I've tried, since SR4 was announced, to stay current on all the important discussions, but alas, I have recently fallen behind. So, if an idea like what I am about to suggest has already been brought up, my apologies.
Some people seem worried that the new system will remove the possibility that existed in SR3 to always be theoretically able to succeed, no matter how hard the task. This is because the Rule of Six made it so that no matter how high the TN was, there was a theoretical chance that at least a single die could hit it. With the new system, it is feared, this will go away. If you need at least X successes against a TN of 5 and you have less than X dice to roll, tough luck.
I don't think it necessarily has to be this way. The "always have a shot" feel of SR3 can be maintained in SR4 with a simple reworking of the Rule of Six. Basically, if you get a 6 on a die roll, not only do you have a success, you can roll that die again. This process does not terminate, so if you roll a 6 again, you get another success and another re-roll. If, on a re-roll, you fail, then you stop and have as many successes as the roll generated so far. Same with a 5 on a re-roll, except you get another success before stopping.
In this way, even a single D6 has a theoretical chance of generating any number of successes. It seems a very basic idea, so I'd bet someone else has thought of it; so again, sorry if this is redundant. But I figure this idea is worth a(nother) mention, at least for a potential house rule.
Full Version: Modified Rule of Six
It has been mentioned in several places, and I think it's a very good idea. It's especially nice that it's occurring to a number of different people. It has nice statistical properties, too--smoothing out probabilities so you don't jump from an action being only "kind of hard" to "utterly impossible" when the number of successes needed goes up by one or a one-die penalty is applied or someone with one lower skill tries the same task.
Yep, at least by mfb, Eyeless Blond and me, and probably quite a few others as well, in several threads. It's easy to miss these things, though, because there's dozens of rather long threads in which everything about SR4 is discussed. 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 -- so much so that I really, really hope the developers haven't decided on a different method of getting rid of impossible tests yet (unless this different method is absolutely brilliant and nobody here has yet come up with it).
Every method is fucking brilliant until it meets reality.
Yes, I'm bitter, cynical, and dissatisfied with life.
Ergo, I play RPGs.
Yes, I'm bitter, cynical, and dissatisfied with life.
Ergo, I play RPGs.
Honestly I don't know how it could work any differently. If the vast majority of tests have a Threshold, then rolling 1 or 2 dice would mean you have no chance of actually pulling off most tasks under such a mechanic without such a method of (possibly) netting more than one success per die.
I really think it'd be a great mechanic to add to such a system; in fact it's one I'd like to add to the current SR system for Success and Opposed Tests. Each die roll nets an extra success each time you beat the TN by 6. One of the things I think is odd about the current system is how you're limited in successes by the number of dice you roll; this would go toward changing that.
I really think it'd be a great mechanic to add to such a system; in fact it's one I'd like to add to the current SR system for Success and Opposed Tests. Each die roll nets an extra success each time you beat the TN by 6. One of the things I think is odd about the current system is how you're limited in successes by the number of dice you roll; this would go toward changing that.
| QUOTE (Eyeless Blond) |
| Honestly I don't know how it could work any differently. |
Neither do I, but then I'm not an RPG designer.
| QUOTE (Eyeless Blond) |
| [...] in fact [the multi-success /w 1 die mechanic is] one I'd like to add to the current SR system for Success and Opposed Tests. |
Guess who already has.
Eyeless - The Aeonverse version of the nWoD rules actually had a variant on that which restricted it further but I can't for the life of me remember how it worked (have to dig out the books) - so my point? There are other ways of handling it.
I personally think it's a very bad idea, as it seems to open up a magnitude of success that isn't reasonable (as opposed to the current exploding-dice method, in which a success is still a single success even if the TN is 24 and the roll was a 61).
~J
~J
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:
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.
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
~J
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.
| 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.
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
~J
Um... so?
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
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
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?
As my ninja-edit states, the part that gets to me is that it was extra work for reduced accuracy.
~J
~J
| 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?
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
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
| 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.
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
~J
Er, see my latest edit. I was deciding how I wanted to phrase what I said.
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
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
| 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.
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
~J
| 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.
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( N*p(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 n*x^(n-1). If we note that dx^n/dx = n*x^(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) = 2*n/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%
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( N*p(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 n*x^(n-1). If we note that dx^n/dx = n*x^(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) = 2*n/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%
| 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.
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:
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.
AH!! For the love of Nuyen!!
Let's get back to the topic of the post....
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...
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.
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.
Dammit. Who got her started? Who was it?! You've doomed us all! The math-speak, it consumes my soul !!
she blinded me with advance probability maths.
Having a "10 again" rule in nWoD is very different than a "6 again" rule in SR, for a reason none of you have mentioned yet: time.
You will reroll a D10 less often than a D6. The developers of SR4 are trying to streamline the game and make it faster/more fluid. Rerolling dice takes time. You roll, then figure out which ones are the 6, then roll again, and probably roll again once. That considerably add time to the whole process. For that reason only, I think they might avoid using that rule. I could be wrong though.
You will reroll a D10 less often than a D6. The developers of SR4 are trying to streamline the game and make it faster/more fluid. Rerolling dice takes time. You roll, then figure out which ones are the 6, then roll again, and probably roll again once. That considerably add time to the whole process. For that reason only, I think they might avoid using that rule. I could be wrong though.
It's hard to make sense of the statistics without knowing how the successes will be used. However, given what I know of both Shadowrun and Exalted, I will make the guess that there are (at least) two possible uses for successes: a threshold system in which you must get X or more successes to pass a test, or a more open ended system where the number of successes you get determines how well you do.
Given these assumptions, let's look at three results: One success, which is either an easy task in a threshold test or just scraping by in the open-ended system; three successes, which is either a moderately difficult task as a threshold or a pretty good result on an open-ended success test; and five successes, which might be a very hard or near-impossible threshold or a superlative performance on an open-ended test.
The goal of the nRo6 is to make it theoretically possible to succeed at any threshold/level with any number of dice while still not breaking the system of difficulty by granting a glut of extra successes. I think it's safe to say, looking at Ellery's charts, that nRo6 works as it should for the three conditions I outlined. Let's look at the average roll (6D6) and the elite roll (12D6).
Without exploding dice, an average roller can get an easy task/just passing result (one or more successes) in 9/10 rolls, and an elite roller only fails once out of hundred rolls. This is reasonable. For a moderately challenging task/good performance (three or more successes), an average roller can get it a little under a third of the time, while a elite roller gets it 4/5 times. This is a little steep for the average roller, but we are talking about good results or a fairly high bar, which one would expect only someone very well trained and/or talented to get with any regularity. For a near impossible task/outstanding performance (five or more successes), the average roller can get it maybe once every fifty rolls, while the elite roller gets it a third of the time. Five is supposed to be very hard to achive, so someone who's just average should only get that many successes rarely, while for someone of extraordinary ability, getting it should be possible but not routinely expected, and this is what we see.
Now, let's look at rolls with the nRo6. An easy/basic result is achived with the same regularity by both average and elite rollers as without exploding dice. For a challenging/good result, average rollers succeed 2/5ths of the time, while elite rollers fail only three times out of every twenty rolls (i.e., 17/20 times they succeed). This is only modestly more than without rerolls, and presrves the "infrequent for grunts, common for experts" balance. For a near impossible/extraordinary result, average rollers get it 1/10th of the time and elite rollers get it half the time.
That's where we see some difficulty. The chance of an average roller getting an outlandishly hard task has been quintupled, and elite rollers see a 20% increase. This follows the theme from my results on the first page that show that the average successes increases noticably with very high dice. This may or may not be a bad thing, however: 10% is still fairly rare, and when the situation is critical, elites would probably rather not be flipping a coin to see if they succeed. Still, if it is a problem, the very hard/excellent performance bar can be moved to six or more successes, which has results very close to those for five or more successes without exploding dice.
Given these assumptions, let's look at three results: One success, which is either an easy task in a threshold test or just scraping by in the open-ended system; three successes, which is either a moderately difficult task as a threshold or a pretty good result on an open-ended success test; and five successes, which might be a very hard or near-impossible threshold or a superlative performance on an open-ended test.
The goal of the nRo6 is to make it theoretically possible to succeed at any threshold/level with any number of dice while still not breaking the system of difficulty by granting a glut of extra successes. I think it's safe to say, looking at Ellery's charts, that nRo6 works as it should for the three conditions I outlined. Let's look at the average roll (6D6) and the elite roll (12D6).
Without exploding dice, an average roller can get an easy task/just passing result (one or more successes) in 9/10 rolls, and an elite roller only fails once out of hundred rolls. This is reasonable. For a moderately challenging task/good performance (three or more successes), an average roller can get it a little under a third of the time, while a elite roller gets it 4/5 times. This is a little steep for the average roller, but we are talking about good results or a fairly high bar, which one would expect only someone very well trained and/or talented to get with any regularity. For a near impossible task/outstanding performance (five or more successes), the average roller can get it maybe once every fifty rolls, while the elite roller gets it a third of the time. Five is supposed to be very hard to achive, so someone who's just average should only get that many successes rarely, while for someone of extraordinary ability, getting it should be possible but not routinely expected, and this is what we see.
Now, let's look at rolls with the nRo6. An easy/basic result is achived with the same regularity by both average and elite rollers as without exploding dice. For a challenging/good result, average rollers succeed 2/5ths of the time, while elite rollers fail only three times out of every twenty rolls (i.e., 17/20 times they succeed). This is only modestly more than without rerolls, and presrves the "infrequent for grunts, common for experts" balance. For a near impossible/extraordinary result, average rollers get it 1/10th of the time and elite rollers get it half the time.
That's where we see some difficulty. The chance of an average roller getting an outlandishly hard task has been quintupled, and elite rollers see a 20% increase. This follows the theme from my results on the first page that show that the average successes increases noticably with very high dice. This may or may not be a bad thing, however: 10% is still fairly rare, and when the situation is critical, elites would probably rather not be flipping a coin to see if they succeed. Still, if it is a problem, the very hard/excellent performance bar can be moved to six or more successes, which has results very close to those for five or more successes without exploding dice.
| QUOTE (otaku mike @ Apr 8 2005, 02:03 PM) |
| Having a "10 again" rule in nWoD is very different than a "6 again" rule in SR, for a reason none of you have mentioned yet: time. You will reroll a D10 less often than a D6. The developers of SR4 are trying to streamline the game and make it faster/more fluid. Rerolling dice takes time. You roll, then figure out which ones are the 6, then roll again, and probably roll again once. That considerably add time to the whole process. For that reason only, I think they might avoid using that rule. I could be wrong though. |
Getting six on a D6 is more common than getting ten on a D10, but I've thrown around 12D6 at a TN of 6 and gotten nothing more times than would seem possible in a just universe. So I can tell you, it's not at all common in any objective sense.
It happens one out of every six times on average on an unweighted die, so it's pretty darn common.
~J
~J
| QUOTE (Kagetenshi) |
| It happens one out of every six times on average on an unweighted die, so it's pretty darn common. ~J |
One out of six is signifigantly more common than one out of ten? I don't know about that. I'm also pretty sure that in practice, the frequency of getting sixes is inversely proportional to how important it is that you get sixes; at least, that's been my experience. I understand that this is an appeal to chance and anecdote and not a mathematical truth, but I also saw a lot of people refering to their poor "dice luck" on one thread, and they didn't seem to be kidding. I knew their pain.
In any case, it's never slowed down any of my games to look for sixes and reroll them. The developers have already streamlined the game to an incredible degree by having a fixed TN. They have a good amount wiggle room for potentially slowing it a tiny bit more.
Er, I think the 6-dice person having as much as a 10% chance to get 5 successes is a feature of the exploding dice system. Also, let's compare apples and apples--a 6d6 normal roller will have as many successes on average as a 5d6 exploding roller. So there it's a little under 2% vs. a little over 5%--a little kinder for achieving that extraordinary result.
| QUOTE (GunnerJ) |
| One out of six is signifigantly more common than one out of ten? |
It is. Over half again as common, in fact.
~J
1/6 = 16.666...%
1/10 = 10%
Again, I'm not so sure.
Look, all I'm saying is, if otaku mike is concerned that having to reroll those incredibly common sixes again, and then again, and then quite possibly again is going to make SR4 a slovenly, cartankerous junker of a game crawling at a molasses pace compared to the sleak, sexy, sonic assassin that is the D10 nWoD system, then I can safely say that he has no reason to fear. In my three years playing Shadowrun, rerolling twice happens every so often in a game, it's a rare game where a die is rerolled three times, and I can count on one hand how many times I've seen a die rerolled more than four times. It was usually an event of some awe and celebration, and I feel that if nothing else, if SR4 doesn't have rerolls, it will be a poorer gaming experience for not having those events.
1/10 = 10%
Again, I'm not so sure.
Look, all I'm saying is, if otaku mike is concerned that having to reroll those incredibly common sixes again, and then again, and then quite possibly again is going to make SR4 a slovenly, cartankerous junker of a game crawling at a molasses pace compared to the sleak, sexy, sonic assassin that is the D10 nWoD system, then I can safely say that he has no reason to fear. In my three years playing Shadowrun, rerolling twice happens every so often in a game, it's a rare game where a die is rerolled three times, and I can count on one hand how many times I've seen a die rerolled more than four times. It was usually an event of some awe and celebration, and I feel that if nothing else, if SR4 doesn't have rerolls, it will be a poorer gaming experience for not having those events.
| QUOTE (Ellery) |
| Er, I think the 6-dice person having as much as a 10% chance to get 5 successes is a feature of the exploding dice system. Also, let's compare apples and apples--a 6d6 normal roller will have as many successes on average as a 5d6 exploding roller. So there it's a little under 2% vs. a little over 5%--a little kinder for achieving that extraordinary result. |
I agree, actually. I merely put forward the option of increasing the threshold to six to show how easy it would be to correct if it is indeed necessary to correct it.
| QUOTE (GunnerJ) |
| In my three years playing Shadowrun, rerolling twice happens every so often in a game, it's a rare game where a die is rerolled three times, and I can count on one hand how many times I've seen a die rerolled more than four times. It was usually an event of some awe and celebration, and I feel that if nothing else, if SR4 doesn't have rerolls, it will be a poorer gaming experience for not having those events. |
Indeed, this is one thing I love about exploding dice; on occasion it makes the die rolls themselves memorable, which is something you rarely see in non-exploding systems. It's just fun seeing everyone's eyes pop out (the roller's included!) when you hit something like a TN 25, and talk about it even days afterward ("Dude, remember when you rolled that 34? That was so weird!") Maybe it *is* a little slower, but IMO it's time well spent, and it's rare enough to not bog down the game.
Oh, I agree that it's not going to slow the game down. I really don't know what people are talking about when they complain about rerolling sixes or rolling a lot of dice, that always seemed to be the best part to me.
~J
~J
Wasn't the exploding 6 system was first implemented in SR1?
Yes, why?
~J
~J
| QUOTE (Kagetenshi) |
| Yes, why? |
Just commenting that it would be more retro Shadowrun than pirated WOD as some people semi-aluded to above.
No one's complaining about the exploding dice, look at SR3. It's the fixed-TN, exploding-dice-for-extra-successes that start looking like WoD.
~J
~J
| QUOTE (Eyeless Blond) |
| Indeed, this is one thing I love about exploding dice; on occasion it makes the die rolls themselves memorable, which is something you rarely see in non-exploding systems. It's just fun seeing everyone's eyes pop out (the roller's included!) when you hit something like a TN 25, and talk about it even days afterward ("Dude, remember when you rolled that 34? That was so weird!") Maybe it *is* a little slower, but IMO it's time well spent, and it's rare enough to not bog down the game. |
Maybe I'm a bit masochistic when it comes to rolling dice - at the moment I like high target numbers and the "rule of 6" - especially in combat!
(from one of the last gaming sessions..)
GM: "Cougi, you're next, but from where you're standing you can't get a safe shot at an enemy."
Me: "So I'll make an unsafe shot. Gimme my TN."
GM: "But two of your chummers are in the way."
Me: "Yepp, double figure TN - gimme!"
GM: "If you miss you'll probably hit your chummers."
Me: "I know.. so what's my TN?"
GM: (sighs) "It's 22."
Me: "Ah.. now that's what I call a target!" (starts rolling dice)
Chums (nerveously stare at dice)
GM (stares at dice)
Me: "Bang.. a success!"
(IP the already injured opponent goes down to my shot)
Chums (start breathing again)
GM: "Okay, you got lucky that time.. who's next?"
Me: "Waitaminute.. it's still my turn! Who else can I shoot?" (grins like a maniac)
Chums (stare at me then at the GM)
GM: "You can't be serious!"
Me: "I sure am.. so next opponent and TN?"
...
I don't try to pull stunts like that all the time.. but now and again, I like the idea of beating seemingly unbeatable odds. It's what makes us PCs special - not that we can do it, but that we try.
Here's to hoping there is something similar in SR4!
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