L5R 5th Edition Probability Results

[lord4571]

so i keep trying to look up the way to calculate these Fudge dice yet it seems every article or probability i dont see the calulation that i want or at least not recognizing it.

their are 2 sets of dice. one being Ring dice, and the other Skill dice, but you can only keep results of X dice = to the total Ring dice you roll. im currently not worrying about which you can keep and just the end results.

i will use R to mean Ring, and S to mean skill.
their are 4 different symbols and blank sides.
- Success is 1 point to completing the task.
- Strife stresses the person.
- Exploding Success is 1 point to completing the task and provides an extra roll of that dice. so if it explodes on a R you roll another R. if it explodes on a S you roll another S. you will always be able to Keep the result of the additional rolled dice.
- Opportunity is an extra benefit to the task but doesn't count as succeeding

R dice are 6 sided
1 -
2 Success
3 Success with strife
4 Exploding Success with strife
5 opportunity
6 opportunity with strife

S dice are 6 sided
1 -
2 -
3 Success
4 Success
5 Success with opportunity
6 Success with strife
7 Success with strife
8 Exploding Success
9 Exploding Success with strife
10 opportunity
11 opportunity
12 opportunity

now im just wanting to be able to figure out xR with yS vs z DC if you could only take success without strife (up to 8),
then the same but with or without strife (up to 8),
opportunitys without strife minimum of 3 or more,
and lasty opportunity's with or without strife minimum of 3 or more.

[Telok]

I like doing brute force analysis of this sort of stuff. It's easier than stats, although you do need to use stats to validate at least a couple test runs too. It may be a few days before I could start on it (RL stuff), but if you can link me to a good explanation of the system I can write a light-weight java program or something that would do it.

I presume that the S dice are d12, yes? Your post says d6 but the table you give has 12 entries.

[Knaight]

Fudge dice are something else entirely, so if you're looking for statistics there you'll get things that are off. Similarly which dice you keep is absolutely essential to die distributions; it's why Advantage in D&D 5e behaves differently than just a d20 roll.

The rest of this is bionomial probability, though success with some numbers of stress but not zero is two probability distributions deep and generally kind of gross. I'd probably do a brute force and count situation here.

[lord4571]

but if you can link me to a good explanation of the system I can write a light-weight java program or something that would do it.

I presume that the S dice are d12, yes? Your post says d6 but the table you give has 12 entries.

Yes the S are D12s and rings are D6s.
http://i.4pcdn.org/tg/1508300409327.png
that is the dice sides.
the highest TN (target number = difficultly check) is 8. the lowest is 1.

Besides this their are several other little affects, but idk what you need from the system. this system is brand new so finding free content isnt that easy. there is a beta book but a lot has changed since the offical release. so if you can tell me what info you need i can give you the answers.

[Telok]

Besides this their are several other little affects, but idk what you need from the system. this system is brand new so finding free content isnt that easy. there is a beta book but a lot has changed since the offical release. so if you can tell me what info you need i can give you the answers.

Ah let's see. Numbers of dice rolled, numbers of dice kept, what counts as a success. That sort of thing. Particularly how the output should be set up to convey the correct information.

For example I did a statistical roller for Dungeons the Dragoning 40K 7th ed. 1.6e. The dice system for that game is roll X d10s, keep Y d10s, explode on 10. If you go over 10 rolled dice add 1 to the kept dice per 2 over. If you go over 10 kept dice add 5 to the total per extra die. There are circumstances where you can reroll dice that get a 1 (or more sometimes) on the first roll, and there are circumstances where the dice can explode on 9s and 10s. You can add/subtract various amounts to the final number and there is a circumstance where it matters how many10s were rolled.

I ended up with a little program that took WkXrYxZ{+/-}N which parses out as roll W, keep X, reroll on Y, explode on Z, add/subtract N. With everything after WkX being optional. The output is a list of the totals split into tenth percentiles with a list of how many 10s were rolled how many times following the total.

So my understanding is that you're rolling X d6(R) and Y d12(S), keeping X number of from those results. An exploding die counts as itself and you can keep the additional die or not, as you wish. We're counting successes versus TN, opportunities, and strifes. A particular interest is success with and without strifes, and 3+ opps with and without strifes. Do you need successes or successes >= TN for the 3+ opp rolls? Are there minimums or maximums forth numbers of dice? Is there any point at which you stop rolling explosions? A max number of anything?

So I'm thinking taking X r-dice and Y s-dice as input, output 0 to 8+ successes as a percentage as one column, add #s of strifes with % occurrence following each TN. Repeat the table with opportunities instead of strifes. Since there are only 9 TN outcomes I could output 9 tables with the % success rate per table and a strife/opp matrix as the table.

[lord4571]

So my understanding is that you're rolling X d6(R) and Y d12(S), keeping X number of from those results.

i think you got it. you roll Xd6(R) and Y d12(S), Keeping the amount of R dice you rolled. (so if you roll 3 ring dice, you can keep 3 from the total rolled.

An exploding die counts as itself and you can keep the additional die or not, as you wish.

if you keep an exploding dice you can roll 1 more of that type of dice, you can chose to keep that additional dice or not. exploding symbol counts as a success.

Do you need successes or successes >= TN for the 3+ opp rolls?

you need successes to be the match or beat the TN, but sometimes when you roll you arn't trying to succeed. sometimes you are just trying to get Opp.'s without succeeding. Opp.'s can be spent on effects even if you fail. normally to activate those effects you need 1-3 Opp.'s. the more Opp.'s you have the more effects you can activate if kept.

Are there minimums or maximums forth numbers of dice?

You will always have 1 ring. you can have a Max of 7 ring dice by spending a Void point, getting assistance, and having the max ring stat of 5.
You can have 0 skill dice. you can have a Max of 6 skill dice by getting assistance, and having the max skill stat of 5.

Is there any point at which you stop rolling explosions? A max number of anything?

If you keep an explosive dice, you roll an additonial dice. if that dice explodes and you keep it, you roll again. this can go on until either it doest explode or if you chose not to keep it.
last session i rolled an explosive that exploded 5 more times.

So I'm thinking taking X r-dice and Y s-dice as input, output 0 to 8+ successes as a percentage as one column, add #s of strifes with % occurrence following each TN. Repeat the table with opportunities instead of strifes. Since there are only 9 TN outcomes I could output 9 tables with the % success rate per table and a strife/opp matrix as the table.

8 is going to normally be the Max TN, and even then we have only ran into a TN 6 by interacting with characters in an improper way due to personality, status, and other reasons.

TN 1: An easy task for most people
TN 2: An average task
TN 3: A difficult task
TN 4: A very hard task
TN 5: An extremely hard task.
TN 6: An extraordinary task
TN 7: A heroic task
TN 8 (or higher): A legendary task

and like i said Opp.'s do have effects. the most general one is if you keep an Opp. and you kept a dice with strife, you can negate the strife with the Opp.'s. there are other effects that are generally far more useful but sometimes its the best thing to do.

the reason why strife is so important is if you get to much strife at any point (strife builds up). you cant keep any dice with strife on it until you lower your own strife. which their are only a few ways that arn't so reliable in some situations. especially if you get like 20 strife built up. the only way to cure all of that at any given point is to outburst. though that will come with major repercussions (including character death).

[Telok]

I've got it rolling and sorting. I'm working on formatting, stats, and display.

Unfortunately family woke up and progress has tanked because people.

I'll try to get them to shut up and let me work sometime.

[lord4571]

I've got it rolling and sorting. I'm working on formatting, stats, and display.

Unfortunately family woke up and progress has tanked because people.

I'll try to get them to shut up and let me work sometime.

thats perfectly fine. RL always comes first and while math is fun its one of those things that just might take some time and a quiet spot. so dont worry to much. i appericate that your doing this for me (and others who notice this and play L5R.

the system is really punishing so its really useful things to note. as one slip up could be your death.

[meschlum]

Since R is limited to 7 at most, the probability of having 8 successes to reach TN 8 is 0 (since you can keep at most 7 successes).

Working out the math is messy otherwise, with a lot of apparent complexity for the sake of obscurity. I'll see what I can crunch out in the abstract, for the math fun if nothing else.

[Floret]

Since R is limited to 7 at most, the probability of having 8 successes to reach TN 8 is 0 (since you can keep at most 7 successes).

Working out the math is messy otherwise, with a lot of apparent complexity for the sake of obscurity. I'll see what I can crunch out in the abstract, for the math fun if nothing else.

It's not impossible, since explosive successes (and consecutive explosions) are a thing. There is no theoretical maximum number of successes, even if the likelyhood quickly approaches zero (while not quite getting there).

[lord4571]

Correct exploding sucesses does allow someone to get 8 or higher.

In my last session rolling 3r and 3s. I was able to get 7 successes cuz of exploding. While low chance its probable.

[meschlum]

My misunderstanding, from text stating that you have to keep an exploding die to reroll it, and that you can then choose to keep the reroll (or not, presumably) - so since the reroll can be kept, it seemed like another kept die.

Difficulty = 1

This is easy. It works unless you don't get a success, so:

P(no success without strife) = (5/6)^R * (2/3)^S

Hence, P(success without strife) = 1 - (5/6)^R * (2/3)^S

P (no success) = (1/2)^R * (5/12)^S

Hence, P (success with strife possible) = 1 - (1/2)^R * (5/12)^S


Difficulty = 2

Slightly more challenging, with a few limit cases.

R = 1, S = 0: Probability is 0 without strife (R die always has strife when it explodes), and 1/12 with strife (1/6 of exploding the die, 1/2 of getting a second success)

Without strife, an R die gets a success with probability 1/6, and a S die gets 1 success with probability 1/4, and an exploding success with probability 1/12

With only R dice, you get 0 successes with probability (5/6)^R, 1 success with probability R / 6 * (5/6)^(R - 1), so you fail with probability (5/6)^(R - 1) * (5 + R) / 6 (With R = 1, this gives 1, as anything to the power 0 is equal to 1)

If you add S dice, whether they are used depends on your results with the R dice - if you have enough from the R, the S are irrelevant.

With 1 success from the R, you have a probability of (2/3)^S of getting 0 successes from the S, so that means you have an overall probability of failure of R / 6 * (5/6)^(R-1) * (2/3)^S

With 0 successes from the R, you need 2 successes (or more) from the S.

You get 0 successes from the S with probability (2/3)^S
You get 1 success from the S with probability S / 3 * (2/3)^(S-1) if you ignore the exploding die.
In fact, you have S / 4 * (2/3)^(S-1) of rolling a single success and S/12 * (2/3)^(S-1) of rolling a single exploding success. In the latter case, you have a 2/3 chance of not getting another success, so your total odds of getting one success out of the S are (2/3)^(S-1) * S * (1/4 + 1/18), or S * (2/3)^(S-1) * 11/36

So you have a total probability of failure of (5/6)^R * ((2/3)^(S-1) * 11 S / 36 + (2/3)^S) + (5/6)^(R-1) * (R / 6) * (2/3)^S

Which can be simplified to (5/6)^(R-1) * (2/3)^(S-1) * (5/6 * 11 S / 36 + 5/6 * 2/3 + 2/3 * R/6), or

(5/6)^(R-1) * (2/3)^(S-1) * (120 + 55 S + 24 R) / 216

Testing: S = 0 gives (5/6)^(R-1) * 3/2 * (120 + 24 R) / 216 = (5/6)^(R-1) * (180 + 36 R)/216 = (5/6)^(R-1) * (5 + R)/6

And from there, you can work out the math for higher difficulties: get the odds of meeting the difficulty with R dice, see how many success you need from S dice when the R aren't enough, and factor in exploding dice. Remember to use the components you've created - in this case, we have the odds of getting 0 and 1 success with S dice (it's easy with R dice), and can use that to work out the probability of getting 2 successes with S dice, etc. Limiting the number of dice means we have a limited number of scenarios to process, which also helps.

[meschlum]

Approaching the issue from a different direction!

R die

An R die is worth the following:

No strife

5/6: 0 Successes (or with strife)
1/6: 1 Success

5/6: 0 Opportunity (or with strife)
1/6: 1 Opportunity (or with strife)

With strife

1/2: 0 Successes
1/3: 1 Success (no rerolls)
1/6: 1 Success and a reroll
1/2: 1 Success
1/3: 2 Successes (no rerolls)
1/6: 2 Successes and a reroll
1/2: 2 Successes
1/3: 3 Successes (no rerolls)
1/6: 3 Successes and a reroll
...

So:
1/2: 0 Successes
5/12: 1 Success
5/72: 2 Successes
5/432: 3 Successes
...

Which boils down to: 1/2 * (1/6)^N * 5 of getting N successes with N >= 1, and 1/2 of getting 0 successes.

Since the maximum used is 8 successes, you can replace the last term (5) with a 6 when N = 8.

Rerolls do not add opportunities, but let you roll again to have another chance. So you have:

3/5: 0 Opportunities
2/5: 1 Opportunity

S die

An S die is worth the following:

No strife

8/12: 0 Successes (or with strife)
3/12: 1 Success
1/12: 1 Success and reroll
8/12: 1 Success
3/12: 2 Successes
1/12: 2 Successes and reroll
...

So:
8/12: 0 Successes
44/144: 1 Success
44/1728: 2 Successes...

Which boils down to 1/12 * (1/12)^N * 44 for N >= 1 Successes, and 8/12 for N = 0. Since the maximum is 8 successes, you can replace the last term (44) by 48 for N = 8.


When dealing with opportunity, there is a 1/12 chance of getting a reroll, giving another chance at opportunity without strife, and 4 results that grant opportunity without strife, so you get:

7/11: 0 Opportunity
4/11: 1 Opportunity

With Strife

5/12: 0 Successes
5/12: 1 Success
2/12: 1 Success and reroll
5/12: 1 Success
5/12: 2 Successes
2/12: 2 Successes and a reroll

So...
5/12: 0 Successes
35/72: 1 Success
35/432: 2 Successes
...

Which boils down to 1/12 * (1/6)^N * 35 for N >= 1 successes, 5/12 for N = 0, and you can replace the 35 with a 42 when N is 8, since difficulties over 8 don't matter.

For opportunity, you still have 4 results which give you 1 Opportunity, but two rerolls that don't contribute to the final result, so you have:

3/5: 0 Opportunity
2/5: 1 Opportunity

Which is equivalent to what you get with an R die and ignoring Strife. S dice still give you better odds when you avoid Strife, but it's worth noting!

Bringing it together

You now have the odds of getting any number of successes on your R and S dice, so you can break up a DC of X into parts: 0 to X successes on R dice, and X to 0 successes on S dice. Then you can sum them up, and done!

For opportunity, with the goal being 3 Opportunity, you can do the same - since the dice are simpler, it's easier to compute, too. Note that since each die is worth 1 Opportunity at most, you need R >= 3 to get to 3 Opportunity - you have to be able to keep 3 or more dice!

[meschlum]

Opportunity Math

You can fail to have 3 Opportunities in the following ways:

0 Opportunities on R, 0 - 2 on S
1 Opportunity on R, 0 - 1 on S
2 Opportunities on R, 0 on S

So...

No Strife

You always have 0 dice from R, so you want the odds with S (this also means you need at least 3 S dice, as well as 3 R dice so you can keep them).

0 Successes with S: (7/11)^N, where N is the number of S dice you have
1 Success with S: 4/11 * N * (7/11)^(N-1)
2 Successes with S: (4/11)^2 * N * (N - 1) * (7/11)^(N-2) / 2

So you have in all: (7/11)^(N - 2) * (7^2 + 4 * N * 7 + 4^2 * N * (N-1) / 2) / 11^2

(7/11)^(N - 2) * (49 + 20 N + 8 N^2) / 11^2 as your chance of failure (1 when N = 2 or less), which can be solved for N = 3, 4, and 5.

N = 3: ~ 95% chance of failure, so 5% chance of getting 3+ Opportunities without Strife (R >= 3)

N = 4: ~ 86% chance of failure, so 14% chance of getting 3+ Opportunities without Strife

N = 5: ~74% chance of failure, so 26% chance of getting 3+ Opportunities without Strife

This could be reproduced fairly closely by rolling a D20, where you have a DC of 26 (impossible) and +2 to your roll per S die (so 1/20 = 5% with S = 3, etc.)

Strife

R and S dice are equivalent in this case, you you just care about the total number of dice rolled (and need 3 or more R dice).

0 Successes: (3/5)^N
1 Success: 2/5 * N * (3/5)^(N-1)
2 Successes: (2/5)^2 * N * (N - 1) * (3/5)^(N - 2)

Where N is the total number of dice you're rolling (R and S).

So you get (3/5)^(N-2) * (3^2 + 2 * 3 * N + 2^2 * N * (N -1) / 2) / 5^2

Which is equal to (3/5)^(N-2) * (9 + 4 N + 2 N^2) / 5^2 as your chance of failure (1 when N = 2 or less)

Since this involves a pool of 3 to 12 dice, there are a lot more cases to check.

N = 3: 93.6% chance of failure (6.4% chance of success)
N = 4: 82% chance of failure (18% chance of success)
N = 5: 68% chance of failure (32% chance of success)
N = 6: 54.4% chance of failure (45.6% chance of success) (minimum number of dice to be able to succeed without Strife, with 3/3)
N = 7: 42% chance of failure (58% chance of success)
N = 8: 31.5% chance of failure (68.5% chance of success)
N = 9: 23% chance of failure (77% chance of success)
N = 10: 16.7% chance of failure (83.3% chance of success) (guaranteed to able to succeed without Strife, as R is 7 or less, so S is 3 or more)
N = 11: 11.1% chance of failure (88.9% chance of success)
N = 12: 8.3% chance of failure (91.7% chance of success)

Which is not quite mapable to a D20, but you can come fairly close with a D20 and a D6 (a D60 would come very close)

[Telok]

Sorry I've taken so long. There's a new baby in the house, I still have a game to DM, job, and... Yeah, life.

Any ways the program is at https://bitbucket.org/JCamp8/l5rdiestatter/src/master/ which should be marked as a public repo and thus be accessible. It in java, for simplicity, and includes a .jar file in the /dist directory that you should be able to run anywhere.

It currently only does one selection process, minimum strife > maximum successes > maximum opportunity. Since the logic behind choosing which dice to keep changes with what outcomes you want each different outcome needs to be it's own function.

I'll work on the other six outputs off and on again, updating the repository and this thread as I do. Working on this makes a nice break from other stuff at times.