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"My new idea of a dice mechanism, and helps are needed" Topic


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349 hits since 4 Dec 2018
©1994-2018 Bill Armintrout
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Personal logo Frank Wang Sponsoring Member of TMP04 Dec 2018 11:24 p.m. PST

Recently I have some ideas of a new dice mechanism. I called it Multipurpose Dice Mechanism. It was inspired by the exploding dice and works like this:

A normal D6, 1-3 means fail, 4-6 means success. And if you rolled a 6, it can, and only can explode one time(roll 1 more die).

For example:
A character's ability is 4, roll 4 dice, result:2, 3, 4, 6.
2 failure and 2 success. 6 can explode so you roll another die, result 6 again. This time the 6 only count as a success but no more explode. So the final result is 1 failure and 3 success.

What I need to be helped is, I really don't know how to use anydice.com to calculate the probability of success. So if anyone good at program language, can you please help to calculate that?

Thank you!
Frank

Andrew Walters05 Dec 2018 8:08 a.m. PST

First step is creating a die with the successes and failures on it, like this:

output 4d{0,0,0,1,1,1}

That rolls four dice which have three successes (1s) and three failures (0s). The 4 is the number of dice, "d" just means die, and the numbers in the brackets are the difference sides of the dice. AnyDice will let you have a die with any number of sides and any numbers on those sides using this notation. And it's recursive, so…

output 4d{0,0,0,1,1,{1,1,1,2,2,2}}

You see again we are rolling 4 dice (change this to what you like), and on each of those dice the first three sides are 0, failure. The fourth and fifth sides are 1, representing a single success. The final, sixth side is another die, effectively *inside* the first die. It's first three results are 1s, because they are a success on the first die followed by a failure on the second for a net of one. The last three sides of the second die are 2, because they represent a success on the first die followed by a success on the second die.

Try it!

This should allow you to see the distribution curve of successes for whatever number of dice you roll in this fashion.

Good luck!

Personal logo Frank Wang Sponsoring Member of TMP05 Dec 2018 9:54 p.m. PST

@Andrew:
Thank you for your carefully answer! This language sentence seems all right but the result is completely different from what I did with pen and paper. I think anydice has a problem in the calculation. In other words, maybe this sentence does not express the thing I need correctly. But it's very close. I'm trying to modify it.
Thank you anyway! You did helped!

Frank

Personal logo Frank Wang Sponsoring Member of TMP06 Dec 2018 12:59 a.m. PST

I made it with AnyDice, perfect!

function: basic ROLL:s {
if [count {1,2,3} in ROLL] { result: 0 }
if [count {4,5} in ROLL] { result: 1 }
if [count 6 in ROLL] { result: 1d2 }
}

output [basic d6] + [basic d6]

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