Dice pool game design woes

[mamba]

So it's still not right. And if the single die row is wrong, they probably all are.

it is right, while it explodes once, the second roll only has a 50% chance to add a point. That is why the 16.666 turns into 8.333

 

[aramis erak]

Haivng played at least a dozen (only 3 being WoD)...

if you're going count successes, WoD's method (explode on 10 if skilled, granting an extra die) seems pretty easy.

Playtesting the current Robotech (we dropped out), all of my players hated the dice mechanic, where 2×4's equaled a success, 1×5 was a success, and a 6 was 2 successes. I don't know if that continued on into release. They flat out said "No more" when they found out that modifiers could make 4's a full success and 5's two, or make 4's nothing, 5's half, and 6's 1 success... THey didn't like 4 result states, let alone modifying those. It also didn't come to intutiveness.

excepting symbolic pools (FFG, proper Modiphius effect dice), having more than 3 states for a die is way too many, unless some of them are pretty mnemonic...

FFG's effect dice are sufficiently mnemonic for me
1-2 result points = roll
5 and 6 are 1 and a special
mediocre dice (3-4) get nothing.

My favorite count success d6 dicepool is that of Arrowflight 1e. stat dice, rolling for skill or less. if 3 or more are 1's, roll another full pool's worth and add. If more than half the dice roll 6's, and the action fails, it's a fumble. Difficulties modify skill; when the TN goes over 5, the TN stays 5, but the difference from counted and 5 is a number of bonus successes gained as long as at least one non-6 is rolled.
The att range is roughly 2-6 with occasional 7s or 8's, skills start 0-4, no unskilled penalty. It's also not available in PDF... 2E uses a 2d6< (Stat+skill) with the same att and skill ranges...

 

[aramis erak]

This calculates 6s adding an additional die.

AnyDice

AnyDice is an advanced dice probability calculator, available online. It is created with roleplaying games in mind.

anydice.com

Not exactly; it limits the iterations and presents only a limited precision result, thus providing a pretty good approximation.

And yes, the pun was calculated.

Still, it's one of the better tools.

 

[soviet]

Instead of pitching a dice pool against a fixed target, you could instead pitch dice pool against dice pool. If each dice represents a 50% chance, then doubling the DC should provide the right number of dice for this new method (ie the second set of dice were the ones that failed). This removes the hard ceiling on which rolls can succeed, although the probabilities will mostly enforce the same outcome it won't be guaranteed.

 

[Morrus]

Instead of pitching a dice pool against a fixed target, you could instead pitch dice pool against dice pool.

Well yes, I mentioned opposed rolls in my OP.

 

[aramis erak]

Instead of pitching a dice pool against a fixed target, you could instead pitch dice pool against dice pool. If each dice represents a 50% chance, then doubling the DC should provide the right number of dice for this new method (ie the second set of dice were the ones that failed). This removes the hard ceiling on which rolls can succeed, although the probabilities will mostly enforce the same outcome it won't be guaranteed.

which increases roll-handling time and reduces reliability of results, IE, more swingy.

 

[Sacrosanct]

Instead of pitching a dice pool against a fixed target, you could instead pitch dice pool against dice pool.

that's exactly what I do in the link I provided upthread. I like it a lot better in practice compared to just counting successes. It makes things more swinging and unpredictable, which I like.

 

[Walt B]

I don't know if anyone will read this, but i'm throwing my hat in the ring in defense of the exploding dice pools.
I am running a system that uses an open-ended dice pool, calculating the probability is actually not that hard, only the math is quite extensive, you're basically working with really big polynomials, because that's all there is, a polynomial that refers back on itself, you don't need to calculate infinite rolls, just enough for the number of hits you want. That will require quite the elbow grease, patience and a really large blackboard though (or you can use over-complicated math to make a general formula that gives the probability of h hits).
The basic math behind is as follows:
I believe that each dice has 3 situations basicaly: miss, hit (but not crit) and crit (which resets the function AND adds a success) so... in a six sided dice, with success on 3 faces and explosion on 1 face... it would make the Probability Generating Function like this:
f(z) = 3/6 + 2z/6 + zf(z)/6
just solve the iteration a couple of times and the probabilities begin to appear, as each coefficient represents the total probability for that specific number of "hits".
so z^0 represents 0 hits (50%)
z¹ represents 1 hit
z² represents 2 hits... and so on
this equation basically means we are rolling just one dice, if we want more... just make the equation to the respective power
f(z of 2 dice) = (3/6 + 2z/6 + zf(z)/6)²
f(z of 3 dice) = (3/6 + 2z/6 + zf(z)/6)³ and so on, using wolfram alpha it becomes easier to get the coefficients.
Solving the first equation for the 2 iteration leaves us with:
f(z) = 3/6 + 2z/6 + z(3/6 + 2z/6 + zf(z)/6)/6
f(z) = 3/6 + 2z/6 + 3z/36 + 2z²/36 + z²f(z)/36)
f(z) = 0,5 + 0,417z + (2/36)z² + z²f(z)/36
50% for 0 successes
41,67% for 1 success
doing it again will show about 7% for 2 sucesses... just keep going.
if anyone is interested in trying to obtain the general formula, just try to develop the following using infinite harmonic series and binomials (trigger warning - high level math and frustration - do at your own risk):
f(z) = (3/6 + 2z/6 + zf(z)/6)^n

To use the equation on wolframAlpha write it like this, just repeat the equation putting a D on the last term when youre satisfied with the precision you want, then ignore any result in which the D appears:
f(z) = (3/6 + 2z/6 + zD/6)^n
f(z) = (3/6 + 2z/6 + z(3/6 + 2z/6 + zD/6)/6)^n
f(z) = (3/6 + 2z/6 + z(3/6 + 2z/6 + z(3/6 + 2z/6 + zD/6)/6)/6)^n
I believe wolframAlpha caps at a certain point if you try and write this over there, but there's some clever math tricks using binomials and hyper-harmonic series that give a "formula" for any h number of hits. I wont go in to anymore details, as that math can truly be a monster (i've done it and suffered a whole lot), but here's the table i came up with using Smath and excel

What people where discussing before is true, the probabilities begins to spread out towards the higher end of the dice count as you add more dice to the pool.
The zero's on the table arent actually zero, just extremely small... (for context, rolling 2 20's on a roll with advantage in d&D is about 0,25% in chance of happening, and we know how rare that is in on itself)

here is a more useful version of the table, wich shows cumulative probabilities...

is easier to see your margin of success going up as the number of dice increases!
if you take this table on a D&D context it becomes easier to determine a similar version of this

​

As you can see, when you roll 10 dice, youre as likely to to roll 7 or more successes on a given test as a character with a +13 modifier is to hit a roll of 26 to 30 or more on a d20 system without advantage (advantage makes the numbers go slightly larger for each ability level) (basically as likely as a dificult test, the rest you can interpret however you like).

In conclusion, if you interpret the numbers right you can do the same as D&D and give the DM and players a table on what is considered possible or impossible at each level, instead of using 5/10/15/20/25 as parameters, use 1/2/3/4/5/6! Give a percentage for each level of dificulty, let's say, 75% is easy, 50% is medium, 25% is hard, 15% is very hard, 5% is extremely difficult (call of cthulhu d100 vibes) and anything under 1% almost impossible. don't mention the actual probabilities, they tend to scare people.

people usually THINK they can grasp probabilities, but in reality they are just intuitively applying linear behaviours and patterns everywhere, if you manage to make this shift in perspective, the behaviour of the dice pool becomes sort of linear and easier to predict. If any anomaly happens. It's just sheer luck and probability! And that's actually possible!

Just to cap it of, in one of my campaigns using open-ended dice pools there is a player, A, who is EXTREMELY lucky and manages to roll extremely high results using 2-4 dice (rolling almost always above the 1% to 0.05% mark), and another player, B, who always rolls extremely low inspite of having a lot of dice (sheer bad luck). This just became a running gag every game. That doesn't mean the system is less predictable or more "broken", it simply is math, just because you have a 99% chance of something happening, its still random! and you may end up having 8 streak of losses on 99% chance os success or 8 streak of successes on a 1% chance of success

Edit: Took some time and organized everything a little bit better, hope the display wont explode just like the dice this time

 

[pemerton]

in a six sided dice, with success on 3 faces and explosion on 1 face... it would make the Probability Generating Function like this:
f(z) = 3/6 + 2z/6 + zf(z)/6

I think that is success on 2 faces, failure on 3.

50% for 0 successes
41,67% for 1 success

My maths is weaker than yours. The way I would work this out is that there is a 1/3 chance of 1 success from one die, plus a 1/6 chance of an explosion with an ensuing 1/2 chance of no more successes, equalling a 1/12 chance of 1 success via that pathway.

And then 1/3 plus 1/12 = 5/12, which is 41 2/3 % (approx 41.67%).

So I agree with you that it's not that hard to compute the likelihoods of particular outcomes, and generally the more remote outcomes don't matter all that much to understanding the general shape of the "outcome space".

 

[Walt B]

My maths is weaker than yours. The way I would work this out is that there is a 1/3 chance of 1 success from one die, plus a 1/6 chance of an explosion with an ensuing 1/2 chance of no more successes, equalling a 1/12 chance of 1 success via that pathway.

When dealing with smaller outcomes these logical aproacches work best! Use a boot to kill a bug, not a granade!

I did something similar while in the initial stages of exploring this problem. Used a huge blackboard and wrote down the logical approach step by step with the help of a teacher of mine to not lose track of any term by accident, after some fiddling the patterns start to appear and math starts to become a more pratical and less spacial consuming approach.

Usually when using logical approaches things will get messy when dealing with more than 1 dice and higher values. If we use logic as an approach in addition to the rest, we can get away with not doing the bulk of the mathematical proccessing by focussing our efforts on the members of the equation that actually matter.

For example, i didn't need to make a super complicated equation to find out that the chance os scoring a 0 decreases by a factor os 2 with every dice added to the pool, we simply look at the term that does not have a z assigned to it, if we multiply it anything other than another one without a z, you get 1, 2, 3 or more hits, so the only way to find 0 is by taking the powers of 3/6.

Looking at it this way makes the equation something that serves more as a way to make sure you don't forget that, in multiplication and distribution, lots of variables affect other variables simply by existing.

I think that is success on 2 faces, failure on 3.

A crit is still a success, that's why we break the 3/6 in to two parts, 2/6 and nothing happens, 1/6 and the dice explodes while still giving you a success. The only ways to get a single success are either by rolling 2/6 (success but no crit) or by 1/6 of critting and still missing on the explosion dice