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In Search of a Fortune Mechanic

Started by JMendes, February 17, 2003, 10:24:58 AM

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Hey, guys, :)

Yes! Good stuff. Hmm... I'm gonna be taking all of these in order.

Quote from: M. J. YoungI take #1 to mean this. If I have a chance of success of X, the value of {(X+1)-X} should be identical in probability to the value of {X-(X-1)}. That is, you're saying that for any X, +1 and -1 should be the same offset in probability.

I take #2 to mean this. If I have a chance of success of X, the value of {(X+1)-X} should be lower in probability than the value of {X-(X-1)}. That is, you're saying that for any X, -1 should be a larger step than +1 in terms of offset in probability.
Your interpretation of #1 is almost correct and your interpretation of #2 is also approximate. The problem only became clear to me upon reading Walt's followup. The corrected statements should be:
Quote#1. If I have a chance of success of X=50%, the value of {(X+1)-X} should be identical in probability to the value of {X-(X-1)}. But, for any X other than 50%, +1 and -1 would not be the same offset in probability.

#2. If I have a chance of success of X>50%, the value of {(X+1)-X} should be lower in probability than the value of {X-(X-1)}. That is, for any X>50%, -1 should be a larger step than +1 in terms of offset in probability. Conversely, for any X<50%, -1 would be a smaller step than +1.

Or at any rate, this is what I originally meant, though I readily admit I stated it badly. See my response to Walt below for possible further insight.

Quote from: ThreeGeeThe bonuses/penalties are straight percentage. They do not modify the skill itself, but rather the converted skill level. In other words, a +5 bonus represents a +5% better chance to succeed, regardless of skill.
Actually, with a skill of 10, the +5 bonus is utterly meaningless, whereas with a skill of 1, that same bonus is going to multiply your chance of success by 6. Under a good decay behaviour, the same +5 bonus should have smallest effect at the extremes (i.e. skill=1 or skill=10 in your example) and greatest effect near the mid-range (i.e skill=5 or skill=6 in your example).

I might also point out that skill (and attribute and whatnot) is really just another type of bonus and does not necessarily have to be treated differently. This in itself is not a requirement, and if a solution presents itself that treats the skill value and the pluses/minuses in wildly different fashions, great. However, given the original specs, intuitively, I find that unlikely.

Quote from: wfreitagI think your definition of symmetrical is being misinterpreted. As Jack pointed out, adding a modifier and taking the same modifier away again will always have equal and opposite effects. I think what you really mean is that the system has a center point and relative to that center point, the probability of success at a given positive modifier is the complement of the probability of success at negative modifier of equal magnitude (that is to say, they add up to 1.0) or equivalently, that the probability of success at a given relative modifier is equal to the probability of failure at the opposite relative modifier.
You are correct. My original statement of spec #1 was ambiguous at best and it took several people pointing out various inconsistencies for me to realize it. Your restatement accurately depicts my original intent.
Quote from: wfreitag alsoWhat makes the symmetry in Symmetry actually useful is that the effect of a bonus is exactly the same as the effect of a penalty of equal magnitude applied to the opposition. [...] But without the ability to say "my penalty is exactly equivalent to the other guy's bonus," [...] this property seems to lose a lot of its singificance.
This, however, I disagree with. I'll restate it as 'this property would have a very different significance'. :) The main point behind this requirement is precisely that I can cancel out bonuses and penalties as they apply to me at will, and (as combined with spec #2) stack them at will, without regard to which point in the distribution I start with. I'll also note that, if the mechanic works as specified, then, in the particular case of a two-way contest, addind a penalty to me will be the same as adding a bonus to my opponent.
Quote from: Lastly, wfreitagSince that limitation exists anyway, you might drop requirement #1 and instead use a well-behaved dice pool mechanism [...] with the constraint that all modifiers add dice to pools. A penalty is applied by adding extra dice to all opponents, which will have equivalent effects as long as all rolls are opposed, and you won't have the granularity effects of removing dice from small pools.
This one, however, is not consistent to your reiteration of Jack's point that at its simplest, a penalty should simply cancel out a bonus. I think that the basis of the Simmetry mechanic is that everything is a bonus/penalty. Thus, the beauty in its simplicity. If I take this route, this particular aspect is lost. I mean, you have to start the minimum pool at some value. :)

I hope I have shed further light into my devious inner mind. :) I know that I am gaining understanding of what my intent is. I don't think I'll be remiss in thanking you guys yet again for the continuing debate. :)


João Mendes
Lisbon, Portugal
Lisbon Gamer


Hey J,

Oh! That's completely different. You want a bell curve, not a straight line. No wonder you threw out the d20 mechanic. I had started along those lines, ala GURPS, but realized it violated my interpretation of #1 rather badly.

Let's see, skills between 3 and 30, centered around 16-17, roll 3d10, roll equal or under skill, apply bonuses and penalties directly, divide the successful roll by some arbitrary constant (2 or 3, probably), compare successes all around. There you go. Raise the number of dice to lower the standard-deviation of the curve (increase the funkiness of the decay), and raise the die-sides to increase the width of the curve (increase the fineness of the granularity).


Mike Holmes

Quote from: JMendesThis one, however, is not consistent to your reiteration of Jack's point that at its simplest, a penalty should simply cancel out a bonus. I think that the basis of the Simmetry mechanic is that everything is a bonus/penalty. Thus, the beauty in its simplicity. If I take this route, this particular aspect is lost. I mean, you have to start the minimum pool at some value. :)
But this method is much simpler, actually. Instead of subtracting penalties from bonuses, you just add dice to the appropriate pool. This serves to perform the subtraction for you in rolling. Let's look at Synthesis, with the adding, and a simple combat example.

I could take a 6 die pool, and subtract 2 dice for slippery and roll against my two opponents each with a 6 die pool. What's the net effect? An average drop from zero successes against each to them each scoring one success. What happens if I add 2 dice to each of their pools? The exact same effect. The only difference is that they now have a larger range of possible successes. Which, given that they're fighting against a disadvantaged opponent makes sense. In fact, this is how the system allows players to exceed their normal limits when attempting something easy. Either it's a bonus for the PC, or it's a penalty for the target. In either case, the game effect is the same, a higher average roll, and a higher max roll.

Or, IOW, what Walt said.

If you use Synthesis with only adding bonuses, there may be no problems at all. Looking back at #2 I forgot that the odds on a result do shift on a curve (expected value is not odds). I think that adding a bonus to yourself versus adding a bonus to your opponent might have not quite equivalent effects, and therefore fail #1 technically, OTOH. But the effects are, in fact, naturalistic which is what the goal is about, I'm betting. That is, it seems that the only reason that you wanted to have them be symetrical is that you could then have them cancel. Since this is simpler, that's no longer neccessary.

I think I have a winner (or very close). No surprise since I was using pretty similar criteria when I adapted the Story Engine mechanic.

The only downside is potentially large dice pools to cover the fine granularity, and all bonuses thrown in. But again, I'm all for that, and you did say that low handling time wasn't important. ;-)

Also, keep in mind that there are options to the multi-player problem. It's not very realistic to have a character roll his entire pool separately against all comers simultaneously (as in my little example). In TROS, the player has to split his pool between opponents. This is visciously real. A more "heroic" option (but less so than rolling evenly against all comers) is to have characters only be fighting one opponent aided by his allies. In this case, there is only one addition to the opposing pool for penalties (or, possibly, with aiding rolls required, the bonus might be added to the aiding roll). Yet another functional model is to have multiple attackers recieve a bonus for superior numbers. The last is my least favorite as it's pretty arbitrary. Anyway, which of these models are you using, as it will affect which system is superior?

BTW, your definition of open-ended needs to be stated as clearly as your clarification above. For many people saying open-ended means that any number (from negative infinity to positive infinity) of successes is possible with any roll. Most exploding die pools are an example of this.

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Hey, :)

Quote from: Mike HolmesI could take a 6 die pool, and subtract 2 dice for slippery and roll against my two opponents each with a 6 die pool.

Hmm... What if I am subtracting 2 dice for slippery but adding 1 for supperior equipment? Am I gonna roll 6 dice against 7 or 7 against 8? If the latter, the penalties and bonuses aren't cancelling. Assuming the former, where does the 6 come from?

About the multiple opponent thing. Things keep getting boggled down, here. I don't mean A and B against C, I mean A against B against C. There is no bonus for multiple attackers, there's a three-way there-can-be-only-one all-out shootout. Hmm.. Ok, I'm gonna come out and say it. :) Spec #7 also means that the mechanic should also work in the limiting case of a one-sided conflict. (This is the part that I said earlier would go against your opposed/unopposed roll...)

Ok, now, to clarify open-ended: open-ended pluses/minuses means a task can get arbitrarily easier without ever actually getting to 100% probability and can get arbitrarily harder without ever getting to 0%. Open-ended effect numbers means a character can generate an arbitrarily large positive result as well as an arbitrarily large negative result. Keep in mind that these are not absolute requisites and spec #6 will be the first to be tossed out if the whole spce set becomes unwieldy.

In light of these clarifications, your solution fails to meet spec #6, but if nothing better comes along, I might just have to take it, or some combination of it and other possibilities pointed out in the thread, as a final solution.

The only viable alternative I've come up with on my own so far revolves around players using some sort of log table (necessarily not open-ended) or log calculator (ugly) to figure out the bonuses, then rolling one die at a time in a slow decay fashion to figure out the effect number.


João Mendes
Lisbon, Portugal
Lisbon Gamer


Hey J,

I am fairly certain your definition of 'open-ended' is impossible. In order to allow for an infinite number of modifiers, an infinitely finely-grained system is required. Otherwise, at some point, the output falls below the granularity threshold and that is the end of it.

I really wish I could draw you a picture, but try to imagine a bell curve drawn on graph paper. At some point on both tails, the curve drops below the last horizontal line above zero. Even using a logarithmic conversion on the bonuses, you run into the same problem, just at a different step.

Can anyone provide a counter-example? It has been a while since I have done any real math, so I could easily be overlooking something.


Rob Donoghue

Hrm.  It's less mathematically elegant, but since you specifically don't care about how ugly the system is, how feasible would a percentile-step (choose an arbitrary center point, and give each "step" a diminishing value, either arbitrarily or by a formula) be?

Say the steps are 10/10/5/5/5/3/3/2/2/1/1/1....... from a 50% center, we end up with a chart that looks like:

1%... (-14+)
1%   (-13)
2%   (-12)
3%   (-11)
4%   (-10)
5%   (-9)
7%   (-8)
9%   (-7)
12% (-6)
15% (-5)
20% (-4)
25% (-3)
30% (-2)
40% (-1)
50% (0)
60% (+1)
70% (+2)
75% (+3)
80% (+4)
85% (+5)
88% (+6)
91% (+7)
93% (+8)
95% (+9)
96% (+10)
97% (+11)
98% (+12)
99% (+13)
99% (+14...)

Apply bonuses and penalties as steps rather than percentile changes, and in the extreme (trans-14 cases) differences become apparent with the successive application of penalties or bonuses (thus, suppose the rank equates to skill: the best swordsman in the world (Rank 17) has almost no difference (under controlled circumstances) from the second best (rank 14), but in an extended contest where both accumulate penalties, the 3 rank difference becomes more pronounced.

Margin of Success calculations are easily derived from the chart to whatever degree the user wants, with the simplest method being the gap between the effective rank and the rank acheived by the roll (which also helps out with our sample swordsmen)

How effective it will prove in a multi-element conflict is pretty subjective - I'm comfortable with the combination of ranks to represent combined effort, but I'm not certain that would suit everyone's needs.

Anyway, It's something of a brute force attempt at a solution (when I use somethign similar, it tends to be rather simpler) but I'm curious where it's flawed. :)

-Rob D.
Rob Donoghue
<B>Fate</B> -


Hey, :)

Rob, yes, your solution falls in line with what I was envisioning, and is of course a direct illustration of Grant's point that infinite open-endedness is a problem.

Grant, I just wanted to note that ultimate open-endedness is not entirely impossible. For instance, my own klunky approach is completely open-ended at all steps:

a) T = (sum of all bonuses)-(sum of all penalties)
b) I = sign(T) * log( |T| ) (where sgn(0)=0)
c) roll d6: 4-6 means R=I and increases, 1-3 means R=I-1 and decreases
d) roll d6: 6 means you are done. stop rolling. R is your final result
e) R=R+1 if increasing, R=R-1 if decreasing
f) repeat from step d

There is no limit to the input into the log function, so there is no limit to the open-endedness of how hard/easy a task can get. There is also no theoretical limit on how many times you can roll a d6 and not come up 6, so there is no limit to how high or low an effect number can be generated.

Of course, you could use a table instead of step b. It makes it easier on the players, but now the open-endedness is limited to how big a table you want to make. You can also use a single linear roll against a table, like Rob suggested, in which case, you are correct, the open-endedness is limited by the granularity of the roll. It is, however, humongously easy on the players.

By the way, Rob, I did not find your solution to be at all inelegant, as you claim. Quite the contrary, in fact. I'd be interested to know why you said that. Oh, and I am also comfortable in combining ranks to show cooperative effort.

Yet again, cumulative thanks for all the input.


João Mendes
Lisbon, Portugal
Lisbon Gamer

Rob Donoghue

Quote from: JMendes

By the way, Rob, I did not find your solution to be at all inelegant, as you claim. Quite the contrary, in fact. I'd be interested to know why you said that. Oh, and I am also comfortable in combining ranks to show cooperative effort.

That was probably more reflective of a personal bias than anything else.  I tend to use a system like that to reflect the ease of learning the basics of a skill but the difficulty of mastering it, (so the progression tends to look like 40/60/75/85/90/93/95/96/97/98/99/99....)  so switching it over to a more balaced curve felt like I was complicating something simple. :)

Mostly though, it's because  while for the table based (rather than formula based) solution pays of in ease of use, the curve is definately mathematically inelegant, which was my main reason for choosing that particular turn of a phrase.

- Rob D.

PS - I'm actually fine with infinitely extending 99%'s.  Einstein isn't going to be able to answer a basic physics problem beter than j. Random Doctor of Physics, so having them operate ona  similar percentile level for 0 difficulty tasks is cool with me, since when a very difficulty task comes along and drops them both 7 ranks or so, Einstein is still in the high 90s, while the doctor is now somewhare around 60%.  (or at least, that's my thinking in using the model - really, it was designed to allow for lots of very good duellists whose differences in skill only really showed itself in the face of adversity)

PPS - I should also note, I got the idea originally from a game revolving around Art Auctions that I cannot remember the name of. They used Log formulas, and I remembered liking the reasoning but being paralyzed by the prose, so I simplified it to a table.

[Edited because I type with flippers ]
Rob Donoghue
<B>Fate</B> -

Walt Freitag

It looks to me like a table will suit you just fine, and it does not limit the granularity if you allow rerolls. Here's one system:

percentile   25  30  35  42  50  59  71  84  100
score       8+r   7   6   5   4   3   2   1    0

Roll percentile dice, reading 00 as 100. Find the lowest (leftmost) percentile in the table that's greater than or equal to the number rolled. The score in that column is the magnitude of the result. (So, for example, the result is 5 if you roll 36, 37, 38, 39, 40, 41, or 42 on the percentiles.) Simultaneously roll any 50-50 chance to decide the sign of the result, positive or negative.

On a throw of 25 or less, the result is "8+r" meaning reroll and add 8 to the result (but do not reroll the sign). Multiple rerolls (adding 8 each time) are possible.

The result for a character's action is the score from the die roll (positive or negative) plus and minus any modifiers for skill, conditions, etc. Degree of success (or failure, if it comes out negative) is the result minus the difficulty (if unopposed) or minus the opponent's result.

In the case of ties (degree of success 0), a positive sign roll (even if the magnitude was zero) breaks the tie and succeeds against a static difficulty score or against an opposing roll whose sign roll was negative. Similarly, in a tie a negative sign roll represents failure against a static difficulty score or against an opposing roll whose sign roll was positive. In all other cases for ties, a partial success occurs or the tie must be broken with another roll.

At the cost of greater relative rounding error in the probability curve, you can widen the table somewhat to reduce the number of rerolls needed:

percentile   12  14  17  21  25  30  35  42  50  59  71  84  100
score      12+r  11  10   9   8   7   6   5   4   3   2   1    0

I believe this meets all your specifications. One caution, though: the distribution of degrees of success/failure decays exponentially, but not always so as to make results of lower magnitude the most probable. For degree of success when the odds were against success, and for degree of failure when the odds were in favor of success, the exponential decay is perfect and the results of lower magnitude are the most probable. But for degree of failure when the odds are against success, and degree of success when the odds are in favor of success, the behavior is different. The most likely outcome is a degree of success or failure equal to the overall numerical advantage or disadvantage in modifiers the character had going in, with the probability of other degrees of success or failure decaying from that maximum in both directions.

Makes some intuitive sense. If you have a 20 point advantage in modifiers, and you get very unlucky and fail despite your advantage, you're most likely to have a degree of failure of 1. Only half the time will the degree of failure exceed -3; only one fourth of the time would it exceed -7, and so forth in the normal exponential pattern. But when you succeed, you're most likely to succeed with a degree of success of 20 and very unlikely to succeed with a degree of success of 1 or 2 (just as unlikely as to succeed with a degree of success of 39 or 38).

If you don't like that, and want the lower magnitude degrees of success and failure always to be the most probable, then make the degree of success or failure a separate roll on the same table (and it doesn't matter which side rolls).

- Walt
Wandering in the diasporosphere


Hey, :)

Walt, your table hits the spot. I am left wondering why you decided to build it high-roll-low instead of the more intuitive high-roll-high, but other than that, the concept is very nice.

Now that the roll part has a satisfactory answer, I am pondering on the fixed part. I've been musing around the following formula:F = LOG   (0.1*|T| + 1) * SIGN(T)
rounded nearest, where T is the sum total of all bonuses and penalties including skills, attributes, equipment, conditions and whatnot. This yields a rather nice curve with F=T for -4<=T<=4, then having diminishing returns to the point where if T=30 then F=15.

My issues with it at the moment are:

a) Should I add the result of the roll (say from Walt's table) to T or to F for the final result? A good argument can be made for both cases.

b) What if there are two cooperating agents? Should I add their Ts together or their Fs? How many rolls should be made? Likewise for opposing agents. [Note that an opposing agent is not the same as an opponent side. The first is, say, an NPC that doesn't want you to succeed at what you are doing. The second is, say, someone who is attempting to do the same as you, only better (like in a race), or someone who is attempting to do to you what you are attempting to do to them (like in combat).] Again, good arguments for the various possible combinations can be built.

c) Is there any alternative to extending this table other than calculating its continuation?

Anyways, I think this thread has been yielding good results. Comments on the above points are now also most welcome, as well as any further alternative mechanics anyone comes up with. :) Again, thanks for all the participation.


João Mendes
Lisbon, Portugal
Lisbon Gamer

Jason Lee

Maybe I'm mistaken, but it seems you want the curve of a 1d+/1d- system  (as per FUDGE) with the decay of a dice pool system (as per Sorcerer).

Would just slapping the two together yield the desired result?

For example (you may have a different scale in mind than the dice I picked):
Roll:  1d10-/1d10+ + pool:2d20 (take the highest)
A bonus would add an additional 1d20 to the pool.

Again, I didn't check the math, but the bonus dice should up the center point of your curve and decay while doing so.  Starting with a dice pool of two ensures you always have a curve on the bonus dice.  The dice aren't open-ended (which I think you were really going for), but you could perform some sort of trick on the bonus dice to make it open-ended.  For example, rolling doubles on the bonus dice allows you to add the appropriate bonus dice together.  You'd still need to stick with opposed rolls to get the uniform bonus/penalty effect.
- Cruciel

Walt Freitag

J, I think you'd be getting into overkill territory by adding a complex procedure to modify the "fixed part." The exponential decay probability curves already implement diminishing returns, in that a point of advantage or distadvantage relative to the target number or opponent's ability becomes less likely to swing the outcome the farther away from an equal match (a 50% chance of success) you are to begin with.

Cooperating and interfering agents first need a model (even if only a general idea in mind) for the kinds of effects you want them to have. No resolution system including this one has such effects built into the math. Should someone who's cooperating be able, or likely, to decrease instead of increase the effectiveness of the one they're trying to help? Does it depend on the task? (Can one person help another to pick a lock?) You have to decide these issues first.

That said, the most conventional way to do it would not be to add the T's together beforehand, but rather resolve the helper vs. the task, determine the number of successes or failures, and add (or subtract) those from the other character's T.

- Walt
Wandering in the diasporosphere


Hey, guys, :)

Wow. I've been away a couple of days and now spent all of four hours just catching up... Anyway, apologies on the late reply.

Walt, my musings for the past few days have led me to similar conclusions, i.e., that a general support/oppose mechanism probably needs more detailed considerations on what is actually being undertaken.

Since I don't have those considerations at hand, I guess we're done for now... :/

Anyway, again, to all you folks that jumped in, thanks for getting my brain strightened out.



PS A friend and I got together and pondered on that table you posted. I'll be sharing what we came up with in a few, but that's for another thread. :)
João Mendes
Lisbon, Portugal
Lisbon Gamer