As I understand it, the standard method for rolling a d3 is to roll a d6 and read it thusly:
1-2=1
3-4=2
5-6=3
My question is, how on Earth did this insane method ever become standard? I for one refuse to use it, instead preferring to read 1-3 as is and subtracting three from 4-6.
Discuss.
Both those methods are completely insane and overly complicated. The obvious answer is to roll three sets of percentile dice of different colours. Assign the values 1, 2 and 3 to the different colours. Whichever set rolls highest gives you the result of your d3. Simple, quick, and so easy even a baby could figure it out.
Roundup[1d6 / 2].
Roundup[1/2]=Roundup[0.5]=1
Roundup[2/2]=Roundup[1]=1
Roundup[3/2]=Roundup[1.5]=2
Roundup[5/2]=Roundup[2.5]=3
Roundup[6/2]=Roundup[3]=3
Another way you could do a 1d3 would be to roll 1d4, and reroll all 4s.
The generalised way to roll any 1dy would be to roll 1d(10 * z), and reroll all results that are equal to y+1 or more. z is the number of decimal digits found in y. Thus, to roll a 43, you'd generally need to roll 1d100 (roll 2 1d10s, 1d10 for the 10s and 1d10 for the 1s) and reroll all results that are higher than 1d43 (reroll all 1d10s that are higher than 4 for the 10s, and if the 1d10 for the 10s turned exactly 4, then reroll all 1d10s that are higher than 3 for the 1s).
Assuming that the d6 is "fair", with opposite faces adding up to 7, the "fair" way to convert is:
1 or 6 becomes 1 2 or 5 becomes 2 3 or 4 becomes 3
This ensures that at each of the 8 corners of the die, there is a "1", a "2" and a "3", so all corners are equal.
The rule is: "If the roll is 1, 2 or 3, keep it. If the roll is 4. 5. or 6, subtract it from 7."
And, no, I don't do it that way, either. 
I just read whatever number is on the top ridge of my d3... don't know why people bother with all this roll a d6 and do math business.
Personally, I don’t think the d12 gets enough love. Roll a d12 and 1-4 = 1; 5-8 = 2; 9-12 = 3. Show your love for d12s.
It came from simple division.
1d6 / 2 (round up).
Though cycling works too.
1d6 base 3.
I switch between the 2 methods and can't decide which i like better. (using a d8, since i lost my d4).
I just say, "Let's not have D3s."
I use d20.
1-3 use that number
4-20 roll again
It takes a while sometimes, but I at least I don't have to use any math.
Wuzzard wrote:
Matau99 wrote:
I use a d5 (roll a d10 and divide by 2), and just reroll results greater than 3.
Wuzzard wrote:
ChrisCarlson wrote:
Okay, I can't think of anything funny, so on a more serious note.
One of our players has a couple of "d12's" that are actually a D3 and a D4. Kinda clever... (The D3 has 12 sides, 1 2 3 1 2 3 1 2 3 1 2 3)
I have a few "blank" white d12s. I use a permanent marker, to write whatever I want on each side, to make my own dice. I think i'll do that.
Once each, on three of the sides, I'll write a "1", a "2" and a "3". Then on the remaining nine sides I'll write "REROLL".
Brilliant idea! Thanks!
ChrisCarlson wrote:
You could roll 2d6 and use this look-up table:
Roll: 2 3 4 5 6 7 8 9 10 11 12
===============================
Result: 3 1 2 3 1 2 3 1 2 3 1
The probabilities work out right. If you like formulae, it's (2d6 mod 3)+1
Why roll two dice when you could just roll one? Because it feels more random.
Matau99 wrote:
BoldItalic wrote:
Coredump00 wrote:
ChrisCarlson wrote:
Gah! Of course. I completely failed to include you demon-possessed left-handers. I'd apologize, but lefties don's deserve a fair shake.
ChrisCarlson wrote:
Failure to catch that the entire purpose of the post was a delivery a cheesy pun.
-1
ChrisCarlson wrote:
Just flip a coin.
In the set of dice that come in the Starter Set box, the d6 has numbers on the faces, rather than pips. The "6" has a line under it, so you know it's a "6" and not an upside-down "9". Think about that one ...
BoldItalic wrote:
Here you go.
Torg_Smith wrote:
BoldItalic wrote:
I prefer using my 5-sided die. Its hard though because all the sides have the same area without any repeated digits.