Sacred Geometry and Arithmancy.

[Yanisa]

Now that the minimum ranks for 100% success seem to have all been found, how about the minimum ranks for 99% success? I think a 1/100 fizzle chance is a more than acceptable risk for free metamagic under most circumstances.

Based on a 1000 rolls (Credits), the magic number seems to be 10. You only really have to be afraid to roll the same number a bunch of times, because you can easier reach even 9th level spells with 4 numbers, this gives you already 6 dice for wiggle room, which is most cases is enough.

[Sith_Happens]

Okay, so according to that if you keep your skill rank maxed, then from 8th level and onwards you'll have a >99% success chance with any effective spell level you can attempt.

[Vhaidara]

We should import this to 3.5. It doesn't have enough free metamagic for casters to keep up with those bloody Toke of Battle weeabos, and the extra 3 ranks, and therefore 3 dice, would make this completely balanced for them to use with a fair chance of success.

[oxybe]

We should import this to 3.5. It doesn't have enough free metamagic for casters to keep up with those bloody Toke of Battle weeabos, and the extra 3 ranks, and therefore 3 dice, would make this completely balanced for them to use with a fair chance of success.

Seriously. Even just sticking to core, did you see that Fighter class and all the free feats it gets?! This, at least, starts to even the playfield.

[grarrrg]

Seriously. Even just sticking to core, did you see that Fighter class and all the free feats it gets?! This, at least, starts to even the playfield.

Yeah, but the Lightning Warrior Fighter doesn't even get a Familiar. So that helps out a little.

Not quite: at the dangerously overpowered levels, you are a bit behind: you only need 3 dice to succeed with "all 1s" for a level 1 spell.

Level 1 is quite dangerously overpowered at that...


Long drawn out math added to the 3-die-bye-bye post.

[DarkSonic1337]

We should import this to 3.5. It doesn't have enough free metamagic for casters to keep up with those bloody Toke of Battle weeabos, and the extra 3 ranks, and therefore 3 dice, would make this completely balanced for them to use with a fair chance of success.

Besides, it's not like 3.5 even has metamagic feats that are as good as Pathfinder's to apply this to anyway, and you get less of them. So the feat would definitely be weaker for a 3.5 Sorcerer than a PF Sorcerer

[Miss Disaster]

Funny thing though ... Knowledge: Engineering is one of the least popular Knowledge skills (as it doesn't ID any creature types). So at least Sacred Geometry helps promotes a sorely underloved skill to the mass populace. C'mon people, work with me here.

[Starbuck_II]

Funny thing though ... Knowledge: Engineering is one of the least popular Knowledge skills (as it doesn't ID any creature types). So at least Sacred Geometry helps promotes a sorely underloved skill to the mass populace. C'mon people, work with me here.

But Red Hand of Doom uses it. Okay, only in one specific spot. It really should come up in modules/APs more often.

[OldTrees1]

Funny thing though ... Knowledge: Engineering is one of the least popular Knowledge skills (as it doesn't ID any creature types). So at least Sacred Geometry helps promotes a sorely underloved skill to the mass populace. C'mon people, work with me here.

Honestly, Knowledge Dungeoneering can be substituted for a lot of Knowledge Engineering. Maybe they should be combined?

Or pull a Roy

[Khosan]

I've ran the numbers, and this works for all combinations of 3 dice (will post if needed, or if I get bored later).
If 3 dice do not reduce to a 0 or a 1, they WILL let you "convert" any other number into itself.

Spoiler: EDIT: BORED LATER!

Show

Start with all Unique 3-die combinations (56 unique), then remove any with a duplicate number, as these easily reduce to 0 (20 non-doubles). Finally, remove all the combinations that reduce to 0 or 1. This leaves 5 combinations:
1, 2, 5
1, 2, 6
3, 4, 5
3, 5, 6
4, 5, 6
We already proved 4 ,5, 6 in the example

1, 2, 5, 1 > 5-2-1-1 = 1
1, 2, 5, 2 > 5-2-2+1 = 2
1, 2, 5, 3 > 5-3+2-1 = 3
1, 2, 5, 4 > 5-4+2+1 = 4
1, 2, 5, 5 > (5*2-5)*1 = 5
1, 2, 5, 6 > (5+1)*2-6 = 6

1, 2, 6, 1 > 6/2-1-1 = 1
1, 2, 6, 2 > (6-2-2)*1 = 2
1, 2, 6, 3 > 6/3+2-1 = 3
1, 2, 6, 4 > (6-4+2)*1 = 4
1, 2, 6, 5 > (6-1)*2-5 = 5
1, 2, 6, 6 > (6*2-6)*1 = 6

3, 4, 5, 1 > 4+3-5-1 = 1
3, 4, 5, 2 > 5+3-4-2 = 2
3, 4, 5, 3 > 5*3-4*3 = 3
3, 4, 5, 4 > (5-3)*4-4 = 4
3, 4, 5, 5 > 5*4-5*3 = 5
3, 4, 5, 6 > 5+4+3-6 = 6

3, 5, 6, 1 > 5+3-6-1 = 1
3, 5, 6, 2 > 6+3-5-2 = 2
3, 5, 6, 3 > (5*3-6)/3 = 3
3, 5, 6, 4 > 6+5-4-3 = 4
3, 5, 6, 5 > 6*5/3-5 = 5
3, 5, 6, 6 > 6+6/(5-3) = 6

Apologizes if I typed a wrong symbol and missed it.

The shorter way to do this is finding combinations of 4 with no repetition (since any repetition can be brought to 0 with just those two). There's only 15 unique combinations then.

1, 2, 3, 4
1, 2, 3, 5
1, 2, 3, 6
1, 2, 4, 5
1, 2, 4, 6
1, 2, 5, 6
1, 3, 4, 5
1, 3, 4, 6
1, 3, 5, 6
1, 4, 5, 6
2, 3, 4, 5
2, 3, 4, 6
2, 3, 5, 6
2, 4, 5, 6
3, 4, 5, 6

The following sets of 3 (there are 20 unique sets of 3 with no repetition) all reduce to 0:

1, 2, 3 > 3 - 2 - 1
1, 3, 4 > 4 - 3 - 1
1, 4, 5 > 5 - 4 - 1
1, 5, 6 > 6 - 5 - 1
2, 3, 5 > 5 - 3 - 2
2, 3, 6 > (6 / 2) - 3
2, 4, 6 > 6 - 4 - 2

And at least one of these is contained within all but one of the sets of 4 listed above. The only one that doesn't is 3, 4, 5, 6, which still reduces to 0 (6 + 3 - 4 - 5), you just need all 4 numbers to do it.

So for any given set of 4, there always exists a solution that brings it to 0. There are 126 possible sets of 4 (counting repetition this time), of those 111 contain a pair (or triplet or quadruplet) and can be solved with 2, of the remaining 15, 14 can be solved with 3 and the remaining 1 can be solved with all 4.

EDIT: Oh, and for sets of 3, there are 56 unique rolls. 36 of those contain a pair and can be reduced. 7 of the remaining 20 can be reduced to 0 using all 3. The last 13 can't be brought down to 0 without the addition of a 4th roll.

So if you roll 3 dice there's a ~75% chance you'll be able to equate them to 0.

[grarrrg]

MORE NUMBERS!

I ran the numbers for 1st level success with various numbers of dice.

Spoiler: Table Below for results, Spoiler here for calculations

Show

Note: I will try to list results that DON'T work, on the chance I missed something. If there are very few that do work, I will list those instead.
Also note: It's hard/weird to write out strings of numbers, as I'm really not sure how to separate them, hopefully it's understandable.

1st level spells (Primes: 3, 5, 7)
1d6 has a 1/3 (2/6) chance to succeed. You must roll a 3 or a 5
2d6 has a 5/9 (20/36) chance to succeed. All "doubles" fail (1ea.) as do 24, 35, 45, 46, 56 (2ea.)
3d6 has a 11/12 (198/216) chance to succeed. Failures are 155, 166 (3ea.) and 145, 156 (6ea.)
4d6 always succeeds

2nd level spells (Primes: 11, 13, 17)
2d6 has 1/18 (2/36) chance to succeed. You must roll a 5,6 (2ea.)
3d6 has 19/36 (114/216) chance to succeed. Fairly even split, 24 unique combo's work, 32 unique combo's fail.

Spoiler: 3d6 for 2nd level Failures

Show

Anything with a Pair of 1's or a Pair of 2's
All Triplets
Combo's where all numbers are Even
123, 124, 133, 135, 145, 255, 336, 366, 455, 456, 556

4d6 has 1247/1296 chance to succeed. Failures: 1111, 1112, 1113, 1114, 1122, 1123, 1222, 1555, 2222, 2224, 2444, 3333
5d6 has 7753/7776 chance to succeed. Failures: 11111, 22222, 55555 (1ea.), 11112, 11113 (5ea.), 11122 (10)
6d6 has 46649/46656 chance to succeed. Failures: 111111 (1), 111112 (6)
7d6 suspect the only failure will be all 1's.

3rd level spells (Primes: 19, 23, 29)
4d6
5d6 has 1921/1944 (7684/7776) chance. Fails with all combinations of 1's and/or 2's (32 total), 11113, 11114, 11115, 22224 (5ea.), 11123, 11124 (20ea.)
6d6

4th level spells (Primes:)

5th level spells (Primes:)

6th level spells (Primes:)
4d6 85/648 (170/1296) meschlum post.

7th level spells (Primes:)
4d6 41/648 (82/1296) meschlum post.

8th level spells (Primes:)
4d6 has 5/108 (60/1296) meschlum post.

9th level spells (Primes: 101, 103, 107)
4d6 has 5/108 (60/1296) chance. Succeeds only with 4456, 1455, 3455, 1366, 3566 (12 ea.)

For completing the table (if we get that far), I'd recommend starting with the bare minimum number of dice (the "Hail Mary"s from last page, and the lowest level you could cast the spell (i.e. 5th spells would be 9 ranks).
Then fill in where the bigger gaps are. If, as an example, 10 dice with 9th level spells is a 99% success, then it doesn't really pay to check 11, 12, or 13 dice, as they're all going to be fractions of a percent different.

Spaces with a "?" are recommended start points, although any/all data will be accepted.
Spaces to the left of 0% are also 0%, spaces to the right of 100% are also 100%.

	1d	2d	3d	4d	5d	6d	7d	8d	9d	10d	11d	12d	13d	14d
1st	33.33	55.55	91.67	100%
2nd	0%	5.56	52.78	96.22	99.70	99.98		100%
3rd		0%	16.67		98.82				100%
4th		0%	5.56				?				100%
5th			0%	?					?			100%
6th			0%	13.12			?				?		100%
7th			0%	6.33				?					100%
8th			0%	4.63					?					100%
9th			0%	4.63					?					100%

The shorter way to do this is finding combinations of 4 with no repetition (since any repetition can be brought to 0 with just those two)....
EDIT: Oh, and for sets of 3, there are 56 unique rolls. 36 of those contain a pair and can be reduced. 7 of the remaining 20 can be reduced to 0 using all 3. The last 13 can't be brought down to 0 without the addition of a 4th roll.

So if you roll 3 dice there's a ~75% chance you'll be able to equate them to 0.

I think you missed part of the point here. We are not just looking for a 0.

If we can get a solution and have some number of leftover dice, can we automatically "ignore" those extra dice, or not?
There are 3 ways:
Make the extras equal 0, then we can just "+0" our solution and we're good.
Make the extras equal 1, then we can just "*1" our solution and we're good.
Make the extras "condense" into an existing number.

It was already proven that 4 was the minimum to get a 0 or 1 and just disappear outright. I went through and showed that 3 dice either disappear outright, or condense into an existing number.

[Khosan]

I think you missed part of the point here. We are not just looking for a 0.

If we can get a solution and have some number of leftover dice, can we automatically "ignore" those extra dice, or not?
There are 3 ways:
Make the extras equal 0, then we can just "+0" our solution and we're good.
Make the extras equal 1, then we can just "*1" our solution and we're good.
Make the extras "condense" into an existing number.

It was already proven that 4 was the minimum to get a 0 or 1 and just disappear outright. I went through and showed that 3 dice either disappear outright, or condense into an existing number.

I just prefer reducing to 0. Feels cleaner to me, less mathy. It's almost purely addition and subtraction (there's one division in there).

I did leave out 1s in that last bit, which would be a bit misleading. Adding in reductions to 1, the chance of rolling 3 die and being able to reduce them to 0 or 1 is...51/56. Just about 90%. The remaining 5 combinations would require one extra die, like you mentioned. And given that this is just what's left over, there's at least a small chance you could get the same result by just rearranging the actual answer so you end up with a different set of 3.

[grarrrg]

I just prefer reducing to 0. Feels cleaner to me, less mathy.

Were all about the "more mathy" here.
We're trying to find the difference between "probably works" and "definitely works", thus MATH is involved, and simple shortcuts are frowned upon, unless they have been tested and can be relied upon.

...The remaining 5 combinations would require one extra die, like you mentioned. And given that this is just what's left over, there's at least a small chance you could get the same result by just rearranging the actual answer so you end up with a different set of 3.

"Probably" getting rid of 3 extra dice is nice and all, but "Knowing" you can get rid of 3 extra dice is much better.
And we have proven you KNOW you can get rid of 3 extra dice.

the chance of rolling 3 die and being able to reduce them to 0 or 1 is...51/56. Just about 90%.

[MATH!]
Technically 31/36 (or more accurately 186/216) for 86.11% chance to reduce to 0 or 1.
As some unique combinations are truly unique (1/216 for the triples), with 2 groups being more common (3/216 for the "pairs" and 6/216 for the "all different")
[/MATH!]

[meschlum]

So let's look at three dice, then. Mostly because it's easier.

Spoiler: You roll three dice and what do you get?

Show

Level 3 requires 19, 23, or 29, none of which are possible via sums of 3 dice.

If a division is used, there is at most one 'free' die left, with a maximum value of 6. Since we're looking for primes, the divided term needs to be added to the remaining die (multiplying won't help), so this gives a maximum of 12.

Therefore, there is a multiplication and an addition (or a subtraction) (two multiplications won't give a prime), and the addition (or subtraction) happens after the multiplication.

19 + 6 = 25, which can be reached via 5 * 5 only, so 5, 5, 6 is a solution
19 + 5 = 24, which can be reached via 4 * 6 only, so 4, 5, 6 is a solution
19 + 4 = 23, which is a prime and cannot be reached
19 + 3 = 22, which cannot be reached with two dice
19 + 2 = 21, which cannot be reached with two dice
19 + 1 = 20, which can be reached via 4 * 5 only, so 1, 4, 5 is a solution
19 - 1 = 18, which can be reached via 3 * 6 only, so 1, 3, 6 is a solution
19 - 2 = 17, which is a prime and cannot be reached
19 - 3 = 16, which can be reached via 4 * 4 only, so 3, 4, 4 is a solution
19 - 4 = 15, which can be reached via 3 * 5 only, so 3, 4, 5 is a solution
19 - 5 = 14, which cannot be reached with two d6.
19 - 6 = 13 which is a prime and cannot be reached

Since solutions for one prime are solutions for all, the range covered when starting at 29 is 23 to 35, so it covers all options for 23 as well.

26: not possible with two dice
27: not possible with two dice
28: not possible with two dice
29: prime
30: 5 * 6 - 1 is a solution
31: prime
32: not possible with two dice
33: not possible with two dice
34: not possible with two dice
35: not possible with two dice

So the solutions for level 3 are:

1, 3, 6 (6 values)
1, 4, 5 (6 values)
1, 5, 6 (6 values)
3, 4, 4 (3 values)
3, 4, 5 (6 values)
4, 5, 6 (6 values)
5, 5, 6 (3 values)

So there are 36 outcomes where level 3 is possible, so a probability of 1/6 (16.7%)

Level 4 uses 31, 37, 41, so the range of numbers covered is 25 to 47.

There are two solutions for values in the 25 to 35 range (5, 5, 6 and 1, 5, 6).

The maximum with two dice is 36, so this adds a single extra solution (1, 6, 6).

The rolls that succeed for level 4 are therefore:

1, 5, 6 (6 values)
1, 6, 6 (3 values)
5, 5, 6 (3 values)

So Level 4 succeeds with 3 dice with probability 12 / 216 = 1 / 18 (5.56%)

We get, with three dice:

Level 3:1/6 (16.7%)
Level 4: 1/18 (5.56%)


And the reasoning used to solve quickly with three dice breaks down when more are used, since division can be more important. On the other hand, my solutions don't use much division besides getting 1 from pairs, so it might work.

[grarrrg]

The 2's are effectively done. 5 dice is a 99.7% chance, and 6 dice is 99.98%
I don't think there is a need to figure out 7 dice.

We get, with three dice:

Level 3:1/6 (16.7%)
Level 4: 1/18 (5.56%)

And the reasoning used to solve quickly with three dice breaks down when more are used, since division can be more important.

I like the method behind this, but it quickly becomes worthless, as you stated.
Applied to 4 dice at 9th level our target values are:
4*4*6 = 96
4*5*5 = 100
3*6*6 = 108

Double 4 just barely gets you to 101 with a 5
Double 5 gets you 101 with a 1 and 103 with a 3
Double 6 gets 107 with a -1 and 103 with a -5
Each of these combinations appears 12 times each for 5/108 (60/1296)

[Gemini476]

So if you were to actually use Sacred Geometry in a game, obviously staying at enough ranks of K(Eng) to auto-succeed and not have to slow down the game with math, what metamagic feats should you take?

Let's have a hypothetical human metawizard who really likes metamagic. His feats are Scribe Scroll, six [Metamagic] feats, and nine Sacred Geometry feats.
That's a total of 24 metamagics, 18 of which he does not need to learn the prerequisites for before he takes them.

So which should he choose? Heighten Spell, obviously, but what else?

And before you say "all of them", I checked and there are actually 39 [Metamagic] feats so you'll miss out on 15 of them.

[Snowbluff]

I said so earlier. Quicken and Echoing. Quicken spell means you won't lose an important action if you screw up, and it's generally hard to use without making it free.

Echoing buffs the crap out of your longevity as a caster. Double all of my spells level 6 and below? Yes please.

[Erik Vale]

Still/Silent [Duh], Quicken [Speed], Highten [Because free DC booster], Echooing [Boost Longgevity]

[DarkSonic1337]

Heighten

Persistent Spell to make everyone save twice against your stuff, Dazing Spell to add Dazes to stuff, Echoing Spell for longevity, Extend Spell all of your buffs, Lingering Spell for the lulz, Quicken Spell for the obvious action economy breakage, Reach Spell because some touch spells would be really good without the "touch" part, Maximize Spell for your Calcific Touch/Enervation/Whatever (empower too), Selective Spell for your party friendly battlefield control, Silent and Still spell for panic buttons, Toppling Spell for more battlefield control, Umbral Spell+Shadow Grasp for fun darkness stuff and every spell you cast entangles.

Maybe add on Widen and Enlarge too.

[Kantolin]

Do note that quicken, in this case, only allows you to use the spell in its normal activation time rather than as a swift action or the normal full-round action. Which would thus be fairly critical for a spontaneous caster, but not so helpful for a prepared caster; for a wizard I'd probably either skip it or take it as a normal feat.

(...unless I'm misreading Sacred Geometry, or unless you have the metamagic down to 0 in which may as well)

[Snowbluff]

Do note that quicken, in this case, only allows you to use the spell in its normal activation time rather than as a swift action or the normal full-round action. Which would thus be fairly critical for a spontaneous caster, but not so helpful for a prepared caster; for a wizard I'd probably either skip it or take it as a normal feat.

(...unless I'm misreading Sacred Geometry, or unless you have the metamagic down to 0 in which may as well)

*shakes head*
1) You negate the increased casting time, as per Sacred Geometry.
2) You apply Quicken Spell, which works with full round action spells anyway, making it a swift action.

Quicken works with Spontaneous Casters in the feat text.

[Kantolin]

Quicken works with Spontaneous Casters in the feat text.

Wow, you learn something new every day. Definitely a positive change.

1) You negate the increased casting time, as per Sacred Geometry.
2) You apply Quicken Spell, which works with full round action spells anyway, making it a swift action.

Are... you sure this works? It seems to imply 'applying quicken spell negates the increased time'. But then again, I could see it interpreted, 'And then works normally', it just doesn't seem like what it's saying.

I mean Sacred Geometry is pretty whacked out with or without quicken, so I can't say what RAI is from a balance standpoint as there isn't any of this 'balance' stuff. And it's perfectly reasonable that my knee-jerk reaction is to aim weaker.

Either way, if it does work then totally quicken. If it doesn't work, then spontaneous people will want to quicken so it's not two full round actions, while prepareds can probably go without it.

[Snowbluff]

*ahem* If a human with the Human Heritage Feat become a Necropolitan, what happens to it when you try to rebuke it?

Nothing! It's not Human, but the feat says it still counts as such. The same applies here. Sure, it may say one thing, but that does not mean that the other feat or ability still doesn't work as normal.

[Raven777]

I'm pretty sure that the intent of the feat is for the player to calculate the prime combinations from his roll in real time and show it to the table, not just point to a proof.

[Snowbluff]

I'm pretty sure that the intent of the feat is for the player to calculate the prime combinations from his roll in real time and show it to the table, not just point to a proof.

Oh, it is.

1) Intent is useless.
2) We know it works.
3) Since we know it works, why not just skip the slow, tiresome task of running the numbers in real time?

[Raven777]

I don't know... I mean... I feel... I love me some Paragon Surge, but this? This just feels... wrong.

[Snowbluff]

I don't know... I mean... I feel... I love me some Paragon Surge, but this? This just feels... wrong.

But it's so right, Raven. Join me, and as Paladin and Pony we shall rule the realm! Let me show you the true power of the dork side!

[grarrrg]

I'm pretty sure that the intent of the feat is for the player to calculate the prime combinations from his roll in real time and show it to the table, not just point to a proof.

The main issue with this is that the "genius" designer at Paizo apparently couldn't be bothered to run the actual numbers, and thus we wind up with a Feat that is a huge power-boost for the classes that are already the strongest.

In a game with a level cap of, oh, 10, the feat is "ok". Meta-boosts up to 3rd level are 100%, but there's still a chance to fail on your highest level slots (not much of a chance...but a chance).
In a game that goes to level 13 or higher, it's just pure stupid-overpowered-ness, as you can not fail with 14 ranks.

The real 'intent' was "Hey guys, I have a stupid idea for a feat that requires lots of numbers and stuff! The players will probably look at it and go 'I'm not doing math', so let's throw in a HUGE bonus if they actually do!"


Also, why'd they skip 2? 2 is a perfectly good Prime Number.

[Snowbluff]

The real 'intent' was "Hey guys, I have a stupid idea for a feat that requires lots of numbers and stuff! The players will probably look at it and go 'I'm not doing math', so let's throw in a HUGE bonus if they actually do!"

Players not doing math... I can't even... grarrrg help me! D:

Also, why'd they skip 2? 2 is a perfectly good Prime Number.

Maybe the designer doesn't know what a prime number is?

[Kalaska'Agathas]

Maybe the designer doesn't know what a prime number is?

But then how did they get the rest of them? I mean, if they're picking from a list of primes, it's not like two isn't going to be on there.