Listen up you little dice monkies

Christian said:
For example, suppose you need to generate a roll of 6-29 (for God-knows-what reason). That's a 23-point range. Looking down our list of dice, it's not hard to see a couple ways to get that-d12 (11 points) + d8 (7 points) + d6 (5 points) = 23 point range. That ends up being 3-26; so d12 + d8 + d6 + 3 gives us the 6-29 roll we were looking for. Easy! :)


Of course, as noted, this will not give you a uniform distribution.

If we wanted a uniform 6-29, I'd use a d6 and a d4. A 1 on the d4 means we read the d6 as it stands. A 2 means we add 6 to the d6 result. A 3 means we add 12, and a 4 means we add 18.

This gives us a uniform spread from 1-24. Reroll any 24s.

Add 5 to the result; a uniform spread from 6-29.

For a fair d13, I'd do the same with 2d4, rerolling 14s, 15s, and 16s.

-Hyp.
 

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Brandigan said:
What's the best way to generate a number from 1 to 13 using dice without having to whittle a d13? Thanks!

Assuming you want a uniform distribution (every possible number is equally likely) I'd use a simulated d40, say by combining d4 and d10. Read the d4 as the tens digit (treat 4 as 0) and the d10 as the ones digit. If you roll (4,0), reroll; otherwise, you have a random number from 1 to 39. Divide by 3, and round all fractions up, to get a random integer from 1 to 13 inclusive.

The general idea: you can simulate an n-sided die with any multiple of n; here, we're simulating a d13 with a d39. You can simulate an n-sided die with a slightly larger number, by rerolling the invalid rolls; I don't like to do this with too many rerollable numbers (I prefer not to use a d20 with 7 rerollable numbers, as other posters have recommended) because it might take too long. Here, I note that 39 is almost 40, so we simulate a d39 with a d40. If the desired number of possibilities factors into numbers for available dice (here, 40=4x10), you can combine dice (basically the same way percentile dice are simulated).

So any time you want an n-sided die, look for numbers slightly larger than multiples of n which are easily factored.
 

The easiest way is definitely just to roll d20 and reroll anything that's too high. Most of the time you'll only need to roll once. that's pretty much always the easiest thing to do... ignore some values on dice that are just a little bit too high.

For things above 20 and below 100, roll d100, except use a die for the 10's digit of close to the approximate size. So 1-60, use 1d8 as the 10s digit, treating 7 as 0 and rerolling 8s.

-The Souljourner
 

The easiest way is to get a dice rolling program (assuming computer nearby). Of course, that's assuming you trust pseudo-random algorithms. Being a programmer, I don't. :)
 

ThirdWizard said:
The easiest way is to get a dice rolling program (assuming computer nearby). Of course, that's assuming you trust pseudo-random algorithms. Being a programmer, I don't. :)

You telling me its easier to turn my computer on, start a program, and type in the appropriate instruction, than to pick up a die or two and toss it on the nearest surface? You're far more slave to technology than I. I'll use a spreadsheet for lots of simultaneous rolls, but not for just one.

BTW, the reason I prefer the simulated d40, reroll 40s to the actual d20, reroll 14-20 is that with the latter, it can get annoying if you get a long string of rerolls. My way always requires two dice, but hardly ever more.
 





Roll d20 Discard 14+

Why would this not give you an even spread? Your numbers would average 7 just as if it were a d13 (1 + 13 = 14, 14 / 2 = 7).

Just because, over time, a d20's rolls will average 10.5 does not mean you are rolling a dispo\roportionate number of 10s and 11s. It just means that the numners average that. Each time you roll a die (unless it is loaded of course) you generate a random number within the range of that die. They do not tend to roll at the middle of the range.
 
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