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Message: <blockquote data-quote="Morrus" data-source="post: 6739878" data-attributes="member: 1">I have a small design problem.  In line with the toolkit nature of my game, I want to include a small sidebar with a couple of alternative dice resolution mechanics. The probabilities need to remain the same; just the dice and how you read them need alternative versions.The current system: roll a d6 dice pool, add totals together, roll over a target figure.  The target figures look like this:[TABLE="width: 500"] [TR] [TD]Trivia[/TD] [TD]-[/TD] [/TR] [TR] [TD]Easy[/TD] [TD]7[/TD] [/TR] [TR] [TD]Routine[/TD] [TD]10[/TD] [/TR] [TR] [TD]Challenging[/TD] [TD]13[/TD] [/TR] [TR] [TD]Difficult[/TD] [TD]16[/TD] [/TR] [TR] [TD]Demanding[/TD] [TD]21[/TD] [/TR] [TR] [TD]Strenuous[/TD] [TD]25[/TD] [/TR] [TR] [TD]Severe[/TD] [TD]29[/TD] [/TR] [TR] [TD]Herculean[/TD] [TD]33[/TD] [/TR] [TR] [TD]Superhuman[/TD] [TD]37[/TD] [/TR] [TR] [TD]Impossible[/TD] [TD]40[/TD] [/TR] [TR] [TD]Mythical[/TD] [TD]45[/TD] [/TR] [/TABLE]    Here are some alternatives I'm considering including in that sidebar:[h=4]Count Dice Rolling 4+[/h]A number of systems do this. This makes each dice 50% chance of rolling 0 or 1, so the average dice rolls 0.5.The average roll on the default method is 3.5. Does this mean all I need to do is divide target numbers by 7 to retain the correct probabilities?  You'd have to round up, I guess, meaning difficulty numbers range from 0-7.[h=4]Count Dice Rolling 5+[/h]Assuming the above method is correct, this results in dividing target numbers by 10 instead of 7.  Difficulty numbers range from 0-5, so it lacks granularity.[h=4]Count Dice Rolling 6+[/h]I don't think this one is viable. It's useful because you can buy d6s with a different "six" face, so it's visually easy. But it reduces granularity massively, dividing difficulties by 21.8 (call it 20).  This makes for only a handful of difficulty numbers (0-3, basically). I think this can be discounted.[h=4]1d20 + bonus[/h]A bit of a departure down D&D Highway. I'm not sure how to work out the equivalent here.If we assume 1d20 rolls 10 on average (rounding down for simplicity).  So that's our baseline for no bonus, or our "routine" entry in the table above.Each d6 needs to convert to a bonus.  Each rolls 3.5 on average. Let's call it 3 (rounding down again).  So the roll is 1d20 + (3 for each die).  Using this rolling method, the difficulty targets remain the same.Is that correct?[h=4]Percentile Dice[/h]I find percentile dice can add a more technical/scientific feel to a game.  So, how to convert here? Average roll is 50.  So that's our Routine entry.Ugh.  I'm stuck.  My brain won't go where it needs to go!  So 100 is five times 20, and on d20+mod each d6 converts to +3.  So on d100, each d6 converts to 15?  And the difficulties are five times higher?  So 5d6 vs. Difficult [16] converts to d100+75 vs. 80?  Nah, that's not right at all!</blockquote>

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