Less Dice?

[RamblingMan]

Any ideas for reducing the number of dice?  Rolling tons of dice is fun, but I'm not sure I'll always be able to get together more than 10 of any die higher than d6.  (And I don't want to roll 3 times for every roll, considering I have to roll several times already to build up dice)

[Clinton R. Nixon]

Ideas:

a) Half the number of dice. 1 success = two facts or 1 die

b) Always subtract dice from both sides to reduce the lower side's value to 2. In other words, if you have 12 vs. 8, roll 6 vs. 2.

[Mike Holmes]

b) Always subtract dice from both sides to reduce the lower side's value to 2. In other words, if you have 12 vs. 8, roll 6 vs. 2.

If you do this, consider allowing one quarter or one third of the reduction to be automatic successes. To make up for the successes that you might be missing out on. Or use the 2 for 1 rule from a.

My suggestion: get more dice!

Mike

[RamblingMan]

Sure, more dice!  I'll email you my address, and you can send me some. :-)

[Mike Holmes]

Sure, more dice!  I'll email you my address, and you can send me some. :-)

Deal.

No Dunjon player should have to suffer through not having enough dice.

I hope you don't live in Uzbekistan.

Mike

[Bailywolf]

One entirely untested scheme for die trimming I was toying with was to set a maximum die pool for the game (either a set number- say 10- or something based on donjon level... perhaps 8+DL).  You only roll dice up to the game's Max, dice beyond this are held as Overhead.  If you win, Overhead is counted as Victories, if you loose your Overhead can counter your foe's Overhead 1 for 1, but otherwise won't help you.

The upshot I can see from this is that it makes success for big-pool characters quite wide...unless facing off against another high-dice character.  It could make for sudden death fairly easily, which may be apropriate in some games (especialy ones in which the DM wishes the players to be cautious and crafty).  

To vary the outcome of this sort of tweak, the actual die max can be varied as can the exchange rate on overhead-to-victories.  Instead of 1:1, it could require 2 points of overhead for each additional victory, or 3 even.  I would say that especialy in the case of a 3:1 ratio, that if you have any net overhead left after your oponent subtracts his (if any) you get at least 1 bonus victory from it.

Still another possible aplication for Overhead points might be to allow you to spend them to make your oponent reroll the die of your choice... or for more, to drop that die entirely.  Say like so:
1 Overhead- oponent must reroll die
2 Overhead- oponent must drop best die
3 Overhead- gain bouns victory


So in practice, say Ergot the Uncanny powers up with a Magnificent Blood of Dragons spell (from which he rolls the success into this next spell).  He then casts Terible Dragons Breath with bonus dice from the previous spell.  He has a total or 15 dice to roll, and the Max for this game is 10.  He rolls 10 dice and has 5 Overhead to play with.  He rolls and so does his oponent... who manages to get a single die higher than Ergot's even though he rolls only 5 dice.  Ergot can't stand to be defeated by a mathmatical fluke, so spends 2 points of his overhead to cause his oponent to drop that high die as if he never rolled it.  This allows him to swing 3 Victories easily, and he spends his final 3 Overhead to get him a fourth victory.

[quozl]

Here's an idea:

Decide how many dice you want to use at most and how many at least.  Let's say 12 at most and 3 at least.  Now when getting ready to roll dice, take the higher total and subtract the lower total.  High total rolls difference + 3 dice.  Low total rolls 3 dice.  If the high total is 10 or more higher than the low total, high total gets total success.

How would that work?

[Zak Arntson]

Depending on the sides of the dice, cash in x dice for an automatic one-die-at-maximum-rollage. For six-sided dice, trade in 2 of 'em for a single automatic 6.

Don't know whether that's appropriate, probablity-wise or not, but that's how I approach it in Donjon's weird third-removed cousin, Fighter-D Alpha.

[rafial]

Ideas:

a) Half the number of dice. 1 success = two facts or 1 die

b) Always subtract dice from both sides to reduce the lower side's value to 2. In other words, if you have 12 vs. 8, roll 6 vs. 2.

Actually (b) doesn't quite work.  Being the obsessive-compulsive freak that I am, as soon as I finished reading the rules, I sat down and wrote a program to simulate Donjon dice pools.  What I quickly learned is that the odds of victory are determined by the relative sizes of the pools.  That is, if you are looking at equals odds of success or failure then:

12 vs. 8 == 6 vs. 4 == 3 vs. 2

if you had 12 vs. 9, you could divide each side by 3 and get 4 vs. 3.  And so on.

Bigger pools increase the odds of getting multiple successes if you win.  Basically the bell curve gets flatter and wider.

I also looked at the effect of die size on rolling.  The success/failure odds don't change (as expected), but the bell curve flattens even more the smaller the die.  With six siders, its is quite flat, meaning the odds of getting multiple successes go way up.  Depending on the sort of game you want, that's not all bad.

For those that care, if you have 2:1 odds in your favor, you have a 70% chance of success.  3:1 gets you 80% and 4:1 gets you 85%.  As many have suspected, in Donjon there is no sure thing.

[RamblingMan]

What I might do is divide the dice by a proportion so that the lower rolls two or three dice.  For example, 8:4 dice becomes 4:2.  Or 9:5 becomes 5:3.  It might be inexact, probability-wise, but it's more fun than rolling four dice six times per person per round of combat.

[Hogscape]

How about (again, totally unplaytested), modifiers, abilities and attributes add to form the 'base' to which you add d6-d6 (as in Feng Shui). The net difference in totals is the fact/dice....