Frank Wang  05 Dec 2018 12:22 a.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, 13 means fail, 46 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 
Frank Wang  05 Dec 2018 12:25 a.m. PST 
sorry I posted it to a wrong area 
advocate  05 Dec 2018 3:48 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 46 by 1.17? 
Stryderg  05 Dec 2018 7:32 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

Stryderg  05 Dec 2018 7:55 a.m. PST 
DOH! just had an idea: work up the numbers on rolling 2 dice from the start: Die1 = 1 AND Die2 = 16 That's 6 rolls, all fails Die1 = 2 AND Die2 = 16 That's 6 more rolls, all fails Die1 = 3 AND Die2 = 16 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 = 16 That's 6 more rolls, all successes Die1 = 5 AND Die2 = 16 That's 6 more rolls, all successes Die1 = 6 AND Die2 = 13 That's 3 more rolls, all successes Die1 = 6 AND Die2 = 46 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% 
Stryderg  05 Dec 2018 7:57 a.m. PST 
I really hope that's what you are looking for because that just gave me a headache <smile> 
RudyNelson  05 Dec 2018 7:59 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. 
TGerritsen  05 Dec 2018 8:53 a.m. PST 
Isn't this pretty much the Axis and Allies War at Sea mechanism? 
Frank Wang  05 Dec 2018 7:07 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. 
Frank Wang  05 Dec 2018 7:17 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… 
Frank Wang  05 Dec 2018 7:23 p.m. PST 
@RudyNelson: Yes actually it's a mutated D2 
Stryderg  05 Dec 2018 7:50 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? 
Frank Wang  05 Dec 2018 10:41 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 
Stryderg  06 Dec 2018 6:36 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  06 Dec 2018 7:40 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 13 fail, 45 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 
Frank Wang  07 Dec 2018 12:28 a.m. PST 
@KimRYoung sure I will. Thank you! 
Frank Wang  07 Dec 2018 12:29 a.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] 