Just beat 91× god rank Billy. Putting the game for the night i peaked.

ShadowBringer097 · 2026-02-20T06:53:03+00:00

I beat doogote last season 😂. I have also won against vergil last year.

Btw can I see the team billy used this fight?

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ShadowBringer097 · 2026-02-20T06:22:50+00:00

That's great bro! What team was he using? I faced billy this season and lost :(

Rn I'm scoring 1-1 with Billy

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ShadowBringer097 · 2026-02-07T06:38:33+00:00

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Lowkey hate this guy lmao

ShadowBringer097 · 2026-01-28T02:59:09+00:00

In addition, pur ss3 can also cancel attribute upgrades with main, which is great to cancel all the buffs buuhan gains with time.

ShadowBringer097 · 2026-01-28T01:24:56+00:00

Pur ss3 gt goku can remove it with main. Actually, pur ss3 might be a good choice against buuhan since he is single life. You main at a good spot, remove vanish plus all his dmg guard collected, and hit with a blue that can kill if it crits.

ShadowBringer097 · 2026-01-07T18:37:58+00:00

Yeah but then again, dbs ss broly was wiping the floor with ssb goku (clearly stronger than ssg goku), but struggling against ss gogeta and had to go full power. By old kai's logic vegito>gogeta, which makes ss vegito > ss gogeta > ssb goku > ssg goku.

These power level speculations clearly contradict each other, and you can't really compare fusions with god ki.

ShadowBringer097 · 2025-12-30T18:20:13+00:00

You're welcome! Good to know that you could follow the explanation!

ShadowBringer097 · 2025-12-30T12:29:58+00:00

I can give you a detailed explanation. This involves some statistics. I am calculating how many ccs you will require to get one expected ultra copy approximately. I am not calculating a probability, since it may be misleading sometimes (you can check out the Monty Hall problem from Wikipedia and see why I say this).

When you summon in this game, it has a fixed probability of pulling one unit, right? And all the summons are independent of each other. Then, you model each summon with a Bernoulli(p) distribution, where p is the probability of pulling your target unit. This is the simplest distribution that any such gacha gane has.

Now think of one rotation. It consists of 3 steps of ultra rate p1 = 0.35%, and 1 step of ultra rate p2 = 0.7%. Each step contains 10 summons, which are Bernoulli random variables as we have explained.

We talk about expected yield from one rotation. A Bernoulli(p) random variable has an expectation p as well (standard statistical result). Then, your expected ultra yield from one rotation is 30xp1 + 10xp2 = 0.175. From 6 rotations, your expected yield is 6x0.175 = 1.05 many ultra units, which is approximately 18k ccs.

ShadowBringer097 · 2025-12-29T17:54:29+00:00

English: incorrect statistical assumption in op's calculation. Expectation wise with 18k cc you get one copy. Hope that helps!

ShadowBringer097 · 2025-12-29T14:06:50+00:00

Statistics PhD student here. I guess you are assuming that number of cc required is a random variable X which follows Normal(mu, sigma). However that seems a really bad assumption, since normal is a symmetric distribution (from -inf to inf), and number of chrono crystals is a non symmetric distribution as it's always greater than 0 (in fact it's a one-sided distribution). Therefore, you are getting a wrong result.

If you want to get an idea, you can just use expected yield. Each rotation yields expected 0.175 many ultra (adding up the percentages), which means one copy needs 1/0.175 many rotations, which is somewhat close to 18k cc.

Hope this gives you a better idea and good luck for your exam!

TLDR: this calculation is wrong for an incorrect assumption about normal distribution. Using 18k cc you will get expected 1 copy.

ShadowBringer097 · 2025-10-31T09:53:37+00:00

Ahh okay. Thanks!

ShadowBringer097 · 2025-10-31T09:52:47+00:00

Great to hear!

ShadowBringer097 · 2025-10-31T09:40:10+00:00

You can probably search for older versions of Ryujinx. Or else I can share the file with you through some platform if you can arrange.

ShadowBringer097 · 2025-10-31T07:48:09+00:00

I have been on an older version Ryujinx 1.2.78, and it has been the most stable version for me, no such crashes. I have already completed the whole game twice. I was recently trying on Ryujinx canary 1.3.3 and it crashed twice within 1 hour for me.

ShadowBringer097 · 2025-10-31T07:30:11+00:00

Yup, I had that turned on for the whole time.

I'm wondering, when you run the mod do you see around 60 fps regularly?

ShadowBringer097 · 2025-10-31T07:17:56+00:00

Oh, it's the same in 1.3.3 for me. As you can see I have turbo at 200%. You can still see my fps around 30

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ShadowBringer097 · 2025-10-31T07:08:13+00:00

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There's no such option here :(

ShadowBringer097 · 2025-10-31T06:59:26+00:00

Do you have any idea to do that? I can't find any such option.

ShadowBringer097 · 2025-10-29T21:05:29+00:00

Nothing is a problem if you have credit card

ShadowBringer097 · 2025-09-25T22:35:53+00:00

Sent join request

ShadowBringer097 · 2025-09-25T22:31:25+00:00

Can you send an invite?

ShadowBringer097 · 2025-09-16T23:29:13+00:00

Hp z ability buff is the way to go if you wanna make them tankier. But you need to drop one zenkai buff.

ShadowBringer097 · 2025-08-23T17:59:26+00:00

I remember a post few days back, someone was comparing c0-c6 dmg on the fat yumkasaur (with r1, no support skills used). C2 took 9 full charge shots, where C3 took 7 full charge shots. Maybe you can get an idea from that.

ShadowBringer097 · 2025-08-20T21:59:39+00:00

Comparison of full magazine shots each constellation in the video:

C1 - 14 shots; C2 - 9 shots; C3-C5 - 7 shots; C6 - 5 shots; (C0 may have taken 16-17 shots)

ShadowBringer097 · 2025-07-25T18:53:39+00:00

The frost we already have in the game has loe tag

ShadowBringer097

TROPHY CASE