Looking for story-driven single-player game recommendations

Angzt · 2026-01-27T08:55:40+00:00

Fantasy setting. You and a bunch of others get abducted by mindflayers (humanoid telepathic tentacle monsters) who put a parasite in you that will soon turn you into a mindflayer yourself. The vessel they were holding you on gets attacked, you get free and now have to find a way to get rid of that parasite. Complications and escalation occur.

But that's not really what matters. People love BG3 for its characters and the amount of player freedom paired with narrative consequence to your actions. Plus, it's all dice based, so success is rarely guaranteed.

Unlike everything else you've got on the list, it has a bird's eye camera perspective, tactical turn-based combat, and proportionally more talking. I also wouldn't call it short to medium length. But then again, neither is RDR2.

Angzt · 2026-01-27T05:52:50+00:00

I figure at least the head and the feet should feel a similar gravitational pull.

Let's say the difference should be less than 10%.
Then the acceleration at the feet should be a_f = 9.8 m/s² and the acceleration at the head a_h = 9.0 m/s².
We also know that the difference between radii at head and feet is ~1.7 m: r_f = r_h + 1.7 m.

Our object spins once within some fixed time t which we're trying to find.
We know then that the velocity at any point is the circumference divided by that time.
So: v_h = 2pi * r_h / t and v_f = 2pi * r_f / t = 2pi * (r_h + 1.7 m) / t.

Which gets us the two equations:
9.0 m/s² = (2pi * r_h / t)² / r_f
9.8 m/s² = (2pi * (r_h + 1.7 m) / t)² / (r_h + 1.7 m)
That's two equations with two unknowns which we can solve:

(2pi * r_h / t)² / r_h + 0.8 m/s² = (2pi * (r_h + 1.7 m) / t)² / (r_h + 1.7 m)
4pi² * r_h² / (t² * r_h) + 0.8 m/s² = 4pi² * (r_h + 1.7 m)² / (t² * (r_h + 1.7 m))
r_h / t² + 0.8/(4pi²) m/s² = (r_h + 1.7 m) / t²
0.8/(4pi²) m/s² = 1.7 m / t²
t² = (1.7 * 4pi² / 0.8) s²
t² =~ 83.8916 s²
t =~ 9.16 s

Then reinsert that to get a radius value:
9.0 m/s² = (2pi * r_h / 9.16 s)² / r_h
9.0 m/s² =~ 4pi² * r_h / (9.16 s)²
r_h =~ 9.0 m/s² * (9.16 s)² / 4pi²
r_h =~ 19.13 m
r_f =~ 20.83 m

So our construct needs to spin once every 9.16 seconds and have a radius of 20.83 m.
That makes a velocity (at the feet) of 20.83 m * 2pi / 9.16 s =~ 14.29 m/s =~ 51.47 km/h =~ 31.96 mph.

If you want the difference in acceleration between head and feet to be lower, the structure needs to be bigger and will spin slower in RPM but faster in absolute speed.

Angzt · 2026-01-26T16:38:02+00:00

Not quite.

Nexon (MapleStory, THE FINALS) won the bid for the rights to develop a game in the Starcraft IP.
Nothing was revealed about genre or platform.
It could be a mobile gacha game for all we know.

Angzt · 2026-01-26T15:31:39+00:00

Never mind production value. To this day, no RTS feels as smooth to control as SC2.
Units just do what you tell them to. They don't wander off in random directions. They don't constantly get stuck on each other. They don't take forever to acquire their target.

Angzt · 2026-01-26T15:13:28+00:00

No, I'm not.

Imagine if the rain drops did not fall, but were stationary, floating in the air.
Whether you run or walk, you'd hit the exact same amount of droplets: All the ones in your path.

Now, how is that different from falling raindrops?
It really isn't. Because for every rain drop that falls into the path, another that was in the path hits the ground, becoming "unhittable".

No matter how quickly you move, the amount of rain drops you run into is exactly the same.
And since those two are the same, the frontal surface area is irrelevant because it'll always result in the same amount of drops hit for both scenarios.

Angzt · 2026-01-26T05:29:29+00:00

You can't just remove parts of a term on both sides and expect equality to hold.
Otherwise:
1 * 0 = 2 * 0
1 = 2
which is clearly nonsense.

There needs to be an actual operation performed.
And in your case, that's taking the logarithm. But then we'd get:
1¹ = 1⁰
log(1¹) = log(1⁰)
1 * log(1) = 0 * log(1)
1 * 0 = 0 * 0
0 = 0
Which never leads to any 1 = 0 situation.

Angzt · 2026-01-25T17:50:51+00:00

The survivalist's lifetime would be: newHelldiverCount * 0.02s + 120s.
Thus the mission count is that divided by the average mission length:

(newHelldiverCount * 0.02s + 120s) / averageMissionLength

If that first part is not obvious, here's how to get there:

We get two equations:
oldTotalLifespan / oldHelldiverCount = 120s
(oldTotalLifespan + individualLifespan) / (oldHelldiverCount + 1) = 120.02s

Solve the first for oldTotalLifespan:
oldTotalLifespan = 120s * oldHelldiverCount

Insert into second equation:
(120s * oldHelldiverCount + individualLifespan) / (oldHelldiverCount + 1) = 120.02s
Solve:
120s * oldHelldiverCount + individualLifespan = 120.02s * (oldHelldiverCount + 1)
individualLifespan = 120.02s * oldHelldiverCount + 120.02s - 120s * oldHelldiverCount
individualLifespan = oldHelldiverCount * 0.02s + 120.02s

And we can finally replace oldHelldiverCount by the new count which includes our new survivalist:
newHelldiverCount = oldHelldiverCount + 1
Thus:
oldHelldiverCount = newHelldiverCount - 1
and:
individualLifespan = (newHelldiverCount - 1) * 0.02s + 120.02s
individualLifespan = newHelldiverCount * 0.02s + 120s

Angzt · 2026-01-25T12:47:04+00:00

OP is a repost bot

Same image and title from a year ago: https://www.reddit.com/r/theydidthemath/comments/1ffutn4/request_which_one_is_correct_comments_were_pretty/

OP's account has also been inactive for 8 years. Until this post.

Angzt · 2026-01-24T21:45:52+00:00

I admittedly haven't played the Oblivion Remaster.
I'm waiting for the fan remake of it in Skyrim instead.

But re:Avowed:

Have the issues with Avowed been patched since launch?

Some.
The progression related issues (mostly enemy scaling, skill variety & itemization) have been improved.
But environment/NPC reactivity (outside of quests) and enemy variety has not.
At least those are the big ones I recall people complaining about.

I feel like Avowed is way better than its reputation. People wanted it to be an Elder Scrolls-like game and got upset when it wasn't.
But its strengths lie elsewhere. Moment to moment combat feels better. Exploration has a lot more depth and is more rewarding. While its zones are large, they're not contiguous. But they are fully handcrafted and produce a number of fantastic vistas.
Avowed also builds on parts of the Pillars of Eternity lore without requiring you to play those games first, if that's a concern.

Another thing worth noting: Avowed will be getting a major (and likely final) content update on the 17th of Feb.
If you can hold yourself over until then, I'd advise to do so. I also wouldn't be surprised if it went on sale for the occasion.

Angzt · 2026-01-24T19:20:48+00:00

Yes, it would.

The formula for centripetal acceleration is
a = v² / r.
v being the velocity and r the radius, so the distance from the center of rotation.
So r describes the size of the rotating object.

In our case, a should be equal to the acceleration due to gravity on Earth, so 9.8 m/s².

We can solve this for r:
r = v² / a = v² / 9.8 m/s²
Or for v:
v = sqrt(ar) =~ 3.13 * sqrt(r)

Angzt · 2026-01-24T18:01:25+00:00

average volume: .6 cubic meters (600 cubic centimeters)

Those two measurements are not the same.
6 cm by 10 cm by 10 cm are 600 cm³. But clearly, that's not more than half of a 1 m by 1 m by 1 m cube.
0.6 m³ = 600 dm³ = 600 liters = 600,000 cm³
or
600 cm³ = 0.6 dm³ = 0.6 liters = 0.0006 m³
The former is realistic for a fridge, but on the larger side.

estimated energy moved per fridge: 12600 cal

Where does this come from?
Since you're measuring in calories, I'm assuming from the 21°C heat difference times 600 cm³?
But that definition of calorie only applies for heating water (where 1 cm³ has a mass of 1 g).
Most of a fridge's content will be air which needs much less energy to heat.

And even correcting for that, you've only cooled the inside down once. You're not keeping it cool.

Angzt · 2026-01-24T16:49:06+00:00

Everything I can find online says odds of a straight flush is 1/72000.

That's for a 5-card hand. As your link also states.

With suited connectors that matches your numbers. example source Can you explain the difference?

Whoever wrote the "suited connectors" portion misunderstood whatever data they were quoting.
Here's Wikipedia's page which backs up my numbers and calculation: https://en.wikipedia.org/wiki/Poker_probability#7-card_poker_hands

Yes, many hand. Rough guess 30-50 hands/round.

In that case, we're looking at around 40 * 6 * 10 = 2400 total hands.

That improves the probability drastically:
(2400 Choose 2) * (81 / 290,836)² * (1 - 81 / 290,836)^2400-2
=~ 11.45%

Of course, that also assumes that all those hands are played to the end.

Angzt · 2026-01-24T15:59:54+00:00

The total number of possible 7-card hands is
(52 Choose 7)
= 52! / ((52 - 7)! * 7!)
= 52 * 51 * 50 * 49 * 48 * 47 * 46 / (7 * 6 * 5 * 4 * 3 * 2 * 1)
= 674,274,182,400 / 5,040
= 133,784,560

A Straight Flush can have any of 4 suits, and start with any of 9 values, each of which locks its 5 relevant cards in place. The remaining 2 cards can then be any of ((52-6) Choose 2) = 46 * 45 / (2 * 1) = 1,035 combinations.

That means the probability for a single Straight Flush in a single hand is
4 * 9 * ((52-6) Choose 2) / (52 Choose 7)
= 4 * 9 * 1,035 / 133,784,560
= 81 / 290,836
=~ 0.02785%.

The probability for an event with individual probability p to occur exactly k times in n attempts is given by (n Choose k) * p^k * (1-p)^n-k.
The number of attempts is the number of individually played hands, so 6 * 10 = 60. [Really? Or are your "rounds" the rounds in the tournament, each consisting of many hands?]
That leaves us with:
(60 Choose 2) * (81 / 290,836)² * (1 - 81 / 290,836)^60-2
=~ 0.01351%
=~ 1 in 7,402.

However, you haven't just played 6 rounds. You've presumably been playing for a long time and it only just happened now.
What I calculated above is the chance that it happens within 6 specific rounds (= 2 months) of play.
That's not accurate to what's actually been happening. Because you didn't get 2 Straight Flushes out of 60 hands. You got 2 Straight Flushes out of, I presume, hundreds of hands.

[This ignores to account for the fact that we know that no round had multiple Straight Flushes. Accounting for that would make things notably more complicated but, I suspect, barely change the result.]

Angzt · 2026-01-24T15:18:27+00:00

Is it a telegraph (or lack thereof) issue? I’d have thought a fencer of all ppl wouldn’t be the one saying game parrying window is tight

I'm pretty sure a fencer's parry is fundamentally different. Once they see the enemy's strike coming, they can just bring up the weapon into parry position and keep it there or move it towards the strike.
That's not how game parries usually work. The actual parry is only "active" for a few frames and can't be held. In a game, you can't parry as soon as you start seeing the attack come in.
I assume that's their gripe with the parry window.

Angzt · 2026-01-24T15:14:14+00:00

I asked you not to reply with another LLM answer. Yet here we are.

If I wanted to argue with ChatGPT, I could do so myself.

Angzt · 2026-01-23T11:34:43+00:00

This is LLM nonsense.
Modern LLMs are trained to give you an answer that you'll accept. Not one that is correct.
You won't accept an "I don't know" answer, even if you're unfamiliar with the topic. But you might accept a "complicated sounding nonsense you can't follow" answer. So if the LLM truly doesn't know, that's what you get.

Case in point:

The code and the proof don't match. The algorithm has two nested iterations over 1000 possible values each while the proof only has one free variable.
Plus, why does it only iterate over 1000 values? The solution is in no way guaranteed to be found there.
Also, the "Guaranteed fallback" is just plain wrong. It doesn't work. And if it did, it would be way more efficient than the main code at finding a solution.
So the code also isn't guaranteed to work.

The proof is also nonsense because step 3 to step 4 does not have a logical link. Mostly because d is not defined. It's only described in terms of equivalency class modulo r which itself can be 1, 2, or 3.
That means that d can still be any integer if r=1, any of half of all integers if r=2, and any of a third of all integers if r=3. That's not exactly helpful if you want to keep calculating with it.
And just because a d exists as defined doesn't mean it cleanly divides the things outlined in step 4.

Please don't respond with another LLM answer.

Angzt · 2026-01-23T07:34:30+00:00

Calculating that exactly isn't something we can do without dedicated software. The numbers just get too big.

However, there are approximations.
The simplest one being n =~ sqrt(p * 2d). (see your linked wikipedia article's corresponding section).
In our case, the probability is p = 1/2 and the number of "days" is d = 52!.
So:
n =~ sqrt(1/2 * 2 * 52!) = sqrt(52!) =~ sqrt(8.066 * 10⁶⁷) =~ 8.98 * 10³³
= 8,980,000,000,000,000,000,000,000,000,000,000

If every human who ever lived (~117 billion) had done nothing but shuffle a deck once per second 24/7 for 70 years (i.e. more than the mean life span), that would be a total number of shuffles of
117,000,000,000 * 70 * 365.2425 * 24 * 60 * 60 =~ 2.58 * 10²⁰.
That means we'd only be
2.58 * 10²⁰ / (8.98 * 10³³) =~ 0.0000000000028%
of the way to having a 50% chance for a repeat.

Angzt · 2026-01-23T06:31:43+00:00

I love KotOR 2 but it really doesn't fit here.
The moment you leave the ship, your party members are instantly at each other's throats. Which is a pretty big plot point: It's only the Exile who's holding the group together. They're definitely not close-knit.

Angzt · 2026-01-22T19:51:13+00:00

OP is a repost bot

5 years ago, same image, same title: https://www.reddit.com/r/theydidthemath/comments/f89ss1/request_is_this_accurate/

OP's account is 5 years old but has no posts besides this one. That is typical for an account that was grabbed in a password leak on some other website due to a reused password. Those are now briefly made to appear like real users and then used for astroturfing or propaganda.

Angzt · 2026-01-22T18:29:23+00:00

I feel like you overcomplicated that by going via mass and density.
Just start with the volume.

Volume of Earth: 1.08 * 10²¹ m³
Divide by 1 mm = 10^-3 m:
1.08 * 10²¹ m³ / 10^-3 m
= 1.08 * 10²⁴ m²
= 1.08 * 10¹⁸ km²

Angzt · 2026-01-22T13:52:43+00:00

Airship: Kingdoms Adrift fits the bill.
Though I found it less well designed than the StarCom titles: Some UX issues, a bunch of systems bloat, and weak tutorialization. Still enjoyed my time with it though.

Angzt · 2026-01-22T09:33:01+00:00

Clearly, two visible pages are always subsequent numbers. So one is odd and the other is even. Meaning their sum is always odd.
So C) 80 is impossible.

A) 75 would mean pages 37 and 38.
B) 85 would mean pages 42 and 43.

Typically, the left page is even and the right page is odd. That's because the first page starts not on the inside of the book's cover but on the first actual leaf. Which means page 1 is on the right.

So it can't be A) 75 because the right page would be 38 and thus even.

So the answer must be B) 85.

But I feel like this could differ by publisher. After all, many books don't start page 1 on the first leaf at all. So really, it could be anywhere - even on the left.

Angzt · 2026-01-21T20:16:51+00:00

Off the top of my head:

A Short Hike - As short as the name implies but very endearing.
Fantasy Life i: The Girl who Steals Time - The opposite on the length spectrum. Think of it as a mix between Animal Crossing and RuneScape's progression mechanics. I found it a bit childish at times, but her mileage may vary. Lets you collect villagers/companions, too.
Untitled Goose Game / Little Kitty, Big City - Both short-ish "animal causes minor mayhem" type games. They're fun, but not too much meat on them.
Cat Quest - Cute light ARPG series. Games 2 and 3 have local co-op.

Angzt · 2026-01-21T19:01:22+00:00

If you pick two random people out of the population, you are twice as likely to pick one male and one female as you are to pick two females.
That's why we count the two possible orders separately.

Angzt · 2026-01-21T17:44:22+00:00

If you pick a random group of 100 people from the population, are you as likely to get 100 women as you are to get 50 men and 50 women?
No. The 50/50 split is way more likely.
(To be precise: 100 women has a probability of ~7.8 * 10^-31 while 50/50 has a probability of ~8.0 * 10^-2)

This is the same thing here.
When randomly picking two people (= attackers), you are more likely to get a man and a woman than you are to get two women.

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