deck, the largest such deck would contain 45 cards. You keep the cards face down throughout, so it does not matter which cards are selected. The game begins by dealing the cards arbitrarily into a number of piles: anything from one pile of 45 [say] to 45 piles of a single card each. Play continues as a series of moves. A move consists of gathering one card from each pile to form a new pile. The piles are not in any particular order, but one card must be harvested from each pile during a move. Piles with only one card are destroyed, but each move creates a new pile. The question concerns the end configuration, if one exists, after which any further moves do not change the number of piles nor change the number of cards in the collection of individual piles, without regard for which pile has what number of cards. i.e. four piles with 3 4 7 5 cards or 4 5 7 3 cards are considered the same configuration. First, does the configuration eventually become stable? OK, there wouldn't be a second question if the answer were No; so as a mental exercise imagine, before trying, what the stable configuration would be. Then play a game to confirm it. Second, using a deck of 45 cards, what is the maximum number of steps required to reach the stable configuration, and what initial configuration of piles produces it? It may be easier to work this out on paper using state trees than to deal the cards. Extra credit: provide the second answer for all the triangular numbers possible with a standard deck. Enjoy.

My comment is probably too far fetched to spend much time on:

Well to be fair, Cavalieri's principle was the calculus dodge for the two cylinder calculation. It's a near cousin to the integral calculus. The three cylinder problem might succumb in a similar way.

Exactly right, both. Captain, I held off with congratulations on the off chance someone would succumb to the possible enormity of the ball count and make a case for the other side. When i sketched an array of balls it began with a long line and morphed, successively, into an enormous square, a massive cube, humongous hypercube, and so on. The number of dimensions of the unbounded hyper cube itself is unbounded. It might have overwhelmed one's reasoning. But not here. Not with these puzzle solvers. Bravo. Now for a hydrophyllically encrypted summary of the result; with extra points for explanation: China bests Japan. Clue: not a reference to the Olympic Games.

? The balls in the bucket decrease in number by one. There initially were no 0-numbered balls, so Alex wasn't given an infinite bucket of 0 balls.

What he said.

A while back we discussed the interesting process where each step consisted of placing consecutively numbered balls into a bucket, two at a time, and then removing the lowest-numbered ball. We postulated an infinite number of balls being processed in this manner. Completion was "ensured" by executing each step after an interval of time that was one-half of the interval of the previous step. An Infinite number of steps could thus be executed in finite time, after which the bucket could be inspected. The puzzle was to describe the final contents of the bucket. The unintuitive result is that it would be empty. Even though after each step the number of balls it contains increased by one. The interested reader can enjoy my misguided discussion of the outcome here. We propose here a different kind of very long task, also involving placing and removing balls into and from a bucket. Completion is again the central issue, and the bucket's final contents if it's the case that the task does in fact complete. Nothing in this task is infinite, but it is unbounded. You are given a bucket that contains an arbitrarily large number N of balls, numbered 1-N. You are to empty it, by a series of steps described below. To make the task interesting, an adversary named Alex tries to prevent you from emptying the bucket. Alex is well equipped to prevent your success. For each numbered ball initially in your bucket, Alex has another bucket with an infinite number of balls marked with the same number. That is, if your bucket initially contains a ball that is numbered 37, Alex has his own bucket with an infinite number of balls also numbered 37. Here is how the steps are executed. You remove a ball of your choosing, say its number is p, and discard it. Alex may then add to the bucket an arbitrarily large number of balls numbered p-1, using his infinite bucket of p-1 balls. An individual step can thus add to your bucket a set of new balls of size equal to the number of electrons in the universe. Moreover, your bucket might have started with that many balls initially. But even if your bucket starts with as few as 2 balls, it is not possible to place a bound on the number of steps needed to empty the bucket. So here's the question: Can you empty the bucket?

I thought about the point you're making, but I didn't come to precisely that conclusion. A fair coin requires ony a single flip to give a p=0.5 outcome. By comparison, this scheme can take an arbitrarily long sequence of flips. But only in a vanishingly small fraction of cases. That's the only "finite N" limitation that I see. Almost every "finite N" sequence of flips is sufficient to produce the desired result. As to whether the scheme produces the exact bias of p, I think it does. You'd have to run the scheme an arbitrarily large number of times to reconstruct [verify?] the value of p that defines its bias, just as you'd have to do with a fair coin to prove that it's fair. But the bias produced by the scheme, each time it's run, is precisely p.

That'll do it. Nice.

I'm locking this thread because I posted it previously in a and where further solutions can be placed. Sic transit gloria memory

To revive this thread by means of a duplicate post was an oversight. Nevertheless it contains the most discussion [the duplicate thread has none.] One condition for the solid is that it be convex. This rules out the zero-volume shapes. The volume that has been put forward as the largest is convex, but in fact it pertains to the convex solid with the smallest volume! If it helps, the shape of the largest solid is the simplest, and has a volume that's larger by almost 1 in3.

Bingo. Nicely done. Now we'll just put red squares anywhere on the table, (i.e. let them move apart,) but not overlapping each other. Can the blue square touch (overlap) even more red squares? This is counterintuitive, which makes it a good puzzle.

Bravo, for accuracy and beautiful symbols. Two coveted stars!

Good going, but so far no one has the answer. Clarification about overlap and touching. Apologies to unrealdon. I used "touch" for red-blue contact, when I meant to say overlap. I'll edit the OP accordingly. Touching at the edges does not count as overlap. The red squares cover the table without overlap. The blue square must overlap a portion of some number of red squares. That is, blue and red can't just touch at their edges.

It's not hard to visualize a convex solid that will snugly pass through each of the 2" holes shown below. The shapes are the outlines of the solid when viewed along three orthogonal axes. In fact there are an infinite number of convex solids that can do this. Describe or draw one of them. What is the maximum volume such a solid? What is the minimum volume? By snugly it is meant that the entire perimeter of each hole is contacted as the solid passes through it.

A classic problem solved by Archimedes before the invention of calculus, concerns two circular cylinders intersecting at right angles. If each cylinder has a radius of 1 unit, what is the volume of space that is common to the cylinders? For warm-up, you can figure out the answer - using calculus if you like. Here's the puzzle of the day: What is the volume of space that is common to three orthogonally intersecting cylinders of unit radius? In both problems the axes of the cylinders intersect.

A table top is tiled with red unit squares, none of them overlapping. A blue unit square is laid on top. What is the maximum number of red squares the blue square can touch overlap for some non-zero area?

After missing the boat on my own puzzle, I am pleased to be able to host this conversation.

A bag contains two blue balls and an undetermined number of red balls. A ball of unknown (red or blue) color is added to the bag. Finally, a ball is drawn at random from the bag. If the drawn ball is blue, what is the probability that the added ball was red?

In a game of chance a player wins his stake with a favorable outcome (W) and loses it otherwise (L). A famous (Martingale) gambling strategy is to double the stake after every loss. A win is thus assured, even after a (finite) string of losses. The series of outcomes W, LW, LLW, LLLW, LLLLW, ... all win the original stake. Nevertheless, the strategy loses, and it's a classic puzzle to show why. Let's change the game a bit and eliminate that approach. The player doubles his stake with every W, but must leave the game after only a single L. Note that is the only penalty for L; the player keeps his accumulated winnings. The player has no strategy: he plays until he must quit. If his initial stake is $1, and W and L are equally likely outcomes, what are his expected winnings?