Using dice to pick 10 numbers randomly

Factual Questions
1

I need to pick a number between 1-10 randomly. I was using A-10 playing cards but they are difficult to put together quickly and efficiently. So, I bought a pair of dice (let’s assume they’re balanced,) and figured I would just roll them separately and add them together, and use 11-12 for 1. However, after some thought, I realized that rolling them one at a time would end up with the same skew towards 6-7-8 as rolling them together.

So, how can I use dice to get an even selection of numbers 1-10? I have two dice if that matters.

2

I’m sure some mathemagicians will show up with the most efficient way, but in the meantime:

A. Roll one die. If it is 1-3 go to B. If it is 4-6 go to C.
B. Roll one die. If it is 1-5, use that number. If it is 6, re-roll
C. Roll one die. If it is 1-5, use that number, plus 5. If it is 6, re-roll.

If your two dice are different colours or whatever you can roll them together to save time (with a nominated role for the two colours).

3

Here’s another way.

Step 1) Roll both dice.

2 or 7 – Call it 2
3 or 8 – Call it 3
4 or 9 – Call it 4
5 or 10 – call it 5

11 or 6 – Call it 1
12 – roll over

Step 2) Roll a single die.
If it is odd, add 5 to the number obtained in Step 1.

4

I am no mathematician but surely, using 11 as 1, and ignoring 12 would do it?

5

Nope, check out any table on craps. Iirc, 7 will come up about half the time, 6 and 8 are next most common, then 5 and 9, etc. 2 and 12 are the least common. Rolling one die will give you a perfect distribution of 1-6, but not two.

6

I’m going to assume this is not a hypothetical math question. The most practical answer I can think of is:

Any decent-sized town will have at least one nerd store with games and comic books. You can find them there.

And if you want to automate the process, try javascript

Math.floor((Math.random() * 10) + 1

7

I tried the nerd store for a 4 sided die, they didn’t have that. I don’t recall if they had a 10 sided, but from what I remember they weren’t replacements for D&D sets, but weird novelty dice.

8

Since there are 36 different outcomes on two dice (assume they are different colours and tabulate all possible results if you doubt this) there is no way to map these easily onto 10 equally likely results. You might as well use the method Mijin suggests, most easily done with two dice that can be distinguished - roll both dice, reroll if the “coloured die” is a 6, otherwise your result is the value of the “coloured die”, +5 if and only if the “white die” is an even number.

9

Find a nerdier store, then. :smiley:

10

This is, in fact, almost. the most efficient way, in the sense that no method is more efficient.

As part of the proof of that, note that there are 36 possible results from tossing two dice. To give 1 through 10 equal chances, you need to assign 3 results to each number in the range, giving 30 results in total, so that 6 results are (in a sense) wasted, if you can only use the first two throws.

To make it slightly more efficient, if the second throw is 6, I’d reuse the first throw to supply more information:
(1) if the first throw is 1 through 5 and the second throw is 6, throw a dice a third time. If the third throw is 1, 2 or 3, use the value of the first throw. If the third throw is 4, 5 or 6, add 5 to the value of the first throw.
(2) if the first and second throws are both 6, start again from the beginning.

11

Well, one third of half the time, or one sixth, on average.

2 & 12 = 1/36
3 & 11 = 2/36 or 1/18
4 & 10 = 3/36
5 & 9 = 4/36 or 1/9
6 & 8 = 5/36
7 = 6/36 or 1/6
Craps tables of course represent this coarsely, with an emphasis on favoring the house.

I find it hard that a d10 cannot be found. Older D&D didn’t use it much in favor of d20, but other systems do. I had a basic D&D (e.g. not AD&D) set as a kid and it had at least d4, d6, d8, d12, and d20, although don’t remember d10.

12

Forget 10-sided die:

You technically only get a certain amount of random numbers per day, but it’s only really limiting if you’re writing hardcore calculations that require random noise for some reason. For just picking a few numbers between 1 and 10, I doubt you’ll run out.

13

The full D&D dice set was one of each die from d4 to d20, including a d10. And of course there’s a lot of systems using d10s nowadays since it is an easy way to roll d100, so yeah they should be fairly common.

14

A d20 can easily be used as a d10 if you want to stick to Platonic solids.

Brian

15

Speaking of which, never actually use a d100, damn things are useless for anything but novelty.

16

The simplest way (I :smiley: can think off )

Don’t add the dice numbers,
just read them.

So 1, 1 means 11
… 1, 2 means 12
… 1, 3 means 13
and so on.

You will get all numbers from 11 to 66.
That must be 56 numbers.
All of them have the same chances of coming up.

Now, you can either ignore 46 of them, and just roll the dice again,
(that’s a lot of useless “rolls”)
or “replace” them with 1, 2, etc.

56/10=5.6

Every 5 numbers you change “digit”.
What the :dubious:… do I mean?

11,12,13,14,15 mean 1
16,17,18,19,20 mean 2
and so on, until…
56,57,58,59,60 mean 10

what’s left (61 to 66), you ignore them, and role the dice again.

17

Yeah, I call d100s ‘golf balls’ because that’s what they look like and that’s how far they roll.

I find it difficult to believe that a real ‘nerd store’ wouldn’t have a supply of both d10 and d4 dice. Whatever kind of store it was, it wasn’t a game store.

18

This seems to work OK.

If you want an actual die - try here

http://www.thediceshoponline.com/dice-sets/394/10-Sided-D10-Specials

19

There are people that don’t own ten sided dice?

20

I played mostly AD&D, but I’m surprised there wasn’t a call for percentile dice in Basic, that seemed to be a big schtick for Gary. Halberds and two handeders also did d10 in AD&D - not sure what they did in basic (d10 also popped up in some spell rolls). One of the biggest uses was it was the per level HP roll for fighters and some of their variants.