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


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468 hits since 4 Dec 2018
©1994-2018 Bill Armintrout
Comments or corrections?

Personal logo Frank Wang Sponsoring Member of TMP04 Dec 2018 11:22 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

Personal logo Frank Wang Sponsoring Member of TMP04 Dec 2018 11:25 p.m. PST

sorry I posted it to a wrong area

advocate05 Dec 2018 2:48 a.m. PST

Interesting. Getting no successes would be based on fail half the time.
If you pass any you get an average of 1 and 1/6 successes per pass. So maybe just multiply the successes on n dice passing on 4-6 by 1.17?

Stryderg05 Dec 2018 6:32 a.m. PST

I was trying to come up with an elegant formula, and failed miserably, so I decided to brute force it:


Die1 Die2 #Success
1 n/a 0
2 n/a 0
3 n/a 0
4 n/a 1
5 n/a 1
6 1 1
6 2 1
6 3 1
6 4 2
6 5 2
6 6 2

Stryderg05 Dec 2018 6:55 a.m. PST

DOH! just had an idea: work up the numbers on rolling 2 dice from the start:

Die1 = 1 AND Die2 = 1-6
That's 6 rolls, all fails

Die1 = 2 AND Die2 = 1-6
That's 6 more rolls, all fails

Die1 = 3 AND Die2 = 1-6
That's 6 more rolls, all fails
That's 18 rolls, all fails and that's 50%
And the rest of the die look like this:

Die1 = 4 AND Die2 = 1-6
That's 6 more rolls, all successes

Die1 = 5 AND Die2 = 1-6
That's 6 more rolls, all successes

Die1 = 6 AND Die2 = 1-3
That's 3 more rolls, all successes

Die1 = 6 AND Die2 = 4-6
That's 3 rolls, all 2x successes

So a total of possible 36 combinations
18 fails = 18/36 = 50%
15 success = 15/36 = 42%
3 2xSuccess = 3/36 = 8%

Stryderg05 Dec 2018 6:57 a.m. PST

I really hope that's what you are looking for because that just gave me a headache <smile>

RudyNelson05 Dec 2018 6:59 a.m. PST

Sounds 50/50 to me which means any type of dice can be used. The 6 result can still mean a cascading effect.

I have been designing over 40 years so there are few new mechanics that I have not seen in the past.

Using d6 is attributed to the commonality of the dice rather than the percentage chance needed.
Good luck with your designs.

TGerritsen05 Dec 2018 7:53 a.m. PST

Isn't this pretty much the Axis and Allies War at Sea mechanism?

Personal logo Frank Wang Sponsoring Member of TMP05 Dec 2018 6:07 p.m. PST

@TGerritsen:
Never know that game and I just went to youtube and watch it. Similar but not the same. In War at Sea a 6 counts as 2 hits. Here a 6 means you have chance to get 2 hits.

Personal logo Frank Wang Sponsoring Member of TMP05 Dec 2018 6:17 p.m. PST

@Stryderg:
Thank you and last night I work out 3 dice with pen and paper.
1 Dice:
0 success:0.5
1 success:0.416
2 success:0.083

2 Dice:
0 success:0.25
1 success:0.416
2 success:0.173
3 success:0.069
4 success:0.00694

3 Dice:
0 success:0.125
1 success:0.312
2 success:0.26
3 success:0.176
4 success:0.0538
5 success:0.0347
6 success:0.0005

calculate 1 die is quite easy but more dice are hell. and the numbers are very weird: 1 die and 2 dice have the same probability of getting 1 success…

Personal logo Frank Wang Sponsoring Member of TMP05 Dec 2018 6:23 p.m. PST

@RudyNelson:
Yes actually it's a mutated D2

Stryderg05 Dec 2018 6:50 p.m. PST

1 die and 2 dice have the same probability of getting 1 success…

That looks right because in both cases the 2nd die roll doesn't matter until you roll a 6 on the 1st one.

How are you getting to 3+ successes?
Are you counting a 6 on the 1st die as 2 successes?

Personal logo Frank Wang Sponsoring Member of TMP05 Dec 2018 9:41 p.m. PST

@Stryderg:
Of course you need 2 basic dice to make a 3+ success. Two situations: Die 1 is 1 success and Die 2 is 2 successes, or opposite.
That's (5/12)*(1/12)*2

Stryderg06 Dec 2018 5:36 a.m. PST

Got it.
The way I'm running the calculations, you have to know ahead of time how many dice will be rolled, then you can figure the chances of each combination happening. Division gives you the probabilities you want.

If you don't know how many dice will end up being rolled, I don't think you can calculate the probabilities.

So, did you get what you need?

KimRYoung Supporting Member of TMP06 Dec 2018 6:40 a.m. PST

This is pretty similar to Scott Pyle's "Goalsystem" which has been around quite awhile. He uses it in his Superhero games and other table top miniature games.

Essentially its roll the number of goal dice with 1-3 fail, 4-5 success, 6 is Two successes.

His Super Mission Force game even has a chart that shows all the statistical breakdowns on the probability of number of success for the number of dice rolled.

You should check it out

Kim

Personal logo Frank Wang Sponsoring Member of TMP06 Dec 2018 11:28 p.m. PST

@KimRYoung
sure I will. Thank you!

Personal logo Frank Wang Sponsoring Member of TMP06 Dec 2018 11:29 p.m. PST

@Stryderg:
I work it out yesterday:
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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