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<blockquote data-quote="Walt B" data-source="post: 9624846" data-attributes="member: 7051927">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&nbsp; 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)/6just 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 &quot;hits&quot;.so z^0 represents 0 hits (50%)z¹ represents 1 hitz² represents 2 hits... and so onthis equation basically means we are rolling just one dice, if we want more... just make the equation to the respective powerf(z of 2 dice) = (3/6 + 2z/6 + zf(z)/6)²f(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)/6f(z) = 3/6 + 2z/6 + 3z/36 + 2z²/36 + z²f(z)/36)f(z) = 0,5 + 0,417z + (2/36)z² + z²f(z)/3650% for 0 successes41,67% for 1 successdoing 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)^nTo 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)^nf(z) = (3/6 + 2z/6 + z(3/6 + 2z/6 + zD/6)/6)^nf(z) = (3/6 + 2z/6 + z(3/6 + 2z/6 + z(3/6 + 2z/6 + zD/6)/6)/6)^nI 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 &quot;formula&quot; 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&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; [ATTACH=full]401133[/ATTACH]&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; 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&amp;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...&nbsp;[ATTACH=full]401134[/ATTACH]is easier to see your margin of success going up as the number of dice increases!if you take this table on a D&amp;D context it becomes easier to determine a similar version of this[ATTACH=full]401135[/ATTACH]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&amp;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 &quot;broken&quot;, 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 successEdit: Took some time and organized everything a little bit better, hope the display wont explode just like the dice this time</blockquote>

[QUOTE="Walt B, post: 9624846, member: 7051927"] 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 [ATTACH type="full" size="1007x287"]401133[/ATTACH] 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... [ATTACH type="full" size="1020x292"]401134[/ATTACH] 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 [RIGHT][/RIGHT] [ATTACH type="full" size="723x295"]401135[/ATTACH] 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 [/QUOTE]

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