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Posted
11 minutes ago, Aspiring Writer said:

You... You guys did what storming game theory does. (Also, pls send these numbers to BS, I would like to see his mind blow up.)

We're not done yet

Posted
Just now, Aspiring Writer said:

From Henceforth, the number of (Whatever we finally decided on) will now be dubbed a Metalbornplex. Suck it Matpat.

How about Mistnumber?

Posted
12 minutes ago, Bejardin1250 said:

So what’s the final number?

3946456273736362783626728372 That’s my guess:P

Add about 19 thousand digits to that, and you might be getting closer. (Not the number 19,000, 19,000 digits to the end of the number.)

But yeah, we’re not done yet. Got a bunch of stuff to figure out how to add still.

Posted
6 minutes ago, Dannex said:

Add about 19 thousand digits to that, and you might be getting closer. (Not the number 19,000, 19,000 digits to the end of the number.)

But yeah, we’re not done yet. Got a bunch of stuff to figure out how to add still.

"BUT I'M NOT DONE YET"

"NOW THAT'S A LOT OF METALS"

Rusts, I love my brain sometimes. :P

Posted

Wait so does the you get one or them all still work. I mean just because we haven't seen one with 2 of the 16 normal metals (a very small percentage) doesn't mean it can't happen with one of the other.......

Posted
2 minutes ago, Retrac said:

Wait so does the you get one or them all still work. I mean just because we haven't seen one with 2 of the 16 normal metals (a very small percentage) doesn't mean it can't happen with one of the other.......

Who can burn Shardmetals/alloys is up in the air.

Posted
35 minutes ago, Frustration said:

Well Atium+Lerasium isn't Harmonium, so Hramonium+Skaisium is different from A+L+S, which would be different again if Harmon took up Dominion.

I have no idea how to even begin calculating that

I know, I've already taken care of that

The 2^16-1 is count the number of individual Godmetal (I'm calling things like Harmonium or the metal that would be created by Harmony taking up Dominion individual Godmetals)

Then I alloy any number (except 0) of these 2^16-1 Godmetal, that's 2^(2^16-1)-1

34 minutes ago, Dannex said:

Does that include triple/quadruple/15ple groupings and all that?

It does

Quote

And does it exclude impossible duplicated shard groupings? (Like a “Preservation-Ambition-Ambition” Shard)

It does too

Posted
4 minutes ago, mathiau said:

I know, I've already taken care of that

The 2^16-1 is count the number of individual Godmetal (I'm calling things like Harmonium or the metal that would be created by Harmony taking up Dominion individual Godmetals)

Then I alloy any number (except 0) of these 2^16-1 Godmetal, that's 2^(2^16-1)-1

It does

It does too

I calculated that in the same calculator I used for the other one and got " 1.001764965 × 1019728 "

Kinda small...

Posted
Just now, Halyo_Alex said:

I calculated that in the same calculator I used for the other one and got " 1.001764965 × 1019728 "

Kinda small...

I got the same result

Yes, compared to the insane 2^(16!) it's small, but it's actually an insanely high number

Posted
18 minutes ago, mathiau said:

I know, I've already taken care of that

The 2^16-1 is count the number of individual Godmetal (I'm calling things like Harmonium or the metal that would be created by Harmony taking up Dominion individual Godmetals)

Then I alloy any number (except 0) of these 2^16-1 Godmetal, that's 2^(2^16-1)-1

It does

It does too

So same thing for just the  total number of possible Shards? (2^16)-1? 

How would you calculate like, the total number of Shard Pairs? I did it manually, it’s 120, (15+14+13+12+11..... +2+1) but is there a way to calculate that in one step? What about just the total number of Shard Triplets or any other number?

Posted

I think there is just infinity metals, with every single alloy and alloys of alloys and stuff like that. All these big numbers are hurting my brain so I'm sticking with a  nice, semi- comprehensible number.

Posted (edited)
3 minutes ago, 2EmLee2 said:

I think there is just infinity metals, with every single alloy and alloys of alloys and stuff like that. All these big numbers are hurting my brain so I'm sticking with a  nice, semi- comprehensible number.

did you just use infinity and compresehsible in the same sentence?

Edited by Frustration
Posted
1 minute ago, Frustration said:

did you just use infinity and compresehsible in the smae sentence?

...Maybe...

Spoiler
Spoiler
Spoiler

:ph34r::ph34r::ph34r:

 

 

 

Posted
46 minutes ago, Dannex said:

So same thing for just the  total number of possible Shards? (2^16)-1? 

How would you calculate like, the total number of Shard Pairs? I did it manually, it’s 120, (15+14+13+12+11..... +2+1) but is there a way to calculate that in one step? What about just the total number of Shard Triplets or any other number?

Yes, there is a way to do it one one step. If the order mattered, for the first Shard you'd have 16 possibilities and 15 for the second, that's 16*15 ordered pairs. Since there's are two possible way to order two Shards, I can just divide by two to get the number of unordered pairs, 16*15/2=120

For the triples I can do a similar reasoning, for the ordered triplets you have 16 possibilities for the first, 15 for the second and 14 for the third and there is 3!=6 ways to order a triplet, therefore 16*15*14/6=560

More generally if you have a set of n distinct object, the number of way to take k among them is n!/(k!*(n-k)!) (note that 0!=1 so it still works for n=k or k=0). If you do the simplification for k=2 you get n!/(2!*(n-2)!)=n*(n-1)/2!=n*(n-1)/2 and for k=3 you have n!/(3!*(n-3)!)=n*(n-1)*(n-2)/3!=n*(n-1)*(n-2)/6

2 minutes ago, Frustration said:

did you just use infinity and compresehsible in the smae sentence?

semi-comprehensible. As a physicist I agree with her interpretation

Posted
1 hour ago, mathiau said:

Yes, there is a way to do it one one step. If the order mattered, for the first Shard you'd have 16 possibilities and 15 for the second, that's 16*15 ordered pairs. Since there's are two possible way to order two Shards, I can just divide by two to get the number of unordered pairs, 16*15/2=120

For the triples I can do a similar reasoning, for the ordered triplets you have 16 possibilities for the first, 15 for the second and 14 for the third and there is 3!=6 ways to order a triplet, therefore 16*15*14/6=560

More generally if you have a set of n distinct object, the number of way to take k among them is n!/(k!*(n-k)!) (note that 0!=1 so it still works for n=k or k=0). If you do the simplification for k=2 you get n!/(2!*(n-2)!)=n*(n-1)/2!=n*(n-1)/2 and for k=3 you have n!/(3!*(n-3)!)=n*(n-1)*(n-2)/3!=n*(n-1)*(n-2)/6

semi-comprehensible. As a physicist I agree with her interpretation

Ok yeah yknow what at this point I'll accept Infinity as the answer until someone gives us a definitive equation...

Posted
Just now, Halyo_Alex said:

Ok yeah yknow what at this point I'll accept Infinity as the answer until someone gives us a definitive equation...

at this point it might as well be.

Posted
15 minutes ago, Halyo_Alex said:

Which means there's a metal that does anything you want.

It doesn't, infinity is so weird that sometimes there's not enough of it

Posted
18 minutes ago, mathiau said:

It doesn't, infinity is so weird that sometimes there's not enough of it

Ok there's Countably Infinite* metals to choose from.

*kind of

Posted

An infinite amount of metals... And we only know what some of them do... A grand total of 20 ( something like that)

Anyone else depressed

Posted
Just now, Bejardin1250 said:

An infinite amount of metals... And we only know what some of them do... A grand total of 20 ( something like that)

Anyone else depressed

I mean just take a guess at what powers you want to exist and pick a metal, you'll probably get it wrong but how would we know

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