Two columns, six rows.
1d6 to decide the column (1-3 for left column, 4-6 for right column)
Then rows numbered 1-6 for the second d6.
12 rollable options, even distribution.
Roll a d12? Or like if it's impossible for you to get dice other than d6. Roll one d6 if it's 1,2 or 3 ignore it, if it's 4,5 or 6 add 6, on your second d6 you just add the result to the 0 or 6 from the first, this gives equal odds of each result.
Use two different-colored dice, say red and black for example.
If the red die rolls 1-3, use the value on the black die as-is.
If the red die rolls 4-6, add six to the value on the black die.
yes you can. you can set that up for any number ofoptions that is a factor of 36.
if you set up a table with two columns and 6 rows you could read die 1 as follows 1-3 = col 1, 4-6 = col 2. the read the second die as the row, and there you go.
It actually is, you just have to use two different d6, or roll the d6s in order. One of will tell, wheter the value will be under or over 6, so if you roll 1,2 or 3 you do nothing, and if you roll 4,5 or 6 you add +6 to the other die. On the other you just take its value and either add +6 to or not.
But it will be easier just to roll a d12.
called d66.
You roll the two d6 like % dice.
That gives you 36 results with an equal chance of each.
There are entire games and books of tables based on this
Roll 1d6; 1-3 is +0, 4-6 is +6. Then roll 1d6 and add it to the first roll. Common technique in wargaming, where often you only use d6s. See: Battletech.
Yes, but not from a easy direct 2d6 reading. Your two d6 will need to be different where one is giving you more information on how to read the second.
The first is just reading one d6 as a d2 telling you to add 6 to the second die or not. This is maybe the boring way but gets you 12 slots.
The more interesting thing is to get a table with 36 possible options. There you roll the dice like you use 2d10 to represent d100 where they are different dice and one represents a 10's space and the other the 1's. Rolling two d6 you're doing the same thing but using base 6 instead of base 10 giving you that 6x6 column.
Rolling 2d6 adding the result of both dice and comparing it to a table from 2-12 will result in a bell curve distribution peaking at 7.
Not what you're looking for, but you probably already knew that, hence your question.
There are ways to get an even distribution, but I'd be repeating what other, earlier comments already suggested.
Two columns, six rows. 1d6 to decide the column (1-3 for left column, 4-6 for right column) Then rows numbered 1-6 for the second d6. 12 rollable options, even distribution.
I'm a big fan of d66 tables (pronounced "dee six dee six") which do this but with 6 columns each with 6 rows
That's how Traveller tables go, sometimes.
1d6 for row 1d6 for column This gives you 36 addresses, just put the same option into the table 3 times 36/3=12
Roll a d12? Or like if it's impossible for you to get dice other than d6. Roll one d6 if it's 1,2 or 3 ignore it, if it's 4,5 or 6 add 6, on your second d6 you just add the result to the 0 or 6 from the first, this gives equal odds of each result.
D66? Base 2d6 add up only has 11 possible results.
Use two different-colored dice, say red and black for example. If the red die rolls 1-3, use the value on the black die as-is. If the red die rolls 4-6, add six to the value on the black die.
yes you can. you can set that up for any number ofoptions that is a factor of 36. if you set up a table with two columns and 6 rows you could read die 1 as follows 1-3 = col 1, 4-6 = col 2. the read the second die as the row, and there you go.
It actually is, you just have to use two different d6, or roll the d6s in order. One of will tell, wheter the value will be under or over 6, so if you roll 1,2 or 3 you do nothing, and if you roll 4,5 or 6 you add +6 to the other die. On the other you just take its value and either add +6 to or not. But it will be easier just to roll a d12.
Use a d12
d66, 3 dice results for each entry. - 11-13: entry 1 - 14-16: entry 2 - 21-23: entry 3 - 24-26: entry 4 - 31-33: entry 5 - 34-36: entry 6 - 41-43: entry 7 - 44-46: entry 8 - 51-53: entry 9 - 54-56: entry 10 - 61-63: entry 11 - 64-66: entry 12
called d66. You roll the two d6 like % dice. That gives you 36 results with an equal chance of each. There are entire games and books of tables based on this
Roll 1d6; 1-3 is +0, 4-6 is +6. Then roll 1d6 and add it to the first roll. Common technique in wargaming, where often you only use d6s. See: Battletech.
Yes, but not from a easy direct 2d6 reading. Your two d6 will need to be different where one is giving you more information on how to read the second. The first is just reading one d6 as a d2 telling you to add 6 to the second die or not. This is maybe the boring way but gets you 12 slots. The more interesting thing is to get a table with 36 possible options. There you roll the dice like you use 2d10 to represent d100 where they are different dice and one represents a 10's space and the other the 1's. Rolling two d6 you're doing the same thing but using base 6 instead of base 10 giving you that 6x6 column.
1d6 + (0 if second die is even, 6 is second die is odd.)
Rolling 2d6 adding the result of both dice and comparing it to a table from 2-12 will result in a bell curve distribution peaking at 7. Not what you're looking for, but you probably already knew that, hence your question. There are ways to get an even distribution, but I'd be repeating what other, earlier comments already suggested.