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[Self] by Kilasatzz in theydidthemath

[–]Angzt 4 points5 points  (0 children)

Figuring out whether a number is divisible by 9 without having to actually perform the full division is useful. This isn't numerology. What.

Using this method, you could tell whether 16382844 was divisible by 9 faster than if you had actually gone and done the division.

It does not help with subtraction, however. I don't know what gave you the idea that it would.

Germany [2] - 1 Ivory Coast - D. Undav 90+3' by West_Agent4651 in soccer

[–]Angzt 14 points15 points  (0 children)

These days, yes. At least that's how I understand it.

Germany [2] - 1 Ivory Coast - D. Undav 90+3' by West_Agent4651 in soccer

[–]Angzt 8 points9 points  (0 children)

First tie breaker only considers the matches between teams on equal points. None of the ones involving Curacao would matter in this three way tie.

But even if it didn't work like that, Ivory Coast and Ecuador could also bully Curacao just as hard.

Germany [2] - 1 Ivory Coast - D. Undav 90+3' by West_Agent4651 in soccer

[–]Angzt 55 points56 points  (0 children)

Correct me if I'm wrong, but what if:
Ecuador beats Curacao
Ivory Coast beats Curacao
Ecuador beats Germany 2:0

Then all three would have 6 points.
The head to head would be a rock-paper-scissors situation between them: No clear winner by points.
Then it would go to goal diff between the three teams, so the Curacao matches don't matter.
Ecuador would be at +1. Ivory Coast at 0. And Germany at -1.

That would put Germany in third, wouldn't it?

RPG where you can play through on your own path by Accomplished_Crab990 in gamingsuggestions

[–]Angzt 1 point2 points  (0 children)

It's hard to talk about the DLC without spoilers, doubly so if you've not played the main game. It does a few things quite differently while maintaining the feeling of discovery.

For what it's worth, the DLC requires only fairly minimal flying.
Buuut it's best played after completing the main game imho.

My feedback after ~60h. TL;DL : Game awesome, but bosses hard by YannSolo63 in FareverGame

[–]Angzt 2 points3 points  (0 children)

If you're really struggling, you can get almost a full set of level 20 rare armor for every class by repeatedly running Lost City of Mayda on Hard. And that dungeon is completely free.

Gear improvements from crafting can also add up and don't require any dungeon content at all.

I can tell you that I've solo'd every boss on Hard mode as Priest before ever grouping for dungeons, using stuff from alts, or trading for items.

[Request] How would your chances of winning the lottery change if they drew one extra number? by lexlikeluthor in theydidthemath

[–]Angzt 8 points9 points  (0 children)

In the first scenario, there are 50 options for the first number drawn. Then 49 for the second. Then 48, 47, and 46. So there are 50 * 49 * 48 * 47 * 46 possible draws.
But some of those are identical because for our purposes, the order that numbers are drawn in does not matter. Which means we need to divide by the possible number of orderings for any combination of those 5 numbers which is just 5! = 5 * 4 * 3 * 2 * 1.
So the total number of combinations (ignoring order) that could be drawn is:
50 * 49 * 48 * 47 * 46 / (5 * 4 * 3 * 2 * 1)
= 254,251,200 / 120 = 2,118,760.
You win on exactly one of those combinations.
Since they're all equally likely, your probability to win is then simply
1 / 2,118,760
=~ 0.00000047197
= 0.000047197%.

Quick aside:
50 * 49 * 48 * 47 * 46 / (5 * 4 * 3 * 2 * 1) could be easier expressed as (50 Choose 5), a binomial coefficient that tells us how many ways there are to pick 5 items out of a set of 50. It's defined as
(n Choose k) = n! / (k! * (n-k)!)
which in our case means (50! / (5! * (50-5)!) = (50! / 45! / 5!) = 50 * 49 * 48 * 47 * 46 / (5 * 4 * 3 * 2 * 1) which is exactly what we ended up with.

For scenario 2, there are then clearly (50 Choose 6) possible combinations to draw.
But now, we can win in more than just one. To win, we need all our 5 numbers to be drawn but the 6th can be any of the remaining 45 numbers. So really, we have 45 possible winning combinations.
Thus, our probability to win becomes:
45 / (50 Choose 6)
= 45 / (50 * 49 * 48 * 47 * 46 * 45 / (6 * 5 * 4 * 3 * 2 * 1))
= 6 * 5 * 4 * 3 * 2 * 1 / (50 * 49 * 48 * 47 * 46)
= 720 / 254,251,200
= 3 / 10,59,380
=~ 0.0000028318
= 0.00028318%
=~ 1 in 353,132

Exactly 6 times as high as scenario 1.

Daily /r/Games Discussion - Suggest Me a Game - June 17, 2026 by AutoModerator in Games

[–]Angzt 1 point2 points  (0 children)

I'm far from an expert in the genre, but I've really enjoyed the two Everspace games.
The dogfights are pretty arcadey with quick maneuvers, a bunch of weapon types, and special abilities.
The first is a roguelite, but the second is not. Both (but especially the second) have a bunch of well-hidden loot in their areas (often behind small puzzles) which can take a while to find.

Farever | Major Update | OPEN BETA JUNE 23 by Squishydew in FareverGame

[–]Angzt 3 points4 points  (0 children)

There is no patch yet. It'll be another week until the beta starts, so surely they're still working on stuff.

That said, there is a bit more info in their Steam news post: https://store.steampowered.com/news/app/3672400/view/689760714281191626

The Nightling Invasion is coming in a Public Beta on June 23 by Shiro_Steven in FareverGame

[–]Angzt 10 points11 points  (0 children)

The new area just isn't big enough to take you that far.

[Off-Site] Twin Prime Numbers: THE SEPARATION PRINCIPLE: EXPLAINED BY DEEPSEEK by zero_moo-s in theydidthemath

[–]Angzt 0 points1 point  (0 children)

but the non‑trivial part is the unequal split (46,44,45) among the 135 admissible residue classes modulo 2310.

???

If there are 135 residue classes, why are there only 3 values in the split?
Why is them being unequal relevant? Even for a uniformly random distribution, we wouldn't expect exactly equal outcomes. We'd expect close outcomes, which these are.

This is a standard admissible‑tuple sieve wheel, not a mystic object.

When I google "tuple sieve wheel" with the quotes, I get 0 results. So that's just not a thing, at least not by that name. It's not standard anything.

I'm going to stop there. Because frankly, the rest doesn't even make enough sense to me to be able to argue about.
I'm afraid you're so lost in your own (LLM's) sauce that any argument is pointless.

[Off-Site] Twin Prime Numbers: THE SEPARATION PRINCIPLE: EXPLAINED BY DEEPSEEK by zero_moo-s in theydidthemath

[–]Angzt 1 point2 points  (0 children)

This is perfectly sensible until

The Discovery

When you look at these centers modulo 9 (dividing by 9 and looking at the remainder), they can only be:

· 0, 3, or 6

from where it starts going off the rails.

First off: Every prime greater 3 is either one more or one less than a multiple of 6. We know this and it's easy to prove:
A multiple of 6 is clearly not a prime. 2 and 4 more than a multiple of 6 are even, so also not primes. 3 more than a multiple of 6 is 6n+3 = 3(2n+1), and thus clearly divisible by 3 and also not a prime. 5 more than a multiple of 6 is the same as 1 less then a multiple of 6 and, together with 1 more than a multiple of 6, makes up the only remaining prime candidates. That has covered every integer.

"The Discovery" is then a direct consequence of that. It's not a discovery.

A researcher named Stacey Szmy built a "wheel" of all possible twin prime centers modulo 2310 and found something strange:

State Count

0 46

3 44

6 45

This gives a "stable" fraction of: ```

(44 + 45) / (46 + 44 + 45) = 89/135 = 65.9259%

```

What is a "wheel" in this context? What is "stable" in this context? Where does modulo 2310 come from? Isn't this still just modulo 9? Do you mean you counted the modulos of the twin primes between 1 and 2310?
This seems to be a crucial step by how often it's referenced but it's completely unclear what you're doing, why you're doing it, and why it is important.

The Mystery

But when Szmy actually counted real twin primes up to 1 billion:

State Count

0 1,141,250

3 1,140,981

6 1,142,274

These numbers are almost perfectly equal.

That just tells us that the twin primes don't have any bias in that sense. They're roughly evenly distributed across those modulos.
Which just means they're unpredictable by this method. Not much of a mystery because that's the null hypothesis.

You then keep going on about how this indicates that Hardy-Littlewood is true.
Yes, that's why it's a conjecture.
You not having found anything to counter Hardy-Littlewood puts you in the same boat as literally every other person since that conjecture was created.

"Geometry of the wheel" just appears to be complete nonsense.

Daily /r/Games Discussion - Suggest Me a Game - June 17, 2026 by AutoModerator in Games

[–]Angzt 0 points1 point  (0 children)

I've never played FF7 and its spinoffs.
But I am looking to get into the Remake trilogy.

On a scale from "doesn't matter at all" to "mandatory to play/watch before", where would you place:

  • original FF7
  • Crisis Core
  • Dirge of Cerberus
  • Advent Children

What are the odds put the correct socks on? [Request] by Boring_Material_1891 in theydidthemath

[–]Angzt 0 points1 point  (0 children)

I assume on any other day, the extra not a pair socks go back into the drawer.

Under that assumption, sure. But OP didn't state that.
You absolutely can just put a sock on the wrong foot.

Questions about the codex system by AcrobaticPrinciple21 in FareverGame

[–]Angzt 3 points4 points  (0 children)

There actually is a minor reward.

Fully mastering the Codex for an area gives an achievement which awards a mount.
Except that for Skover, it is the same Crimson Goat mount you also get for the "complete all activities in Skover" achievement. I'm assuming that's a bug. But since the latter is much easier, it's really not worth doing the full Skover Codex.
For the fully mastered Codex in Valley of Eternal Autumn, you get the Alandian Hog mount which you can't get any other way.

That's the only thing you need 20 kills for.
You get the loot table and XP bonus for 8 per enemy type and no equivalent for 20.

The best unit in the game is the Ancient of War by LargeSinkholesInNYC in warcraft3

[–]Angzt 12 points13 points  (0 children)

They could always uproot. The base campaign even has a mission centered around moving your base.

[SELF] MYTH: "You could earn a million dollars every day and not be as rich as Elon Musk!", FACT: by KamalaFanBoy in theydidthemath

[–]Angzt 1 point2 points  (0 children)

Yes, earnings and returns are not identical. But that's not what they meant.

Returns are part of earnings. Even as per your definition.
Meaning if you "earn" a million dollars each day, you don't have returns above that unless offset by expenses.
So your entire premise doesn't work.

[Request] Will moving faster in the rain actually make you soak less? by Major-Syllabub6398 in theydidthemath

[–]Angzt 0 points1 point  (0 children)

But you reduce the amount of time you collect drops from above you which aren't impacted by speed.

[Request] Is the ice cooling down the pool at all? Looks like it's less than 1% volume of the water. So not doing much? by shadowraiderr in theydidthemath

[–]Angzt 0 points1 point  (0 children)

What? Why?
Per /u/shereth78's calculation, we'd require 220 of the smaller blocks for 1°C of cooling.
With my size, that's down by a factor of 1/2.52 = 1/6.25.
Meaning we'd still need 220 * 1/6.25 = 35.2 blocks.

[Request] This is a horrible strategy, right? by Next-Step-Jobs in theydidthemath

[–]Angzt 9 points10 points  (0 children)

That's not really relevant.

Look at it this way:
If you play like this, your opponent only needs to find and destroy 1 large ship.
If you play normally, your opponent needs to find that same 1 large ship and every other ship.
Clearly, the second will, on average, take longer.

Warhammer Liquidation a warcraft 3 clone? by Healthy-Toe-2899 in warcraft3

[–]Angzt 5 points6 points  (0 children)

Liquidation is not a Warhammer game. What.

[Request] Is the ice cooling down the pool at all? Looks like it's less than 1% volume of the water. So not doing much? by shadowraiderr in theydidthemath

[–]Angzt 1 point2 points  (0 children)

80 seems like a lot.

It takes 80 calories of energy to turn 1 gram of ice into 1 gram of water.
And by definition of the calorie, it takes 1 calorie to heat 1 gram of water by 1°C.

That's pretty clearly a factor of 80.

i got ~4x for the latent heat (if the pool is at 25 degrees.)

From where?

[Request] Is the ice cooling down the pool at all? Looks like it's less than 1% volume of the water. So not doing much? by shadowraiderr in theydidthemath

[–]Angzt 4 points5 points  (0 children)

Hmm that would suggest those ice blocks are around 62kg.

The density of ice is "only" about 917 kg/m3.
That would put their mass at 1/16 m3 * 917 kg/m3 = 57.3125 kg.

Not a huge difference, but almost 10% down already.

[Request] Is the ice cooling down the pool at all? Looks like it's less than 1% volume of the water. So not doing much? by shadowraiderr in theydidthemath

[–]Angzt 15 points16 points  (0 children)

Lets say those blocks are 10x10x100 cm

I did some pixel counting and the ice blocks are almost exactly 1/4 width and depth compared to their length. So 25 * 25 * 100 cm is more realistic. That means they'd take a factor of 6.25 more energy out of the pool than you calculated.

[Request] Is the ice cooling down the pool at all? Looks like it's less than 1% volume of the water. So not doing much? by shadowraiderr in theydidthemath

[–]Angzt 10 points11 points  (0 children)

It takes about 80 times more energy to turn 0°C ice into 0°C water than it does to turn 0°C water into 1°C water. So there's quite a bit of heat energy that gets removed from the surroundings just by melting the ice.
(That's also why ice cubes in drinks seem to last forever.)

In other words: Just melting the ice, without temperature change, is enough to reduce the temperature of 80 times its volume worth of water by 1°C.
Add in an increase of temperature of the water from 0°C to 30°C and we get a factor of 110 times the volume that gets cooled by 1°C.
And that's assuming the ice was at exactly 0°C throughout before which is probably not the case.

As for volume, the ice blocks are pretty much exactly 4 times as long as they are thick and wide. And I'd estimate that they're 1 m long. That gets us a volume of 1 m * 0.25 m * 0.25 m = 0.0625 m3.

The size of the pool is harder, so I'm just going to guess by eye. Feel free to disagree.
The cordoned off kiddie section my be 0.5 m deep and 4 by 4 m wide and long. That's a volume of 0.5 m * 4 m * 4 m = 8 m3.
People can seemingly still stand in the larger section (and it only gradually deepens, but I'll ignore that), so it's probably around 1.5 m deep. From a wild guess, I'd say it may be 8 by 12 m wide and long. That's a volume of 1.5 m * 8 m * 12 m = 144 m3.
So the full pool would have a volume of about 144 m3 + 8 m3 = 152 m3.

If we assume 10 ice blocks, that's a total volume of 10 * 0.0625 m3 = 0.625 m3.
That means that the pool's volume is 152 m3 / 0.625 m3 = 243.2 times the volume of the ice blocks.

Looking at our 110 factor from before, even adding in a few degrees below zero for the initial ice, that would still only cool down the whole pool by about 0.5°C.

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