https://binghsien.tumblr.com/post/797952616610119680

tsumiki-miniwa:

tsumiki-miniwa:

prokopetz:

tsumiki-miniwa:

prokopetz:

Huh. I think I just figured out a relatively elegant way to randomly assign a standard 7/6/5/4/3 array to the Costume Fairy Adventures Facets.

First, roll once on the following table, rerolling doubles:

d66 Strong Facets Weak Facets
11 ———- Reroll ———-
12 Moxie, Focus Craft, Grace
13 Moxie, Focus Craft, Shine
14 Moxie, Focus Grace, Shine
15 Moxie, Craft Focus, Grace
16 Moxie, Craft Focus, Shine
21 Moxie, Craft Grace, Shine
22 ———- Reroll ———-
23 Moxie, Grace Focus, Craft
24 Moxie, Grace Focus, Shine
25 Moxie, Grace Craft, Shine
26 Moxie, Shine Focus, Craft
31 Moxie, Shine Focus, Grace
32 Moxie, Shine Craft, Grace
33 ———- Reroll ———-
34 Focus, Craft Moxie, Grace
35 Focus, Craft Moxie, Shine
36 Focus, Craft Grace, Shine
41 Focus, Grace Moxie, Craft
42 Focus, Grace Moxie, Shine
43 Focus, Grace Craft, Shine
44 ———- Reroll ———-
45 Focus, Shine Moxie, Craft
46 Focus, Shine Moxie, Grace
51 Focus, Shine Craft, Grace
52 Craft, Grace Moxie, Focus
53 Craft, Grace Moxie, Shine
54 Craft, Grace Focus, Shine
55 ———- Reroll ———-
56 Craft, Shine Moxie, Focus
61 Craft, Shine Moxie, Grace
62 Craft, Shine Focus, Grace
63 Grace, Shine Moxie, Focus
64 Grace, Shine Moxie, Craft
65 Grace, Shine Focus, Craft
66 ———- Reroll ———-

Assign values of 7 and 6 to the Facets listed under the “Strong Facets” column. To do so, roll a single die. On any odd roll, set the first-listed Facet to 7 and the second-listed Facet to 6; on any even roll, set the first-listed Facet to 6 and the second-listed Facet to 7.

Next, assign values of 4 and 3 to the Facets listed under the “Weak Facets” column, following the same procedure as above.

Finally, set your remaining Facet to 5.

(There are 120 ways to assign five distinct values to five slots, which has a pesky prime factor of 5 floating around in it, so I believe rerolling one time in six is the best we can do with only d6s and d12s to work with. And no, the Eat God stretch goal for the Costume Fairy Adventures conversion guide hasn’t been met yet, but I’m reasonably sure it will be, so I’m getting the jump on some of the math shit!)

This nerd sniped me pretty hard. Probably would’ve been stronger if I was at a computer.

TLDR: I think 1/6 rerolls to get to 120 is the best you can do. It’s possible that a d6666 or something could do better, but I doubt it.

d66 has 36 options, and accepting 1/36 rerolls, you get to 5*7, which has the 5 factor (yay!) but also a 7 factor that’s hard to remove. Doing even 1/7 rerolls is already inefficient, since (1/7)+(1/36) is greater than 1/6.

Doing d6 with d12 has 72 options. 71 is useless here (prime), and 70 is 35*2, so it has all of the same problems as 35.

Doing d12 with d12 gets to 144. Besides the obvious 140 option (35*4, so 7 issue again) there’s also the idea of going straight to 120 from here. Unfortunately, that rerolls 24/144 of the time, or 1/5, and 1/5>1/6.

The last thing I tried was d666, getting to 216 options. Accepting one reroll gets to 215 options, or 43*5 - which DOES get the 5 factor, but 43 is also prime and needs to be cleaned.

• Rerolling 1/43, to get to 42 options, is 6*7. There’s still no good way to get from 7 to 6 here.
• Rerolling 2/43 gets to 41 options, a prime number.
• Rerolling 3/43 gets to 40 options, or… 8*5. Funny enough, the issue is now that there’s too many 5s, and I don’t see a way to clean this one up.
• Rerolling 4/43 gets to 39 options, or 3*13. 13’s prime, and accepting another 1/13 rerolls at this point is now more inefficient than the target 1/6.
• Rerolling 5/43 gets to 38, or 2*19. Accepting another 1/19 rerolls is now more inefficient than target.
• Rerolling 6/43 gets to 37. Prime.
• Rerolling 7/43 WOULD get to 36, which would theoretically be better, BUT: (7/43)+(1/216) is more than 1/6, making it more inefficient than the original target.

We actually can do a bit better, though it’s ugly as hell.

A lookup table with equally weighted results must have a number of entries equal to some product of the prime factors of the number of faces on the available dice. The smallest number equal to or greater than 120 which has prime factors of only 2 and 3 (i.e., those furnished by a d6) is 128; every number in the range from 120–127 either has an unreachable prime factor or is itself prime.

128 is 2^7, which can be reached through a series of seven consecutive odds-or-evens rolls on a d6; alternatively, if we also allow d12s as stated in the initial post, we can read d12s by threes to use them as ersatz d4s, and reach 128 equally weighted options using a d12,d12,d12 lookup table plus a single odds-or-evens roll on any die.

This gives us 8/128 = 6.25% forced rerolls, which I believe is provably the best we can possibly do.

(I did not consider this option in the original post because I cannot for the life of me think of any way to do this elegantly. I’m open to suggestions if you have any ideas!)

Now that I see it in that format, there is a more efficient method: And my original theory that a d6666 can do better is correct!

To do it, instead treat each d6 as a d3.

d3333 has 243 possible outcomes. Rerolling 3 of these gets to 240 possible outcomes, which can be mapped to 120 easily. This is also 3/240=1.25% forced rerolls, which is better.

Now, as for how to do this elegantly: I… have no idea. Barring some idea of recursive lookup tables where the first table is effectively equivalent to “Go to table X, where X is 1d33” (note that table 33, options 31-33 would all need to loop back), or having a gigantic table that has 243 rows (and therefore also take like 3 pages), I… do not see another answer.

(Unrelated, but now that I see it this way, there’s probably a proof that you can approach 0% rerolls as you add more and more dice.)

Laid down to think about it, I got a proof that you can approach 0% rerolls with enough dice (though of course never reach 0%).

Fair warning though: This is now completely in the realm of mathematics, and any idea of “elegantly translating it into a tabletop RPG book” is completely out the window.

If you can route to any multiple of 120, say k*120 equally likely outputs, then no more rerolls are necessary. Just divide the outputs into 120 equally sized and equally likely buckets, and assign outputs accordingly.

Therefore, if there are n possible outcomes from [a pile of dice], there is no reason to ever reroll more often than probability 119/n. If you were to reroll with probability (120+m)/n, you could’ve just rerolled with probability (m/n) anyway, for the same result.

Finally, note that (in the bounds of this example) there is no limit to how large n could be. You could roll a very large pile of dice, producing a very very large value of n, producing a worst-case reroll frequency of 119/n, which approaches 0 as n quickly approaches infinity.

https://binghsien.tumblr.com/post/797952616610119680