bonanova

The first mailing is not secure. A thief could steal your lock.

In certain set-up configurations, Y-san and taliesin can catch the ant with 2 helper spiders for a cost of $20. With complete generality, flowstoneknight can catch the ant with four helpers for a cost of $40. Is that the cheapest solution? BTW, if you want to set things up, why not just wait until Spidey is on one end of the edge the ant is on and then have him call a buddy to come sit on the other end? That costs only $10. OK, so much for set-ups. General case now. How deep does Spidey have to dig to get that ant?

What the heck, let's rearrange all the letters Ahh but now I'm getting silly.

Nicely worded.

I like answers; I love proofs.

Hi nobody, Here's a way to see past the "may need infinite" difficulty that you allude to. It's counterbalanced by the equally infinite number of trials you will have to perform before the infinitely long trial occurs. That is, the infinite difficulty is tempered by the infinitessimal probability that the difficulty will occur. So averaging over all possible trial outcomes can still give you a finite result even when some outcomes are very unfavorable. Also, take the argument that "if you bet against someone who claims he can show all six numbers in say 20 rolls of the die you will lose money" to show that the average number has to be less than 20. p.s. don't take that bet. Hope that helps. - bn

A spider is trying to catch an ant. They are constrained to walk on the edges of a cube. The ant can move up to three times as fast as the spider can. After trying, and failing, to catch the ant, the spider text-messages some of his buddies for help. The cube becomes swarmed with spiders, and eventually the ant becomes dinner. Poor ant. But suppose his buddies were to charge him $10 each [in spider currency,] and the spider was trying to minimize his expenditures. What is the least the spider would have to spend to ensure the ant is caught?

For those who have difficulty with what "on average" means, use this meaning. Roll a single die until you have seen all six numbers, and write down the number R of rolls it took. Repeat the process a large number of times and take the average of R. That gives you a good estimate of "on average". And you can see from this description that the average [not necessarily an integer] exists and is finite. The average can be computed based on probabilities for a fair die, as Chuck Rampart did. Some responses answered this question "How many times must one roll a single fair die to be certain all six numbers show?" That's a different question. No finite number of rolls will insure that. The die could come up "1" on every roll, for example. The difference between certainty and expectation of an outcome based on chance is basically this. If you bet money on outcomes that have a favorable expectation, then over time you will win, and v.v. In this case, take CR's answer [it's not an integer]. If you bet that you would show all six numbers in the next higher integer rolls of a die, you will win money over time. If you bet that you would show all six numbers in the next lower integer rolls of a die, you will lose money over time. Hope that helps explain logical expectation in practical terms.

The question really has to do with filling the entire plane with zeroes or eights. I made it more visual, with a sheet of paper using and infinitely "sharp" pencil point, allowing the dimensions to be arbitrarily small, as largeneal points out. It doesn't matter whether you think of the paper or plane, so feel free to couch your explanation for either case. Is it possible that the counts for zeroes and eights are not the same?

Hey Ploper - good to see you again. taliesin, Yes, I meant disjoint figure eights: their lines may not intersect nor touch.

On average, how many times must I roll a fair six-sided die before all the numbers 1, 2, 3, 4, 5, 6 show up?

Hi ElSkins, and welcome to the Den!

So I met this girl named Lisa, and we've gone out a couple times. But, being a true puzzle geek, only a couple of times: basically I still have no social life. Which probably explains why I spent last night doodling on a sheet of paper. I started off with zeroes, since those poor fellows sometimes do need a little affirmative action. And I crammed as many zeroes as I could onto my paper. But, even for a puzzle geek, that soon became boring. So, I glued two of them together and made an eight! I thought briefly, but ultimately rejected the notion, of writing up a patent disclosure. And I began to imagine cramming my paper full of figure eights. Then I wondered: can I fit as many eights on the paper as I can zeroes? Now I ask you the same question. Assume your pencil is infinitely sharp: that the lines which draw the figures are mathematical lines - they have zero width. Edit: The figures are disjoint: the lines may not intersect nor touch.

Thought they'd take longer to find - nice going, MJJJ

Yeah - and we used different colors for circles. So maybe we should swap - red is more female and blue more male?

If groups of parts are in, you'd have to include "Man".

CL's list has everything found so far - still three to go!

Edit - Woon and Andromeda list two others, leaving three to find.

Missed it Angle MAB = 60 [equilateral => equiangular => 60] Angle EAB = 70 [given] Angle LAM = Angle EAN = Angle EAB - Angle MAB = 70 - 60 = 10.

That's the solution. The cube is cut into four pieces. Nothing unfair about it ... consider: With 3 cuts I can divide a pizza into six pieces. Now only the first cut actually cut the [original, uncut] pizza. The following cuts cut the pieces that resulted from the first cut. But you'd still say I cut the pizza into 6 pieces wouldn't you? Also, the OP did not say that only two cuts could be made. So no rearrangement was really necessary, if you don't like rearrangements. The first cut makes two pieces. They can then be cut, once each and without rearrangement, into the final four pieces.