"Check my math, die success progression" Topic
13 Posts
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Stryderg  07 May 2021 9:12 p.m. PST 
While looking for a way to increase chances of successful die results on d6's, I came up with this: Base chance: on 1d6, a 1 is successful One step up: on 1d6, a 1 is successful, if a 2 is rolled then roll a second d6 and 1, 2 or 3 is successful Two steps up: on 1d6, a 1 or 2 is successful Three steps up: on 1d6, a 1 or 2 is successful, if a 3 is rolled then roll a second d6 and 1, 2 or 3 is successful Four steps up: on 1d6, a 1, 2 or 3 is successful etc I think each progression is an increase of 8.33% 1st question: Does that make sense? 2nd question: Is my math right? 3rd question: Is there a better way to describe that? 
jwebster  07 May 2021 9:47 p.m. PST 
Note that Pr(1) is twice the others. Not sure whether this then gives 11 levels or 12, so 8.3% may be slightly off. Perhaps you could use a d12 instead You could also roll both dice at once with the second a different colour. The problem I have with this scheme is that you would always be having to look up on chart whether you need the second die or not John 
advocate  08 May 2021 2:51 a.m. PST 
Your figures look correct to me, though your base chance is obviously 16.66%. For me, its too complicated, even with jwebster's good suggestion of two different dice. You could use a D12 and count your base chance as 12 (has the added advantage that you could take your base down to 8.33%). A D10 would approximate the effect with fewer increments. 
etotheipi  08 May 2021 3:13 a.m. PST 
It's not what you described. The outcomes of the second die rolls are independent of the first: Each roll of the second die, halves the probability of a success from the probability of the first. d0 d1 p 1 16.6% 1 13 8.33% 12 33.3% 13 13 16.6% 13 50% 13 13 25% 14 66.6% 14 13 33.3% 15 84.4% 15 16 41.6% 16 100% 16 16 50%
You can't come up with a table that replicates a d12 from 2d6. The 7, 8, and 11 rolls do not have common factors with {1, 2, 3, 4, 5, 6}. You need a procedural solution: 8.3%: 1 on d1 and 13 on d2 16.6%: 1 on d1 25%: 1 on d1 or 1 on d1 and 13 on d2 33%: 2 on d1 41.6%: 12 on d1 or 3 on d1 and 13 on d2 and so on … with this, the steps the way you describe them are out of order. 
GildasFacit  08 May 2021 3:26 a.m. PST 
No etotheipi, you have misinterpreted the instructions. The roll of the 2nd die is dependant on the 1st die being a certain number. It probability of success is not affected but it only counts if the first die is the appropriate number. so … Base – D1=1 > 16.7% Step 1 – D1=16.7% + D1=2 AND D2=1,2 or3 > 16.7%/2 (for the 50% success rate of throwing 13 on the 2nd die after a 2 is thrown on the first) and so on. I'd second using a D12 or even a D20 if you want a linear ramp distribution with smaller increments than a D6 gives. 
Stryderg  08 May 2021 4:21 a.m. PST 
Thanks guys. I thought it would be overly complicated, but was hoping I was on to something. 
Dexter Ward  08 May 2021 5:13 a.m. PST 
Just use a D12, or if you want to stick to d6, roll 2 dice. If the second dice is 4,5,6, add 6 to the first dice. That will get the same effect with much less complication. 
UshCha  09 May 2021 12:04 a.m. PST 
We long since addopted D20. I have to confess it was not my idea. My son is a dedicated RPG man and many of them long since abandoned D6 in favor of D20. We did so on his recommndation and have never looked back. 5% seems to be a sweetspot. Most certainly modelling reality closer than 5% in a wargame is being a bit optermistic. However 5% lets you account sensibly for the key variables. Slavishly sticking to a D6 when clearly inadequate seems to me to be a waste of time. Plus at times its possible for 1 D20 to do two jobs at once if the proabilities are compatible, this makes the game faster. 
Wolfhag  09 May 2021 4:20 a.m. PST 
I use a D20 for almost everything in my game. You can get a 1% to 4% result using a reroll on a "1" with a 14 = 1%, 58 = 2%, 912 = 3%, 1316 = 4% and 1720 = 5%. Wolfhag 
Stryderg  09 May 2021 7:08 a.m. PST 
I agree that d20's are awesome, but I'm kicking around rules for a print and play game. And I figured that the great unwashed masses would have easier access to d6's. 
Wolfhag  09 May 2021 1:40 p.m. PST 
You're probably right about the D6. Wolfhag 
etotheipi  11 May 2021 12:39 p.m. PST 
Just use a D12, or if you want to stick to d6, roll 2 dice. If the second dice is 4,5,6, add 6 to the first dice. That will get the same effect with much less complication. I personally agree with you, however I have seen what I consider to be unaccountable shock in response to gaming concepts like "add six to a number". For things like your recommendation, I often make a d2 from a d6. Take a black d6 with white pips. Use a black marker pen on the corner and side pips. Now you have a binary die that rolls {0,1}. Of you can do {1,2} about as easily. The d3 is also useful for similar concepts, but it always looks a little wonky since you need four odd numbers (center pips) for {1,2,3,1,2,3} but only get three odd numbers in the standard pip layout for {1,2,3,4,5,6}. 
bobm1959  29 Jun 2021 5:58 a.m. PST 
Your maths is correct. Each additional step increases the success rate by 8.333%. However if you're a fan of diminishing returns i.e. each step up after the first contributes less and less you could consider each step contributes a further D6 with a 1 still being the success (the number of 1's being irrelevant). you then get each step increasing by 13.9%, 11.57%, 9.64%, 8.04%. It might suit your rules and is certainly easier to remember where steps =number of D6. 
