### Post by sgingell on Feb 15, 2006 18:41:29 GMT -5

Something that crossed my mind:

There are some things to like about dice, random chance forces people to choose between the sure thing and the risky big win, random chance also allows for freak events that neither the PC's or GM planned to get worked into the story, etc... So if you fancy hauling out your dice bag to play MURPG, here is an easy way to do it.

Any character may 'trade in' stones of effort for die rolls on the following table:

2 stones: Roll a d4

3 stones: Roll a d6

4 stones: Roll a d8

5 stones: Roll a d10

6 stones: Roll a d12

10 stones: Roll a d20

The result on the die is the number of stones of effort you get.

Example: Cyclops shifts 6 stones into Optic Blast but trades them in them for the roll of a d12. He rolls a 9 and therefore makes a 9 stone attack (had he rolled a 2 it would be a 2 stone attack).

Alternately you could trade in just some of your stones. Wolverine shifts 8 stones into defense and trades in 3 stones for a d6 and ends up with a defense of 1d6+5.

Quick, easy, and on average almost exactly as powerful as the deterministic way. A d6 averages 3.5 so is slightly better than 3 stones, but is also slightly worse in being unpredictable. If you think the chance of a high roll supports rounding up, just add one to all the stone costs in the table above.

The only last thing to consider is that the random element only would effect resistance, never difficulty. Cyclops may get lucky and slip past Wolverine's defenses with an Optic Blast, he's not going to get lucky with his Strength and lift 10 tons though...

-Stephen

There are some things to like about dice, random chance forces people to choose between the sure thing and the risky big win, random chance also allows for freak events that neither the PC's or GM planned to get worked into the story, etc... So if you fancy hauling out your dice bag to play MURPG, here is an easy way to do it.

Any character may 'trade in' stones of effort for die rolls on the following table:

2 stones: Roll a d4

3 stones: Roll a d6

4 stones: Roll a d8

5 stones: Roll a d10

6 stones: Roll a d12

10 stones: Roll a d20

The result on the die is the number of stones of effort you get.

Example: Cyclops shifts 6 stones into Optic Blast but trades them in them for the roll of a d12. He rolls a 9 and therefore makes a 9 stone attack (had he rolled a 2 it would be a 2 stone attack).

Alternately you could trade in just some of your stones. Wolverine shifts 8 stones into defense and trades in 3 stones for a d6 and ends up with a defense of 1d6+5.

Quick, easy, and on average almost exactly as powerful as the deterministic way. A d6 averages 3.5 so is slightly better than 3 stones, but is also slightly worse in being unpredictable. If you think the chance of a high roll supports rounding up, just add one to all the stone costs in the table above.

The only last thing to consider is that the random element only would effect resistance, never difficulty. Cyclops may get lucky and slip past Wolverine's defenses with an Optic Blast, he's not going to get lucky with his Strength and lift 10 tons though...

-Stephen