Converting the 2D6 spread to the 1D20 spread.

[Treebore]

Anyone seen or know the "optimal" way to convert 2D6 assumptions to the 1D20? IE what would a result of 6 on 2D6 be equivalent to on the D20? What would an 8? A 10?

[Captain_K]

2d6 would have a centering affect, d20 would be linear.. I would assume something like 7 on 2d6 would have to be something like 10 and 11.. etc. with 2 or 12 being only 1 and 20...

[Treebore]

2d6 would have a centering affect, d20 would be linear.. I would assume something like 7 on 2d6 would have to be something like 10 and 11.. etc. with 2 or 12 being only 1 and 20...

Yeah, I was thinking of having each "target number" on the 2D6 being related to two numbers on the D20 as well.

So in my completely non Mathematical way, I was thinking of having 1,2 and 3 be the "snake eyes" in the 2D6, and 10 and 11 be equivalent to 6, and 13 and 14 being equiv. to 8, 16 and 17 to 10, and 19 and 20 being equiv. to 6/6. I'm just hoping someone far more mathematically savvy than I am can help me tighten up the precision of this relationship. I am not worried about accuracy, since thats kind of impossible as far as I can tell. So I'd just like to get it reasonably precise.

[Aramis]

Some have come before you with the same question:

https://www.google.com/#safe=off&q=conv ... +d6+to+d20

specifically, these tables look relevant

http://www.sarna.net/wiki/User:LRichard ... Conversion

[Treebore]

Some have come before you with the same question:

https://www.google.com/#safe=off&q=conv ... +d6+to+d20

specifically, these tables look relevant

http://www.sarna.net/wiki/User:LRichard ... Conversion

Yeah, I figured as much, and yeah, I think that guys work tells me exactly what I wanted. Thanks!

[Treebore]

So this is the table L. Richardson came up with that I'd want to use, and by having the 20 on a D20 being a "wild card".

Table 3: Modified 2d20 Equivalence
Die Roll (2D6) Cumulative P Equivalent 1d20
"two" 0.028 1
"three" 0.083 2
"four" 0.167 3
"five" 0.278 4-5
"six" 0.417 6-8
"seven" 0.583 9-12
"eight" 0.722 13-15
"nine" 0.833 16-17
"ten" 0.917 18
"eleven" 0.972 19
"twelve" 1.000 20


This table too has an issue however. The odds of rolling a "twelve" or a "two" become 1/20 each, as opposed to 1/36. While this might be acceptable for threshold rolls, such as resolving hits, if this table was used to determine hit locations (for example) the odds of both head-shots and critical center torso hits nearly doubles. While some player groups might welcome this increased lethality, a great many will be less than enthused about their hard earned `Mechs being twice as likely to be taken out by "freak" head-shots.

This can be remedied by making the roll of 20 on the d20 a wildcard result. This is to say that the 20 is rerolled and if the result is 1-10, treat it as a "two"; if the result is a 11-20, treat the result as a "twelve". This has the end result of making head-shots and CT critical rolls slightly less common, something that many players might welcome. The rest of the hit possibilities are then reshuffled in the table and the result is as follows:

[Captain_K]

2d6 d20
TOTAL PROBABILITY
----- -----------
2 0.03 1
3 0.06 2
4 0.08 3
5 0.11 4,5
6 0.14 6,7,8
7 0.17 9,10,11,12
8 0.14 13,14,15
9 0.11 16,17
10 0.08 18
11 0.06 19
12 0.03 20
Sum = 1.00

[Treebore]

2d6 d20
TOTAL PROBABILITY
----- -----------
2 0.03 1
3 0.06 2
4 0.08 3
5 0.11 4,5
6 0.14 6,7,8
7 0.17 9,10,11,12
8 0.14 12,14,15
9 0.11 16,17
10 0.08 18
11 0.06 19
12 0.03 20
Sum = 1.00

Yep, matches up with Richardsons results, just need to add the 20 Wild Card fix to lower the odds of the 1 and 20 results from 1 in 20 closer to 1 in 36.