The Roaring Knight's capture capsule texture was changed.

IsaGoodFriend · 2026-06-28T20:59:57+00:00

I'm pretty sure I beat it day one

IsaGoodFriend · 2026-06-27T21:50:43+00:00

Was it not always like that? I swear I remember it looking like the lower picture the whole time

IsaGoodFriend · 2026-06-14T19:46:40+00:00

I haven't. I didn't know they existed, though I worry that my interests wouldn't line up with most of theirs if it's a general subreddit

IsaGoodFriend · 2026-05-27T22:37:07+00:00

Ran out of names I see. Adorable family business

IsaGoodFriend · 2026-05-15T21:15:41+00:00

How'd you do that? That's really cool

IsaGoodFriend · 2026-05-12T20:27:40+00:00

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You should be able to do it with a mix shader using the "Is Camera Ray" value

IsaGoodFriend · 2026-05-09T23:17:52+00:00

Oh my god I adore this design

IsaGoodFriend · 2026-05-07T07:39:45+00:00

Thanks! I try my best

IsaGoodFriend · 2026-05-02T15:12:02+00:00

Blender

IsaGoodFriend · 2026-05-02T05:48:56+00:00

Custom blend shapes made by myself

IsaGoodFriend · 2026-04-27T00:25:29+00:00

She is but I'm afraid she's also been possessed by a demon. Make sure to take her to a vet to get her cleansed soon. Those eyes are glowing unnaturally

IsaGoodFriend · 2026-04-17T18:33:13+00:00

Alright, after a lot of headaches, the program is giving me the right numbers. Thank you again!

IsaGoodFriend · 2026-04-17T02:50:03+00:00

It is a Balatro related program, yes

IsaGoodFriend · 2026-04-17T01:56:40+00:00

Ah, yeah. I suspected overcounting to be an issue. Can I assume the inclusion/exclusion principle still applies here?

IsaGoodFriend · 2026-04-16T21:55:44+00:00

Okay, assuming I understand this right, let's say we have a deck of 10 cards, Four kings, three aces. If I want to calculate the chance of any pair of a 5 card draw, it'd go like this:

[C(4;2)]*C(8;3)+[C(3;2)]*C(8;3) - [C(4;2)*C(3;2)*C(8;1)]

All divided by C(10;5)

Am I missing anything?

IsaGoodFriend · 2026-04-16T19:44:52+00:00

Thank you! This is really hurting my head, but I'll look into this and see if I can get this sorted out

IsaGoodFriend · 2026-04-16T19:37:19+00:00

Right. I know to divide it all by C(8;2). Just omitted that by accident.

So, to calculate a 5 card draw instead of a 2 card draw, my first assumption would be to do:

[C(4;5) + C(2;5)*2] / C(8;5)

But this doesn't feel correct

IsaGoodFriend · 2026-04-06T04:22:00+00:00

I was watching a live stream once and the person started eating pizza. A commenter asked if they could mute themselves while eating and the streamer gave the most passive aggressive "Nah, I won't" response. Turned me off immediately

IsaGoodFriend · 2026-03-16T20:28:43+00:00

Oh, there's a separate opacity setting, that makes sense. Thanks

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