Math (probability) puzzle

[joft]

Consider the set of all cubes with side length between 0 and 1, or equivalently with face areas between 0 and 1, or volumes between 0 and 1. Suppose we select a cube at random from this set.

1) What is the probability that its side length is between 0 and 1/2?

2) What is the probability that its face area is between 0 and 1/4?

3) What is the probability that its volume is between 0 and 1/8?

4) Can you explain what's going on here?

[Max™]

Wouldn't 3 be nested within 2, which nests inside 1?

If it's for "the set of all cubes with side length between 0 and 1", assuming that the smaller probabilities are within the larger set, then you can adjust the probabilities to represent the side length cases exclusively, can't you?

It would be different if you were asking for the probability of selecting say, points on cube sides with side length 0 to 1/2, OR points on face areas with 0 to 1/4, OR points within cubes measuring 0 to 1/8 in volume.

The actual number of said cubes shouldn't be too odd of a distribution should it?


Is that how it was phrased exactly?

[joft]

Why only nested?

The length of the side of a cube is between 0 and 1 if and only if the area of the face of the cube is between 0 and 1. Seems like equivalent characterizations of the same sample space.

The length of the side of a cube is between 0 and 1/2 if and only if the area of the face of the cube is between 0 and 1/4. Seems like equivalent characterizations of the same event.

[Max™]

Yeah, I was actually reading that backwards, taking it to mean cubes with side length 1/8th, 1/4th, and 1/2, but for the 1/8th volume 1/4th area 1/2 length case, it should be the same cubes, just phrased differently.

[Maxxthepenguin]

So are we supposed to assume that the probability of side length is uniformly distributed between 0 and 1? Without defining the distribution there's no way we can possibly do any sort of analysis and once the distribution is defined, everything is really fucking obvious.

How is this even a puzzle?

[CDF]

1) 1 or 0
2) 1 or 0
3) 1 or 0
4) The cube was selected already.

Waiting for a statement like "all side lengths between 0 and 1 are equally likely."

[Max™]

Yeah, I was wondering how we were supposed to choose a distribution, whether there were more members of any size or not.

[Necronus]

When you're talking about "The set of all cubes with trait between X and Y" that means not just equal distribution but actually just one of each.

If it was "A set of cubes that have trait between X and Y" then theres nothing known about the distribution.

Its a semantics thing to me, but I guess I could be wrong.

[Quicklet]

Seems pretty obvious that the probability is 1/2 for all.

[CDF]

Recall the statement: "the set of all cubes with side length between 0 and 1, or equivalently with face areas between 0 and 1, or volumes between 0 and 1." It is impossible for all three "sets" to be uniformly distributed. As currently perceived this question is not much of a puzzle, so perhaps there is more to it.

[Max™]

That's what I was trying to figure out, hence why I mistakenly read it being 1/8th or 1/4th or 1/2 side length, not volume, trying to find a "puzzle" where there wasn't one.

[joft]

The question is not well-defined, and that's the point. It's supposed to seem counter-intuitive; like the answer should be 1/2, 1/4, and 1/8 respectively even though it's just the same question being rephrased.

I'm still waiting for someone to give a cogent explanation of what's going on here. CDF is closest so far.

[Max™]

Ah, you're asking for probabilities from three different sets then?

If the set is all cubes with 0 to 1 side length AND all cubes with 0 to 1 face area AND all cubes with 0 to 1 volume, then the distribution is different compared to side length OR face area OR volume.

[joft]

A union A union A = A intersect A intersect A = A

try again

[Max™]

Let's use L, A, and V, for Cubes with Length = 0 to 1, Area = 0 to 1, and Volume = 0 to 1.

If considering V alone, it doesn't contain A, or L, though it does intersect both, L contains all three. So the probabilities for the various examples would be different if the sets are given so you sample probability of v=1/8/a=1/4/l=1/2 in V, or A, or L independently. As opposed to sampling L alone, which contains V and A.

V = 0 to 1 giving a sample (v, a, l)
A = 0 to 1 giving a sample of (V) + (v, a, l)
L = 0 to 1 giving (A) + (V) + (v, a, l)

I was trying to overcomplicate it I guess, trying to figure out why you would mention V or A when L contains both?

[joft]

they are all three exactly the same underlying set

edit: all I've done is given 3 descriptions of the exact same underlying set, and 3 descriptions of the exact same event.

[Cletis]

ummm, 42. rite?

[Max™]

Yep, I was definitely overcomplicating it, kept wanting to read them as different sets when they aren't, probably because I kept wanting to sum the faces due to the combination of cube + area + edge length bringing up 6*length^2 in my head.

[Maxxthepenguin]

they are all three exactly the same underlying set

edit: all I've done is given 3 descriptions of the exact same underlying set, and 3 descriptions of the exact same event.

No shit?

This thread is fucking retarded

[joft]

If the answer is so trivial, Mr. penguin, why not give a clear explanation for those of us who aren't as brilliant as you? Why do I want to answer 1/2, 1/4, and 1/8 to the same question stated differently?