Listen up you little dice monkies

[Squire James]

Would a deck of cards (Ace=1, Jack=11, Queen=12, King=13) count as "getting out the d13"?

 

[calypso15]

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).

D20, drop 14+ will give you a uniform distribution. I think the person above was saying that 2d7-1 wouldn't give you a uniform distribution.

Calypso

 

[Hypersmurf]

Why would this not give you an even spread?

It does. It's just annoying to use

It's anything involving adding two or more dice together that doesn't.

-Hyp.

 

[Staffan]

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).

If you're talking about "d20, ignore 14+", ignore what I'm saying below. Your post didn't make it clear what you were replying to (you might want to try the "reply" button at the bottom of each post, instead of the bigger "Post Reply" at the bottom of the page, alternatively investigate the uses of the Quote tag. But that's beside the point).

2d7, or as suggested, d6+d8, do notgenerate an even spread. That's because there's only one way of rolling a 2 or 14, while there are 7 or 6 (depending on method) ways of rolling an 8 (1+7, 2+6, 3+5, 4+4, 5+3, 6+2, and 7+1 if rolling d7s, but not if rolling d6+d8). That's called a bell curve (well, more like a triangle when using two dice, but...).

 

[Hypersmurf]

Would a deck of cards (Ace=1, Jack=11, Queen=12, King=13) count as "getting out the d13"?

That's a winner

-Hyp.

 

[MerakSpielman]

Or, like they did in the old days, use chits!

 

[cordell]

The bottom line is, to avoid a lot of re-rolls when trying to generate a non-standard number range, you have to do Math. The more Math you do, the fewer number of re-rolls there will be.

If you need this for a one-time only situation, then its best just to roll d20 and re-roll 14 or higher. Defining the process will take longer than actually rolling a d20 six times.

If you need to do this d13 on a regular basis, then define a process and write it down. Mine would be:

Roll d100. reroll 99 and 00. Otherwise, divide by 6. Raise any fractions/decimals to the next highest integer. (This assumes you can divide by 6 in your head).

01 thru 06 = 1
07 thru 14 = 2
etc, etc.

Why you might have a need for a d13 on a regular basis is beyond me. Maybe your regular gaming group includes 13 characters and you need to choose randomly quite often...

(Now, if you want to really open a can of worms, try to convince a gamer that human beings cannot tell the difference between random/nonrandom number streams in a six-sided die rolled six times...)

 

[The Souljourner]

I definitely like the deck of cards idea. Brilliant!

-The Souljourner

 

[orsal]

Roll d100. reroll 99 and 00. Otherwise, divide by 6. Raise any fractions/decimals to the next highest integer. (This assumes you can divide by 6 in your head).

Don't you mean to make that d80, rerolling 79 and 80? 13x6=78.

This amounts to the same as my d40 suggestion, except that your way you have to deal with numbers twice as large.

The deck of cards works great if you only need to generate one number. It gets either more confusing or more time-consuming if you need, say, 3d13, since you'd have to either reshuffle with every draw or ignore a bunch of cards. If the first card drawn was, say, the 5 of spades, then for the second draw you'd have to ignore all spades, so that there are the same number of each rank. Unless you want to reshuffle.

 

[Altamont Ravenard]

(Now, if you want to really open a can of worms, try to convince a gamer that human beings cannot tell the difference between random/nonrandom number streams in a six-sided die rolled six times...)

MORBO DEMANDS THAT YOU ELABORATE!