I do want to point out that number of party members is a direct factor in success in the game. Here's a simple equation to model the situation:
Let n=number of party members and x=probability of success and let i be some multiplier that represents a given average level of damage or effect per party members...

n*i*((1+x)^n -1)=y,

Where y is the overall effectiveness of a given party. As n increases, y increases significantly because you are increasing the probability of a successful roll and multiplying the overall output on that roll, assuming the effectiveness of a given party member, i, is held constant. Let us also use the same equation for repeatable rolls in the world, except slightly modified...


Where x is the probability of success on a given roll, and y is the overall success for a worldly encounter for a party. As n, the party size, gets bigger, the overall success of rolls, regardless of how difficult the rolls are, gets more frequent. A very difficult roll with a large party can have the same probability of overall success as a very easy roll for a small party. Therefore, party size is important for game balance in the world and in combat.

Finally, let us consider a situation of one-time rolls, like ones which can or cannot initiate combat in dialogue. Only one person speaks a dialogue option, and party members cannot try again. Still, with a large party, you can stack bonuses and items, such as guidance. We can modify the equation with a new factor m=viable bonuses probability, and v=number of unique bonuses. These factors increase with n, as you can specialize more characters with a larger party. So...

x+mv=y, where both m and v are increasing in n.

Therefore, for all major mechanics, a larger party means an easier game, unless the whole game is rebalanced/reworked.

THEN you get the super-easy effect, where you can repeat rolls with different party members AND use different bonuses...

((1+x+mv)^n)-1=y, such that mv can only repeat for as many spell slots exist and so decrease with every n.

I need a minute to fix this. Most of it is fine tho.

Most of the above holds as a general rule of thumb, but a more accurate mathematical representation would be probability of failure...
Let "b" be probability of failure, and all other variables stay the same.
Let probability of failure with a given party be b^n, and probability of success be x=1-b^n.
The equation for multiple rolls however should be done from the perspective of failure.
1-(b^n)=y is the equation for y as the probability of overall success on a given roll, or what was once equation two. My conclusions still hold, just replace success with failure when reading.
Overall combat effectiveness is then...
n*i*(1-(b^n))=z, where z is overall effectiveness for a given party. As b^n gets smaller as n increases, my paragraphs above still hold.
Then for one-time rolls with situations like guidance...

Last edited by Zerubbabel; 30/01/23 04:05 PM.

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