Modified Rule of Six - Dumpshock Forums

Modified Rule of Six, ...and "always having a shot"

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GunnerJ

Apr 7 2005, 11:12 PM

Post #9

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Just to give a bit more substance, I went and modified my SR4 success calculator to show what would happen with the Rule of Six I outlined.

Here is a sample of the old results, which doesn't use the new Rule of Six:

CODE

For 100000.0 rolls:
Average successes for 1 dice is 0.33368
Average successes for 2 dice is 0.66763
Average successes for 3 dice is 1.00079
Average successes for 4 dice is 1.32859
Average successes for 5 dice is 1.66718
Average successes for 6 dice is 1.99574
Average successes for 7 dice is 2.34018
Average successes for 8 dice is 2.66486
Average successes for 9 dice is 3.00747
Average successes for 10 dice is 3.33277
Average successes for 11 dice is 3.66099
Average successes for 12 dice is 3.99936
Average successes for 13 dice is 4.33072
Average successes for 14 dice is 4.66684
Average successes for 15 dice is 4.99718
Average successes for 16 dice is 5.33722
Average successes for 17 dice is 5.66404
Average successes for 18 dice is 5.99436
Average successes for 19 dice is 6.32825
Average successes for 20 dice is 6.66187
Average successes for 21 dice is 6.99238
Average successes for 22 dice is 7.33322
Average successes for 23 dice is 7.65661
Average successes for 24 dice is 7.99934
Average successes for 25 dice is 8.32793

Here's the results for rolling with the new Rule of Six:

CODE

For 100000.0 rolls:
Average successes for 1 dice is 0.40367
Average successes for 2 dice is 0.79871
Average successes for 3 dice is 1.1997
Average successes for 4 dice is 1.59618
Average successes for 5 dice is 2.00199
Average successes for 6 dice is 2.40159
Average successes for 7 dice is 2.7964
Average successes for 8 dice is 3.2053
Average successes for 9 dice is 3.59774
Average successes for 10 dice is 3.99779
Average successes for 11 dice is 4.40174
Average successes for 12 dice is 4.79707
Average successes for 13 dice is 5.21423
Average successes for 14 dice is 5.6
Average successes for 15 dice is 5.9991
Average successes for 16 dice is 6.40154
Average successes for 17 dice is 6.80289
Average successes for 18 dice is 7.20725
Average successes for 19 dice is 7.59984
Average successes for 20 dice is 8.00396
Average successes for 21 dice is 8.40999
Average successes for 22 dice is 8.80651
Average successes for 23 dice is 9.20198
Average successes for 24 dice is 9.59781
Average successes for 25 dice is 10.00305

For any interested, here's the code. First the old method sans reroll:

CODE

public int getHits(int dice)
{
int result;
int hits = 0;

for (int j = 0; j < dice; j++)
{
result = r.nextInt(6)+1;
if (result >= 5) hits++;
}

return hits;
}

Now, the new method with rerolls:

CODE

public int getRerollHits(int dice)
{
int result;
int hits = 0;

for (int j = 0; j < dice; j++)
{
result = r.nextInt(6)+1;
if (result >= 5) hits++;
int q = 0;
while ((result == 6) && (q < 20))
{
result = r.nextInt(6)+1;
if (result >= 5) hits++;
q++;
}
}

return hits;
}

Note, r is a random number generator. In Java, the NextInt(x) function of Random gets a number from 0 (inclusive) to x (exclusive), hence why result equals NextInt(6)+1. Also, I limited the number of rerolls to 20.

Kagetenshi View Member Profile	Apr 7 2005, 11:15 PM Post #10
Manus Celer Dei Group: Dumpshocked Posts: 17,006 Joined: 30-December 02 From: Boston Member No.: 3,802	Out of interest, why do you sample rolls? It seems to me it would be better (both to avoid statistical anomalies and to avoid any weaknesses in your RNG) to just calculate the theoretical expected successes. There are visible anomalies creeping in there all over the place (look at expectation for one die sans NRo6, for instance; you've been rolling high). ~J

GunnerJ View Member Profile	Apr 7 2005, 11:17 PM Post #11
Moving Target Group: Members Posts: 669 Joined: 25-May 03 Member No.: 4,634	The results that I feel are most important to compare are for 6 dice and 12 dice. 6 dice represents SR4's average ability, and the successes on a 6D6 roll represent what an average man on the street with intermediate training can pull off. 12 represents the most exceptional ability in SR4 from chargen and without enhancements from areas other than attributes and skills. The successes from a 12D6 roll represent what an exceptional individual with expert training can accomplish. I contend a Shadowrunner will roll dice closer to 12 than to 6 in most cases, but for establishing a global norm, 6 dice results are important.

GunnerJ

Apr 7 2005, 11:22 PM

Post #12

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QUOTE (Kagetenshi @ Apr 7 2005, 11:15 PM)

Out of interest, why do you sample rolls? It seems to be it would be better (both to avoid statistical anomalies and to avoid any weaknesses in your RNG) to just calculate the theoretical expected successes. There are visible anomalies creeping in there all over the place (look at expectation for one die, for instance; you've been rolling high).

~J

I sample the rolls because just doing formulaic analysis, at least without the rerolls, is a matter of just dividing the dice rolled by 3. I wanted something closer to what actually rolling the dice would get. As for the reroll method, I lack the mathematical ability to calculate expected averages for it.

What do you mean an anomoly on one die? Look at the results sans rerolling: one die averages 0.33368 successes, which is very close to 1/3, which is what we expect. I'm not sure I follow you.

Kagetenshi View Member Profile	Apr 7 2005, 11:23 PM Post #13
Manus Celer Dei Group: Dumpshocked Posts: 17,006 Joined: 30-December 02 From: Boston Member No.: 3,802	You just noted it yourself, it's off by over three millionths. You're not, paradoxically, going to get closer to actually rolling the dice by actually rolling the dice, at least not while your dierolling doesn't approach infinity. ~J

GunnerJ View Member Profile	Apr 7 2005, 11:24 PM Post #14
Moving Target Group: Members Posts: 669 Joined: 25-May 03 Member No.: 4,634	Um... so?

Kagetenshi View Member Profile	Apr 7 2005, 11:25 PM Post #15
Manus Celer Dei Group: Dumpshocked Posts: 17,006 Joined: 30-December 02 From: Boston Member No.: 3,802	So it's wrong, and for at least that portion it's easily corrected. I'll see if I can figure the formula for the with-reroll side; it shouldn't be difficult, once I get some caffeine in me. Actually, I think what bugs me isn't that it's wrong, but that it was far more work to do what you did than to get the precise value. I'm really not sure what the point of it was. ~J

GunnerJ View Member Profile	Apr 7 2005, 11:25 PM Post #16
Moving Target Group: Members Posts: 669 Joined: 25-May 03 Member No.: 4,634	Look, I'm doing this for fun, because I want to make guesstimates about a P&P RPG. I'm not building a bridge here. Lives aren't on the line. Is there some reason why we need accuracy to the millionth degree?

Kagetenshi View Member Profile	Apr 7 2005, 11:26 PM Post #17
Manus Celer Dei Group: Dumpshocked Posts: 17,006 Joined: 30-December 02 From: Boston Member No.: 3,802	As my ninja-edit states, the part that gets to me is that it was extra work for reduced accuracy. ~J

GunnerJ

Apr 7 2005, 11:27 PM

Post #18

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QUOTE

Actually, I think what bugs me isn't that it's wrong, but that it was far more work to do what you did than to get the precise value. I'm really not sure what the point of it was.

Well, let's see. I'm not a math/stats wizard. I am a (fairly, IMHO) competant programmer. I enjoy programming, further, more than doing statistical analysis, and it's easier for me. Were I able to do the analysis you suggest, I would have, probably. But I am not able to do it. Are you starting to get the picture?

Maybe we could discuss something relevent now? Like the 6D6 and 12D6 results perhaps?

Kagetenshi View Member Profile	Apr 7 2005, 11:36 PM Post #19
Manus Celer Dei Group: Dumpshocked Posts: 17,006 Joined: 30-December 02 From: Boston Member No.: 3,802	Fair enough. Incidentally, unless I messed up my math, you should expect exactly .4 successes per die with exploding sixes (which jibes with your findings). So what about 6 and 12 dice are we looking at? We've got 2.4 and 4.8 successes, so that suggests that any 3+ Threshhold test is fairly hard, any 5+ very difficult indeed. Doesn't seem like a lot of modifier wiggle-room for Threshold… ~J

GunnerJ

Apr 7 2005, 11:38 PM

Post #20

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QUOTE (GunnerJ @ Apr 7 2005, 11:17 PM)

The results that I feel are most important to compare are for 6 dice and 12 dice. 6 dice represents SR4's average ability, and the successes on a 6D6 roll represent what an average man on the street with intermediate training can pull off. 12 represents the most exceptional ability in SR4 from chargen and without enhancements from areas other than attributes and skills. The successes from a 12D6 roll represent what an exceptional individual with expert training can accomplish. I contend a Shadowrunner will roll dice closer to 12 than to 6 in most cases, but for establishing a global norm, 6 dice results are important.

My intention here is to show that while exploding sixes allow the theoretical chance to pull anything off, they don't grant a huge number of average greater successes.

6D6: w/o reroll avg = about 2, w/ reroll avg = about 2.4
12D6: w/o reroll avg = about 4, w/ reroll avg = about 4.8

It's not a huge deal, and it allows for something that I at least think is important, as do several others.

EDIT:

QUOTE

Incidentally, unless I messed up my math, you should expect exactly .4 successes per die with exploding sixes (which jibes with your findings).

The way I see it, they are very much close enough for my purposes.

Kagetenshi View Member Profile	Apr 7 2005, 11:40 PM Post #21
Manus Celer Dei Group: Dumpshocked Posts: 17,006 Joined: 30-December 02 From: Boston Member No.: 3,802	I see that you consider them important, but I'm not seeing what about them we're discussing. They've got an expected number of successes. Are we debating whether they should have more? Less? ~J

GunnerJ View Member Profile	Apr 7 2005, 11:43 PM Post #22
Moving Target Group: Members Posts: 669 Joined: 25-May 03 Member No.: 4,634	Er, see my latest edit. I was deciding how I wanted to phrase what I said.

Kagetenshi View Member Profile	Apr 7 2005, 11:45 PM Post #23
Manus Celer Dei Group: Dumpshocked Posts: 17,006 Joined: 30-December 02 From: Boston Member No.: 3,802	I pity anyone who has to read this after all of the edits :) I think it's a better solution than doing nothing, certainly, but I'm up in the air as to whether or not I like it as a solution in general. I'll have to run some numbers. ~J

GunnerJ

Apr 7 2005, 11:49 PM

Post #24

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QUOTE

I pity anyone who has to read this after all of the edits smile.gif

Seriously, what a mess...

QUOTE

I think it's a better solution than doing nothing, certainly, but I'm up in the air as to whether or not I like it as a solution in general. I'll have to run some numbers.

Well, the idea behind the nRo6 is to allow a neat thing for "mechanical flavor" without breaking the game. An average 0.06666... extra successes doesn't seem like that big a deal; it more or less preserves the statistics that we'd expect for the most common rolls and preserves that "lucky shot" feeling everyone loves from the Rule of Six in SR3.

Kagetenshi View Member Profile	Apr 7 2005, 11:53 PM Post #25
Manus Celer Dei Group: Dumpshocked Posts: 17,006 Joined: 30-December 02 From: Boston Member No.: 3,802	It's true, but providing that extra successes still mean a better overall quality of success, it provides a chance for a better lucky shot, as far as I can tell. I'm going to have to spend some time with Mathematica before I rightly figure out whether or not I care, though. ~J

GunnerJ

Apr 8 2005, 12:10 AM

Post #26

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QUOTE

It's true, but providing that extra successes still mean a better overall quality of success, it provides a chance for a better lucky shot, as far as I can tell.

True. True. I just think that we can fairly modify the word "better" here with either "negligably" or "insignifigantly in most cases" when it comes to the nRo6.

Ellery View Member Profile	Apr 8 2005, 01:18 AM Post #27
Moving Target Group: Members Posts: 778 Joined: 6-April 05 Member No.: 7,298	It is better to do this analytically, but it's not quite trivial. The probability of getting at least N successes on one roll is > q(N) (1/3)(1/6)^(N-1) because you have to roll a 6 (N-1) times and then you can roll a 5 or 6 to finish off. Therefore, the chance of getting exactly N successes is > p(N) = q(N)-q(N+1) = (1/3)(1/6)^(N-1) - (1/3)(1/6)^N = (1/3)(5/6)(1/6)^(N-1). Therefore the expected number of successes E you will get is sum( Np(N) , N=1..infinity), which we can rewrite as > E = (1/3)(5/6)( 1 + 2(1/6) + 3(1/6)^2 + ... + N(1/6)^(N-1) + ... ). This is a standard infinite series of the for nx^(n-1). If we note that dx^n/dx = nx^(n-1), we can rewrite as (1/3)(5/6)( sum( d(x^n)/dx ) )\|x=(1/6). Pull out the derivative to get > E = (1/3)(5/6)* (d/dx)[ sum(x^n,n=1..inf) ] \| x=(1/6) we then note that sum(x^n,n=1..inf) = 1/(1-x) - 1, and that (d/dx)[ 1/(1-x) ] is 1/(1-x)^2, and we have > E = (1/3)(5/6)/(5/6)^2 = (1/3)/(5/6) = 2/5 = 40%. Since each die is independent, the expected number of successes for n dice is > E(n) = 2n/5 In contrast, for the non-exploding case, the expected number F(n) is > F(n) = n / 3 You can also compute the probability distributions of each method to compute the probability of getting a certain number of successes with or without exploding dice. The scope of that computation is beyond what I want to explain here right now, but I'm happy to produce tables upon request. Just as an example, if you're rolling four dice, the probabilities are (without and with exploding dice) > 0 successes: 19.75% 19.75% > 1 success: 39.51% 32.92% > 2 successes: 29.63% 26.06% > 3 successes: 9.86% 13.49% > 4 successes: 1.23% 5.32% > 5 successes: 0% 1.76% > 6 successes: 0% 0.51% > 7 successes: 0% 0.14% > 8+ successes: 0% 0.05%

GunnerJ

Apr 8 2005, 01:32 AM

Post #28

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QUOTE

The scope of that computation is beyond what I want to explain here right now, but I'm happy to produce tables upon request.

That would be pretty rocking, actually.

Ellery

Apr 8 2005, 02:32 AM

Post #29

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Aha, there are code tags. That will make this much prettier. How big do you want the tables? I'm doing 12x8 for now, just to keep things in a reasonable amount of space, and truncating to three decimal places. All values are in percentages, of course.

Chances of getting at least N successes, no exploding dice:

CODE

#0# #1# #2# #3# #4# #5# #6# #7# #8# (succ)
+1+ 100 33.333 0 0 0 0 0 0 0
+2+ 100 55.556 11.111 0 0 0 0 0 0
+3+ 100 70.37 25.926 3.704 0 0 0 0 0
+4+ 100 80.247 40.741 11.111 1.235 0 0 0 0
+5+ 100 86.831 53.909 20.988 4.527 0.412 0 0 0
+6+ 100 91.221 64.883 31.962 10.014 1.783 0.137 0 0
+7+ 100 94.147 73.663 42.936 17.33 4.527 0.686 0.046 0
+8+ 100 96.098 80.491 53.178 25.865 8.794 1.966 0.259 0.015
+9+ 100 97.399 85.693 62.282 34.969 14.485 4.242 0.828 0.097
+10+ 100 98.266 89.595 70.086 44.074 21.313 7.656 1.966 0.34
+11+ 100 98.844 92.485 76.589 52.744 28.9 12.209 3.863 0.882
+12+ 100 99.229 94.605 81.888 60.693 36.848 17.772 6.645 1.876
(dice)

Chance of getting at least N successes, exploding dice:

CODE

#0# #1# #2# #3# #4# #5# #6# #7# #8# (succ)
+1+ 100 33.333 5.556 0.926 0.154 0.026 0.004 0.001 0
+2+ 100 55.556 18.519 4.63 1.029 0.214 0.043 0.008 0.002
+3+ 100 70.37 33.333 11.728 3.412 0.883 0.211 0.048 0.01
+4+ 100 80.247 47.325 21.262 7.773 2.453 0.697 0.183 0.046
+5+ 100 86.831 59.396 31.962 14.053 5.273 1.752 0.53 0.149
+6+ 100 91.221 69.273 42.753 21.821 9.494 3.629 1.251 0.396
+7+ 100 94.147 77.077 52.893 30.488 15.048 6.51 2.524 0.893
+8+ 100 96.098 83.092 61.957 39.467 21.691 10.463 4.511 1.768
+9+ 100 97.399 87.644 69.761 48.264 29.083 15.438 7.324 3.151
+10+ 100 98.266 91.04 76.288 56.518 36.852 21.281 11.003 5.155
+11+ 100 98.844 93.545 81.623 63.997 44.654 27.772 15.51 7.856
+12+ 100 99.229 95.376 85.902 70.585 52.2 34.657 20.745 11.28
(dice)

Chance of getting exactly N successes, no exploding dice:

CODE

#0# #1# #2# #3# #4# #5# #6# #7# #8+# (succ)
+1+ 66.667 33.333 0 0 0 0 0 0 0
+2+ 44.444 44.444 11.111 0 0 0 0 0 0
+3+ 29.63 44.444 22.222 3.704 0 0 0 0 0
+4+ 19.753 39.506 29.63 9.877 1.235 0 0 0 0
+5+ 13.169 32.922 32.922 16.461 4.115 0.412 0 0 0
+6+ 8.779 26.337 32.922 21.948 8.23 1.646 0.137 0 0
+7+ 5.853 20.485 30.727 25.606 12.803 3.841 0.64 0.046 0
+8+ 3.902 15.607 27.313 27.313 17.071 6.828 1.707 0.244 0.015
+9+ 2.601 11.706 23.411 27.313 20.485 10.242 3.414 0.732 0.097
+10+ 1.734 8.671 19.509 26.012 22.761 13.656 5.69 1.626 0.34
+11+ 1.156 6.359 15.896 23.845 23.845 16.691 8.346 2.981 0.882
+12+ 0.771 4.624 12.717 21.195 23.845 19.076 11.127 4.769 1.876
(dice)

Chance of getting exactly N successes, exploding dice:

CODE

#0# #1# #2# #3# #4# #5# #6# #7# #8+# (succ)
+1+ 66.667 27.778 4.63 0.772 0.129 0.021 0.004 0.001 0
+2+ 44.444 37.037 13.889 3.601 0.814 0.171 0.035 0.007 0.002
+3+ 29.63 37.037 21.605 8.316 2.529 0.672 0.164 0.038 0.01
+4+ 19.753 32.922 26.063 13.489 5.32 1.756 0.514 0.138 0.046
+5+ 13.169 27.435 27.435 17.909 8.78 3.521 1.222 0.381 0.149
+6+ 8.779 21.948 26.52 20.932 12.327 5.865 2.379 0.854 0.396
+7+ 5.853 17.071 24.183 22.405 15.441 8.538 3.986 1.63 0.893
+8+ 3.902 13.006 21.135 22.49 17.776 11.228 5.952 2.743 1.768
+9+ 2.601 9.755 17.883 21.496 19.182 13.645 8.113 4.173 3.151
+10+ 1.734 7.226 14.752 19.77 19.666 15.571 10.278 5.847 5.155
+11+ 1.156 5.299 11.922 17.626 19.343 16.882 12.261 7.655 7.856
+12+ 0.771 3.854 9.474 15.317 18.386 17.543 13.912 9.465 11.28
(dice)

From the previous analysis, we know that with exploding dice we expect 0.4 successes per die, and without we expect 0.3333.... This means that the "12" line for non-exploding dice should average 4 successes, and the "10" line for exploding dice should also average 4 successes. Keep this in mind when comparing rolls--it's fairer to compare sets of dice with the same average.

Phantom Runner

Apr 8 2005, 04:14 AM

Post #30

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AH!! For the love of Nuyen!!
Let's get back to the topic of the post....

QUOTE

And yeah, as more and more people are coming up with this themselves, I'm starting to think this would be a very good idea --

Actually this idea is not new at all. Remember it being stated that the new SR mechanic is based on the nWoD dice rolling mechanic except that SR will use d6s?
Well in nWoD they have the 10 again rule, which states 10s count for their success and then the player gets to roll them again. Each 10 that comes up counts again and gets a reroll again....

For SR, "6 again" might not be a bad idea. But I don't know (as few do) if the devs will want to make the SR mechanic so similar to the nWoD mechanic (not that that's a bad thing). Rather I see a player being able to get dice from a convention SR3 already has called a "Karma Pool". Often times a player only had 1 or 2 dice and really needed to get more than 1 or 2 successes. By spending Karma they could increase their dice pool....

...At least I hope Karma Pool is still in the game and works the same way...

Ellery View Member Profile	Apr 8 2005, 04:42 AM Post #31
Moving Target Group: Members Posts: 778 Joined: 6-April 05 Member No.: 7,298	The above stuff shows how much of a shot someone has under different conditions. How is this not exactly the topic of the post? Anyway, no, it's not really new, but then again, it's not new in nWoD because SR was doing it before (albeit structured slightly differently). I'd bet there was some game doing it before SR was doing it. Heck, there are even entries like "20 - roll again twice" in 1st ed. AD&D treasure tables. Maybe we can find some classic dice games where you get rerolls to rack up success.... Regardless of how old the idea is and who thought of it first, it still makes sense here, I think.

Critias View Member Profile	Apr 8 2005, 05:54 AM Post #32
Freelance Elf Group: Dumpshocked Posts: 7,324 Joined: 30-September 04 From: Texas Member No.: 6,714	Dammit. Who got her started? Who was it?! You've doomed us all! The math-speak, it consumes my soul !!

mfb View Member Profile	Apr 8 2005, 06:03 AM Post #33
Immortal Elf Group: Members Posts: 11,410 Joined: 1-October 03 From: Pittsburgh Member No.: 5,670	she blinded me with advance probability maths.

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