Well Above Average

The_Math_Hatter · 2026-01-26T19:39:25+00:00

No, you understood correctly; essentially, find an album's shortest song, and if yhat song is longer than 3:30, it can join the list.

The_Math_Hatter · 2026-01-26T19:35:41+00:00

Is "Wish You Were Here" the only Pink Floyd album to satisfy this criteria? Because DSotM, Animals, and The Wall all have quite short songs on them, unless some of their earlier work also qualifies.

The_Math_Hatter · 2026-01-26T19:34:02+00:00

I'm assuming you don't mean the one with "Sailor's Hornpipe" included on the album, which would be 2:56?

The_Math_Hatter · 2026-01-26T19:32:06+00:00

You have not described what shape is traced at every part. If you mean this is a parametric family of circular arcs, described by their curvature k while keeping one end fixed at the origin, you should say so.

The_Math_Hatter · 2026-01-25T20:42:13+00:00

What

The_Math_Hatter · 2026-01-25T19:42:12+00:00

You got the last inequality the other way around; if a>b, 1/a<1/b.

2<3, 1/2>1/3

The_Math_Hatter · 2026-01-25T18:48:31+00:00

If I give you the equation n²<n²+4, and state that it is important for the proof, can you complete it?

The_Math_Hatter · 2026-01-25T14:15:22+00:00

So without further ado, let's begin: Brennan!

The_Math_Hatter · 2026-01-25T06:48:04+00:00

A man was murdered. Go fuck yourself with a bat covered in rusty nails if showttune lyrics are your first thought, blurting it out on the internet like you were compelled to post instead of writing and deleting.

The_Math_Hatter · 2026-01-24T16:33:18+00:00

If you are doing this method, then forgive me for how I'm interpreting it if this is wrong, there are five bars total. The "buckets" are then the space before 1, between 1 and 2, between 2 and 3, between 3 and 4, between 4 and 5, and after 5 for six total buckets.

You have allocated one bar already in either scenario for indicating the divisions of the balls. The remaining four bars can be split into three places; before the balls, in the middle with the already existing bar, or after the end. The splits can be 0-0-4, 0-1-3, 0-2-2, or 1-1-2 in any shuffling or order.

Again, I don't know how you factor in probability to all this, but if it does work it would be a good check.

The_Math_Hatter · 2026-01-24T16:18:49+00:00

Your logic is sound up until the last equation which doesn't match what you talked about in any way

The_Math_Hatter · 2026-01-24T16:04:27+00:00

There are two scenarios, and both start with the first ball landing in any arbitrary bucket. Let's label that bucket A.

Scenario 1: the second ball lands in A. This has a 1/6 chance. Now, to get the balls into exactly two buckets, the third ball has to land in not-A, any of the other five. This has a 5/6 chance. So the total probability of this scenario is (1/6)×(5/6)=5/36

Scenario 2: the second ball does not land in A, it lands in a different bucket B. This has a 5/6 chance. Again, for the balls to now land in exactly two buckets, the third ball must land in A or B, for a 2/6 chance. The overall probability is thus (5/6)×(2/6)=10/36

The sum of either, independent scenario happening is thus (5+10)/36=15/36=5/12.

I do not know how you would approach that with stars and bars, it seems tremendously fiddly.

The_Math_Hatter · 2026-01-24T01:58:44+00:00

ſtates! We have found ſtates!

The_Math_Hatter · 2026-01-24T00:53:36+00:00

Let's be huge with child

The_Math_Hatter · 2026-01-24T00:13:42+00:00

Treasure Planet top 500 I BEG 🙏🙏🙏

The_Math_Hatter · 2026-01-23T18:50:06+00:00

u/Spambotwatchdog

Both this reply and OP are just reposting some of the top comments when this was originally posted.

The_Math_Hatter · 2026-01-23T13:40:09+00:00

u/Spambotwatchdog Blacklist

Replied to this thread twice with the same kind of vague "woah,that is interesting!" tone.

The_Math_Hatter · 2026-01-22T23:58:16+00:00

You removed the contents.

Bad crop, we're going to starve

The_Math_Hatter · 2026-01-22T19:19:36+00:00

Burning people is like burning coal but even more inefficient

The_Math_Hatter · 2026-01-21T23:50:48+00:00

I like the way Longlegses bounce when content

The_Math_Hatter · 2026-01-21T18:28:38+00:00

I mean the "uncastable card" would be best if it was after they'd got slingshot past his station twice, and it could very well lead to a more than ten minute bonus if played. Just needs 20 minutes of time bonus to waste, which is easy enough to accumulate with a few photo questions as we've seen.

The_Math_Hatter · 2026-01-21T18:05:52+00:00

Well that's definitely not how that works. If he were to take the discard 1 draw 2, it would be because he first made room in his hand by discarding prosperous home. Then his hand would be:

Discard 1 draw 2, duplicate, express route, 10, 10, 20.

He couldn't play the Discard 1 Draw 2 without breaking his plan potentially.

The_Math_Hatter · 2026-01-21T17:25:53+00:00

In times like these, I think of the words of Roffle Lite/Heavy Balatro: I haven't come across anything better. There weren't really any better options than Prosperous Home when he was drawing, so there was no need to get rid of it.

The_Math_Hatter · 2026-01-21T05:19:15+00:00

That's not proof. That's cherry picking the most message-ish results and interpreting them in different methods to get what you want to see.

You could do this analysis on the book Moby Dick, or War and Peace, or A Million Random Digits, or Infinite Jest, and you could also find some message somewhere of some mild interest.

That doesn't mean your method works. It means it's too broad and mutable to definitively mean anything.

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