plasmid

I think that Wilson is closer to the mark, but

has apparently found a bug in the way polling handles poll options with simply a space; unfortunately this seems to be only an annoyance that doesn't have any real potential to be expoited toward humorous effect

the poll must have disappeared after some editing, putting it back up

While that's what it's written to sound like, the title of the riddle rules out anything having to do with a scurvy buckaneer or the like, as is the typical way of my riddling style A mug o' grog for the sea dog with the lance!

Hm, it seems like a fair number of people (Molly, Slick, me, probably others) would like to play but don't have time to log in very regularly, at least not enough for typical role playing. Think it would be possible to design it in such a way that weekend warriors could play a role, either as a semi-recurring character, or a shady bad guy that occasionally looks for a way to throw a wrench into peoples' plans, or a cat that spends most of its time off hunting mice but occasionally intervenes in the affairs of the larger and less furry creatures?

You are correct... in that it doesn't really explain the plank clue

A blade to my nose is applied With plank o'er abyss to my side Urged foward to creep Then finally to leap Pray fins could preserve my poor hide

My earlier solutions showed that a method that distributes points evenly will place them on the x-axis with probability distribution p(x) = 2-2x. The probability that a randomly selected point will have an x-coordinate less than some value 'a' is the integral of this; Int[p(x)] (x=0 to x=a) = (2a - a^2) The important concept here is that while these formulas do not describe the only way to generate such a distribution, they do describe characteristics of a uniform distribution that must be satisfied by any solution that creates a uniform distribution. Now to test the proposed solution: x = -log(xseed) / (-log(xseed) - log(yseed) - log(zseed)) where nseed is the "seed" random number between 0 and 1 that is used to generate coordinates. First, this can be simplified. x = -log(xseed) / (-log(xseed) - log(yseed) - log(zseed)) x = log(xseed) / (log(xseed) + log(yseed) + log(zseed)) x = log(xseed) / log(xseed * yseed * zseed) If this were simply a function of xseed then the probability that this will be less than some value 'a' would be given by solving for the values of xseed that would produce a value of x less than 'a'. Note that as xseed increases, x will decrease, so x will be less than 'a' for any xseed greater than the xseed that generates x=a. So (the probability that x<a) is equal to (one minus the xseed that generates x=a). Solving for the xseed that generates x=a, log(xseed) / log(xseed * yseed * zseed) = a log(xseed) = a log(xseed * yseed * zseed) xseed = (xseed * yseed * zseed)^a xseed^(1-a) = (yseed * zseed)^a xseed = [(yseed * zseed)^a] ^ (1/(1-a)) xseed = (yseed * zseed)^(a/(1-a)) Now since [the probability that x<a] is equal to [one minus this xseed] (if you are given any particular yseed and zseed)... Prob(x<a) = 1 - (yseed * zseed)^(a/(1-a)) Since yseed and zseed are independently generated random numbers, what you really are looking for is the integral of this over the domains of yseed and zseed. Prob(x<a) = Int[1 - (yseed * zseed)^(a/(1-a))] (yseed=0 to 1 and zseed=0 to 1) First integrating on yseed, Prob(x<a) = Int[1 - (yseed * zseed)^(a/(1-a))] over (yseed=0 to 1) = Int[1 - yseed^(a/(1-a)) * zseed^(a/(1-a))] over (yseed=0 to 1) = yseed - zseed^(a/(1-a)) * Int[yseed^(a/(1-a))] over (yseed=0 to 1) = yseed - zseed^(a/(1-a)) * (1-a) yseed^(1/(1-a)) over (yseed=0 to 1) = 1 - zseed^(a/1-a) * (1-a) Then integrating on zseed Prob(x<a) = Int[1 - zseed^(a/1-a) * (1-a)] over (zseed=0 to 1) = zseed - (1-a) * Int[zseed^(a/1-a)] over (zseed=0 to 1) = zseed - (1-a) * (1-a) zseed^(1/1-a) over (zseed=0 to 1) = 1 - (1-a) * (1-a) = 1 - (1 - 2a + a^2) = 2a - a^2 To recap, I've just proven that using your friend's method, Prob(x<a) = 2a - a^2. And I've proven in the previous posts that for a uniform distribution Prob(x<a) = 2a - a^2. So, although this is not a very intuitive way of proving it, it does prove that the proposed method creates a uniform distribution.

Brilliant solution, Captain. I can adapt the arguments from my previous answer to show that it will work.

@plasmid: Can you generalize this to higher dimensions? Well, I think you can generalize the approach to uniformly map random numbers to a particular shape in multiple dimensions, but actually doing the calculations would be Ugly spelled with a capital Courtney Love, so it would probably not be practically useful unless you were to use approximations with analytic integration with a computer or something.

Since the site has been updated, the search engine mechanics have changed. If you simply type the words you're looking for in the search box, you might not find what you're looking for. Instead, I recommend clicking the little gear to the right of the search box to go to the advanced search page. If you remember some keywords from the riddle, type them in the top "Find words" search box, and then be sure to switch the "Display results" option near the bottom from the default "As topic lists" to instead be "As posts". It seems that the search engine will only search through the titles if it's set to "As topic lists", and will search through the actual text of the post if it's set to "As posts". Also, the site redesign made it much easier to search for riddles by a particular author. Just type the author's name in the "Find author" box and have "Display results" set to "As topic lists". But be aware that the author name is case sensitive.

bonanova's comment about being able to do this with only two random numbers as a seed forced me to come up with a general way of shifting one probability distribution into another. This approach might also be useful in other logic problems or paradoxes that rely on quantifying probability distributions. But I still have not had the patience to write an Excel sheet that tests whether that generates a uniform three dimensional distribution. Edit: fixed spoiler... er, well, I tried to

Did you have a different method in mind to convert a single random number into a three dimensional coordinate with uniform probability density? Aside from CaptainEd's geometrical method, I'm having a hard time finding a way of doing it elegantly. I'm also considering bonanova's comment about being able to transform two random numbers (a, b) into a set of coordinates (x, y, z) instead of needing to start with three random numbers. I think I have a way to do it, but I'm having trouble proving it.

After checking Wikipedia and seeing that they can build mounds and river-like trails, I think that this is so similar to what I had in mind that for riddling purposes I can call it a hit. Their bodies are not so clearly split into three segments or as typically red colored as those of an ant to fit the first stanza, but that's more a quantitative than a qualitative difference.

I'm orb of snow from grand nobility Just one among a myriad entire Then cleaved from one and now split into three No longer snow, ablaze in raging fire I've hewn a wretched mountain from the Earth Erupting flowing rivers 'cross the land I bore through skin and rob it of its worth While leaving only dust and grains of sand Approach, if found by pair who stand on guard You'll face a pair of scythes to claim your soul Yet more will rise, will surge, as one bombard So flee, forsake the place where we patrol A nuclear reaction, I am not Nor meteor that screams to Earth so hot

will hopefully come up with a better answer before the hints start

Clarification: do we start off with a series of random numbers {n1, n2, n3, ...} with, say, uniform probability distribution between zero and one and have to find a way to transform that into a set of coordinates {(x1,y1,z1), (x2,y2,z2), (x3,y3,z3), ...} that fulfills the criteria of the OP? That is, find functions f(n) = x, g(n) = y, and h(n) = z to transform between those two sets and end up with a uniform density?

If you like word riddles, just look for any riddle by shakeepuddn to find a pearl. If you're not already familiar with his style, here's one that's not too difficult (at least by his standards) to give you a taste of it. The honestant / swindlecant / randomcant that had me pulling my hair out. A "hats" problem that blew just about everyone's mind.

Aha, I knew I wasn't hallucinating it (although perhaps I was hallucinating about seeing this simple case before). Thanks for finding the hard version, k-man. lawl, I would have accepted that as a valid answer if this were one of my "How would you cross puzzle land" riddles

Indeed, the probability distribution could change things quite a bit. For example, if the host were picking the players' hats out of a bag of N white and M black hats... The intended probability distribution is a 50% chance of receiving either a white hat or a black hat for each of the players, independent of the hat colors of the others.