The dice mechanic is this:

When you roll, either opposed (like fighting) or unopposed (like driving), you count the evens as successes and the odds you discard. Any roll that yields the highest possible value, e.g. 6 on a d6, is counted and rolled again. If every die rolled yields an odd number, you've critically failed.

This is the base mechanic. What I think is original is the evens versus odds part. The explosion and critical failure part, I simply adapted. I like the chance of every die being able to explode, and that the critical failure seems to me to be harder to get as dice are added.

it seems to me that the die type can change depending on the level of realism you want, higher dice for more realism. I'm not sure if this is correct though, with my high school education. For this discussions pruposes, we can assume: Six-sided dice are used; the number of dice you roll is based on the attribute it's linked to, and therefore can be one or many -- although I don't imagine more than 12 is possible; 2d is the basis of a normal man.

I'm curious about two things. First, has anyone ever published a game or even thought of the evens versus odds part, and if so, who? Second, since incorporating the criticals changes the fact that you can't just flip coins (or does it even?), how do I figure out the probability curves and stuff for this, and therefore understand what kind of target numbers for rolls could be?

Thanks,
Ryan Shehee

P.S. I'm new, so If I've missed anything, feel free to let me know.

I think you're setting yourself up for trouble trying to look at either end of your scale (the d4 end vs. the d20 end) as "more realistic" than the other. There are many good arguments that a highly open-ended mechanic, such as the above under a d4 setup, actually doesn't convey "cinematic" (as if it were opposed in some way to "realism", which is taxonomy not even worth getting into). Usually better to discuss the scale of "more luck-dependent" versus "more skill-dependent" instead.

The probabilities of the system are pretty straightforward, overall. You start with a coin-flip mechanic but then state, in essence, that one in every (die size divided by two) successful flips also gets you another coin to roll, which in turn rolls up, and so on. This is essentially a variant on the standard "exploding dice" setup with the value on the die decoupled from the "score" generated by the die.

A die of N sides, N even, will generate an average number of successes, X, which is given by:

X = [(N/2 - 1)/N] * 1 + [1/N]*(1+X)

...where the first part is "all the evens except the high roll, times one success each" and the second is "the high roll only, times one success plus another coin with the same average as this one." The algebra here is even simpler than for a standard exploding die (worth its face value), and gives us in the end

X = N/(2N-2)

...as the average number of successes one die of size N will generate. Thus, for a d6, X = 6/(12-2) = 0.6 successes per coin on average. For pools as small as two dice, however, you can expect a fair bit of variation from the average - the standard deviation will be high. This will be more true, the smaller the die chosen.

It can also be observed that your evens/odds would be equivalent to "upper half of die range is a success, max roll gets you another die." (Such as, on d6, 4-6 is a success, 6 also gets you a reroll.) This format might be more useful to you, honestly, as "high is good" remains a useful baseline if you don't have a darn good reason to do something else. (In your system, mind you, if there were a way to interpret individual dice - such as, the effect of one attempt along the way, or something - then high could be read as "more extreme" whether it's a success or a failure. But this is kind of abstract.)

So you have this:
(1) Fewer dice, each with an average result of X successes, means more variation and fewers rolls which actually hit that average. Obviously you can scale your number of successes required, to suit this; if you want things more averaged out and less chancy, then increase pool sizes and scale up the successes they need by the same proportion.
(2) One such die will generate slightly over half a success, per the above formula, on average. A d6 gets you 0.6 successes per die; a d4 goes as high as two successes per three dice on average (0.6 repeating), a d20 gets only 10/19 = ~0.53 successes per (getting close to a coin flip straight up). The smaller the die, the higher the average successes per.
(3) Smaller dice will also create more variation - luck will have more of a chance to be a factor. However, if you tabulated it out, I suspect that if your die pools are so small (2-4 dice typical), then the effect of just one additional die remains larger than the luck factor, even for d4s, so it won't go all the way to "luck more important than skill." Whether this is good, or not, is up to you; I discussed in another post here a way to actually monitor how much luck should matter for a given character, separate from all other concerns of actual effectiveness.

Overall, it's interesting, but I'd guess the evens/odds thing is more cute than useful, unless it ties into your design in some deeper way.

Hope that helps.

- Eric

OK, I have just a little more to go before I understand probability. But thanks for the help so far Eric. My next question is this:

How do I figure out the probability of rolling two or more dice and then subtracting the lowest number?

Thanks again,
Ryan Shehee

shehee wrote: How do I figure out the probability of rolling two or more dice and then subtracting the lowest number?

To clarify Chris' question, subtracting the lowest number from what? From the highest? From the sum of all other dice? From the sum of all dice (that is, roll four dice and ignore the lowest)?

--M. J. Young

M. J. Young wrote: To clarify Chris' question, subtracting the lowest number from what? From the highest? From the sum of all other dice? From the sum of all dice (that is, roll four dice and ignore the lowest)?

Sorry, I did mean subtracting from the sum of all the dice. Originally I was thinking of subtracting the lowest result from the highest result and then adding the results, but I figured this is the same as simply subtracting the lowest result from the total. I saw this on a website somewhere, in the original highest from lowest example, but can't find it now.

Thanks,
Ryan Shehee

shehee wrote: I'm curious about two things. First, has anyone ever published a game or even thought of the evens versus odds part, and if so, who?

shehee wrote: I did mean subtracting from the sum of all the dice.

First, thanks Talysman for your answer. Ironically, this was a system I thought of to use instead of Donjon's, after I played that game and was blown away by it (like I said, I'm new to whatever this trend is in revising RPGs).

Anyway, as far as the "subtract the lowest" part (which is a totally different mechanic from the even/odds, since I think I'm totally confusing this whole thing--I just figured it was under the same topic of "Dice Mechanics") maybe an example would help:

I roll two dice, then subtract the lower from the higher. I roll three dice, subtract the lower from the highest, and add the other. I roll four dice, subtract the lower from the highest, and add the others--same for five, six, etc.

(By the way, I can see now the difference in saying "lowest from highest", whereas originally I tried to state it more simply and fowled it up. One wouldn't be subtracting the lowest result from the sum of all the dice, because then you're just removing that dice, like Christopher is saying.

Now that I understand the difference, we can explore both, I just originally wanted to look at the "highest from lowest" probability.)

Thanks,
Ryan

shehee wrote: I roll two dice, then subtract the lower from the higher. I roll three dice, subtract the lower from the highest, and add the other. I roll four dice, subtract the lower from the highest, and add the others--same for five, six, etc.

Actually, Christopher, that's exactly what I was thinking, I...um...think. I didn't realize how difficult it would be to explain though. Maybe that's why when I first saw this explained, I didn't get it myself, and only now think I do.

The thing I'm lost on though, is how did you get those results? Where's the "formula", or whatever equation is used? (This is where my math becomes my weakest subject.)

Thanks,
Ryan Shehee

shehee wrote: how did you get those results? Where's the "formula", or whatever equation is used?