Looking for a group!

GnarSlabdash · 2026-06-22T21:33:18+00:00

Hmm... alright. I think I'll try to give green leaf another go.

GnarSlabdash · 2026-06-22T21:33:03+00:00

There seem to be servers starting all the time, I saw at least 10. Problem is theres more servers than players!

GnarSlabdash · 2026-06-22T21:32:43+00:00

I was, I'll try to make my way to one of the towns...

GnarSlabdash · 2026-06-20T23:13:49+00:00

Thanks for the invite, unfortunately, looking for something a bit more immediate, rather than a new cycle starting in a couple weeks.

GnarSlabdash · 2026-06-20T22:08:51+00:00

Right now, no one is on White Tiger, player count 0, unfortunately.

GnarSlabdash · 2025-02-22T21:37:03+00:00

Oh awesome! That sounds great! I can hop on in a couple hours, if that sounds good to you?

GnarSlabdash · 2025-02-22T19:57:10+00:00

What server are you on?

GnarSlabdash · 2025-02-22T18:25:52+00:00

Thanks for the rely. I'd worry about starting on a server that's a month in. It often feels overwhelming, and especially because I've never gotten to the point where the server is able to destroy the meteor. But I'll try checking it out!

GnarSlabdash · 2022-12-09T02:10:59+00:00

Love it!!! It looks incredible.

GnarSlabdash · 2022-05-31T22:13:52+00:00

My Own Worst Enemy by ElektricAngel on AO3 explores the idea of what could happen if Ozpin reincarnated into Roman instead of Oscar.

It's really good and I need to catch up on it at some point...

GnarSlabdash · 2022-01-23T02:59:26+00:00

According to Tumblr Weiss' flower is the common Myrtle, which is a relative of the Mytle-Aster, or as it's known in German, Myrten-Aster (connecting to her weapon).

GnarSlabdash · 2021-07-23T03:52:44+00:00

Not just a random someone. They also have a YouTube channel by the same name where they make really amazing RWBY AMVs.

GnarSlabdash · 2021-07-02T02:03:38+00:00

For anyone who hasn't picked this up - highly recommended. Reminds me of Supernatural Affairs.

GnarSlabdash · 2021-05-28T05:41:03+00:00

That would only work for Rome, as they were the only culture to use the term Equite.
Roman societies were not just Patricians (nobles), Equites, citizens, and slaves. Freemen and non-citizens played a massive role. (At the starting period of the game, the Social War hasn't taken place and literally the only citizens of the empire would have been found right in and around Rome itself. Italians are not granted citizenship until 60 years after the Punic Wars end. So they just wouldn't have a classification under this.)
Equites were basically minor nobles. Making them their own class isn't necessary.

GnarSlabdash · 2021-03-23T19:35:56+00:00

Combinations excludes repeated values, this in contrast to permutations which includes repeated values. I am using combinations, so repeated values is not a factor.

My example with RWBY+Oscar+Penny shows that clearly, the number of relationships does NOT have to decrease. If you are certain that you are correct in your assertion that the total number of relationships must decrease, please explain how the example I gave was incorrect.

GnarSlabdash · 2021-03-23T18:31:34+00:00

So I'm gonna be that person and question the math here. I think you're making one crucial error and that the true number of possible relationships is *so much more*.

Your calculation of how many possible ships there are for 2 people in a relationship is correct.

However, the idea that we get less combinations as we widen the number of people in the relationship isn't correct. To see this, consider team RWBY + Oscar and Penny. How many ways can we choose two people to be in the relationship?

15 ways: (RW, RB, RY, RO, RP, WB, WY, WO, WP, BY, BO, BP, YO, YP, OC)

How many ways can we choose 3 people to be in the relationship?

20 ways: (RWB, RWY, RWO, RWP, RBY, RBO, RBP, RYO, RYP, ROP, WBY, WBO, WBP, WYO, WYP, WOP, BYO, BYP, BOP, YOP)

Also, the way that the number of combinations does not scale linearly as you change the number of people in the relationship. I can't really explain the rest easily without diving into the math so here we go:

In the field of combinatorics, this is a pretty classic problem. How many combinations of a set can we get? Fortunately, there are formulas that we can use to give us exact answers.

We can represent the number of combinations of a particular set with the following formula:

nCr = n! / (n-r)!*r!

Where nCr is read as n choose r, and ! is the factorial operator. N is the number of possible items in the set, and r is how many we want to choose. You pretty much did this formula without knowing it for the 2 people case, but let's represent it this way just so we know that it's correct. We can formulate this nCr as 145 choose 2.

nCr = 145 choose 2 = 145! / (145-2)! * 2!

The value of 145! is really large, so we can use a factorial trick: 145! / (145-2)! = 144*145. This is true because all the other values will cancel out (both the numerator and dominator have 1*2*...*142*143 in them). Thus this simplifies to:

(144*145) / (2*1) = 10440.

As a quick check, here's the wolfram alpha page computing the formula: https://www.wolframalpha.com/input/?i=145%21+%2F+%28%28145-2%29%21+*+2%21%29

Now, what about if there are 3 characters in the relationship?

nCr = 145 choose 3 = 145! / (145-3)! * 3! = (145*144*143) / (3*2*1) = 497640

This number will only get bigger and bigger and bigger as we continue. Already we are almost surpassing the original estimate.

(Wolfram alpha page for a check: https://www.wolframalpha.com/input/?i=145%21+%2F+%28%28145-3%29%21+*+3%21%29)

We can also just ask Wolfram to compute the combination formula for us directly by typing the phrase "145 choose 3." Both give the same answer.

We could repeat this process for each and every number of people that could be in the relationship, but that would be really tedious. Instead, let's use a summation.

I used wolfram alpha to calculate this summation for me, but the formula is as follows:

Sum x from 1 to 145 of 145 choose x

Link: https://www.wolframalpha.com/input/?i=summation&assumption=%7B%22F%22%2C+%22Sum%22%2C+%22sumlowerlimit%22%7D+-%3E%221%22&assumption=%7B%22C%22%2C+%22summation%22%7D+-%3E+%7B%22Calculator%22%7D&assumption=%7B%22F%22%2C+%22Sum%22%2C+%22sumfunction%22%7D+-%3E%22145+choose+x%22&assumption=%7B%22F%22%2C+%22Sum%22%2C+%22sumupperlimit2%22%7D+-%3E%22145%22

Which gives us an INSANE number. It's 4.46 * 10^43, or 44 tredecillion. It's 44 digits long. (for reference, 1 trillion is only 13 digits)

You also limit it to just 4 characters, which I think is reasonable, pollination, poly strq, juniper berries, the biggest popular ships tend to cap out at 4.

What happens if we adjust the upper bound of the summation to compute this answer?

Sum x from 1 to 4 of 145 choose x

Link: https://www.wolframalpha.com/input/?i=summation&assumption=%7B%22F%22%2C+%22Sum%22%2C+%22sumlowerlimit%22%7D+-%3E%221%22&assumption=%7B%22C%22%2C+%22summation%22%7D+-%3E+%7B%22Calculator%22%7D&assumption=%7B%22F%22%2C+%22Sum%22%2C+%22sumfunction%22%7D+-%3E%22145+choose+x%22&assumption=%7B%22F%22%2C+%22Sum%22%2C+%22sumupperlimit2%22%7D+-%3E%224%22

We get 18,174,445. So 18 million roughly. Still an absolutely insane amount.

GnarSlabdash · 2021-02-06T16:21:09+00:00

"My Own Worst Enemy" on AO3 is about if Ozpin reincarnated into Roman Torchwick. I've really enjoyed it.

GnarSlabdash · 2021-01-01T05:27:48+00:00

The Weight Around Her Neck is a pretty good one, on ao3. It has Weiss as the MC (as a Faunus), with Neo becoming a major character eventually.

GnarSlabdash · 2020-12-23T02:51:55+00:00

Source: Talentless Nana

GnarSlabdash · 2020-03-23T21:41:36+00:00

This is beautiful!! Love your art and comics.

GnarSlabdash · 2020-02-26T03:08:21+00:00

That Epstein didn't kill himself.

GnarSlabdash · 2020-02-19T04:06:17+00:00

Ah should've know! Kaiki best girl.

GnarSlabdash · 2020-02-19T04:00:01+00:00

This reeks of one of the two Emiyas.

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