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[Christes]

Yep.  Just finished my first year.

[Vector]

Yep.  Just finished my first year.

Another lucky guess for the Vector-bot!  Which area?

[Christes]

I assume you mean of math...

I haven't decided yet.  I'm liking Algebraic Topology, though.

[Vector]

I assume you mean of math...

I haven't decided yet.  I'm liking Algebraic Topology, though.

Yes.  I'm not an internet-stalker unless you divulge too much personal information about yourself, in which case I'll inform you that you might want to make a few choice edits.

Erm, anyway.  I have almost no experience with that, but Albert's Theorem is pretty nice ._.  I'll, uh, say hi again in a couple years or something.

[Muz]

Um, IMO, that solution wasn't all that elegant. Not from a practical perspective... converting your dice into binary works, but it's too time consuming to be a good solution. I can't see a less complex solution out of it, though. Other than to just use digital dice.

[Vector]

Um, IMO, that solution wasn't all that elegant. Not from a practical perspective... converting your dice into binary works, but it's too time consuming to be a good solution. I can't see a less complex solution out of it, though. Other than to just use digital dice.

Practicality?  Why cares about practicality, when you have binary?

[Muz]

I don't know if you've ever rolled a 10d6, but rolling a 10d6 with 4-sided dice would slow the game to a crawl every time someone casts a fireball

[Neonivek]

Hint: If I roll two d6, the probablity of getting a 12 would be the lowest, but getting a score of 6 would be highest

Actually that is a poor hint if you want to do it properly. Scaling down is easier then scaling up

You see a d20 has only 20 possibilities and a d6 has 6

Two dice have 21 possibilities and so on. (1/1, 1/2, 1/3, 1/4, 1/5, 1/6, 2/2, 2/3, 2/4, 2/5, 2/6, 3/)

Eliminate 1/1 or 6/6 and then create your rolls based on the results

Here we will eliminate 6/6... That will be a reroll.

1/1 = 1
1/2 = 2
1/3 = 3
1/4 = 4
1/5 = 5
1/6 = 6
2/2 = 7
2/3 = 8
2/4 = 9
2/5 = 10
2/6 = 11
3/3 = 12
3/4 = 13
3/5 = 14
3/6 = 15
4/4 = 16
4/5 = 17
4/6 = 18
5/5 = 19
5/6 = 20
6/6 = Reroll

and so on. The order of the dice don't matter

I am not sure this works, something about my knowledge or probability says to me that all numbers in my example are not equal.

Now there is a type of math that does this easily but I forgot what it is called.

"How would you approach the problem of generating a range of any numbers from any die?"

Well with a dice that is lower then the dice you want the result with you want to make a combination of sets that can equal the sets of the larger dice.

The problem can come when you add more then two dice. I believe if I used three 6s... a 1/1/1 has less of a chance then 1/1/2. (Which is why I dislike probability).

[ed boy]

That wouldn't work properly. For example, there are two ways to get 2/3 (dice A is 2 and dice B is 3 or dice B is 2 and dice A is 3), but only one way to get 1/1.

You would need to tell the dice apart (by marking them or by rolling them one at a time) or have a random way of selecting the dice once they are rolled. This would give you 36 possibilities.

[Neonivek]

That wouldn't work properly. For example, there are two ways to get 2/3 (dice A is 2 and dice B is 3 or dice B is 2 and dice A is 3), but only one way to get 1/1.

You would need to tell the dice apart (by marking them or by rolling them one at a time) or have a random way of selecting the dice once they are rolled. This would give you 36 possibilities.

Yep because dice sets, now that I remember, have 2/3 and 3/2. Thanks.

HOWEVER! I am not beaten yet

1/1, or 6/6 = 1
1/2 = 2
1/3 = 3
1/4 = 4
1/5 = 5
1/6 = 6
2/2 = 7
2/3 = 8
2/4 = 9
2/5 = 10
2/6 = 11
3/3 = 12
3/4 = 13
3/5 = 14
3/6 = 15
4/4 = 16
4/5 = 17
4/6 = 18
5/5 = 19
5/6 = 20

Dang that doesn't work either because of 2/2, 3/3, 4/4, or 5/5. They just do split up or eliminate into the number. However between 1/1, 2/2, 3/3, 4/4, 5/5, 6/6 we get 6 or the equivilant of 3 different numbers.

Ignoring them and taking all the sets from 1-15.

Any combination of double numbers gets a special rule.

1/2 = 1
1/3 = 2
1/4 = 3
1/5 = 4
1/6 = 5
2/3 = 6
2/4 = 7
2/5 = 8
2/6 = 9
3/4 = 10
3/5 = 11
3/6 = 12
4/5 = 13
4/6 = 14
5/6 = 15

From here we get 5 numbers missing

Any combination that are doubles gets you ONE more roll of dice. a 6 is a reroll
1= 16
2= 17
3= 18
4= 19
5= 20

There we go!

no wait I don't I am braindead.

You could use the 36 combinations but I wanted to see if I could do it with only 2 rolls while ignoring the order. Apperantly not...

Either way we got the way to do it with d6s to get a d20. Not too tough if you remember that you can eliminate combinations.

Alright so here we go!

Roll one die! Any roll above 4 and you reroll! You also ignore any doubles

1/2 = 1
1/3 = 2
1/4 = 3
1/5 = 4
1/6 = 5
2/1 = 6
2/3 = 7
2/4 = 8
2/5 = 9
2/6 = 10
3/1 = 11
3/2 = 12
3/4 = 13
3/5 = 14
3/6 = 15
4/1 = 16
4/2 = 17
4/3 = 18
4/5 = 19
4/6 = 20

There we go!

So what combination should I do next?

[qwertyuiopas]

With the roll order considered, for a dice of X sides to get a number from 1 to N:
Find Y in X^Y-1 < N ≤ X^Y.
Roll Y dice, noting the order.
Use the rolls, in the rolled order, as the digits in a base X number.
If that number exceeds X^Y - (X^Y mod N), repeat with a new set of rolls.
The result is the number mod N.

X, Y, and N should all be positive integers, and X and N should be at least two. It may work with non-integers and greater ranges, but I am not experienced enough to know at a glance nor determined enough to think about it.

Conversions between bases is reccomended wherever deemed easiest.


Without roll order, it becomes difficult if not impossible to achieve a proper unbiased distribution.

Finally, the easiest way: If N is a factor of X, simply roll a die, mod N.


Or, the internet user's way: Ignore the dice, go to random.org.

[Virex]

If your die has N sides and you want to generate a result R in a continuous range of 0 to M, only caring about integers, a simple way would be:


Convert M to base N, using whichever method is appropriate. Call this O
Roll for the most significant digit of R (the one with the highest value) and subtract 1. Assume the value of all other digits is 0. If you exceed the value of O, re-roll this die.
Repeat that step for every other digit of R.
When finished, convert R from base N to the required base (probably base 10).


To make the range go from 1 to M, do not subtract 1 from the final digit. To generate over a discontinuous range roll over a continuous range, then add to the result to account for the discontinuous. For example if you're rolling in range 1-4, 6-10, you would roll in range 5-9 and add 1 to all values exceeding 4. This can be used to roll over a range starting at an arbitrary number as well.


Note that this method is similar to other methods as well. The trick however is that by using base N and rolling once for every digit, you are guaranteed there is no interference between the different die rolls and thus you get an even distribution. You also need less re-rolls because you re-roll on a per-die base, but this does slightly alter the chances at the far edge of the range. To avoid this, use Qwerty's method and roll R at once, rerolling whenever it exceeds O. This is slower but better if accuracy is needed at the edges of the range.

[Puck]

That might not be all that relevant... but I'm a little bit sick. Nothing major. Not even distracting.

I read the thread and didnt understand ANYTHING. But my IQ scores as well as math grades say it should be otherwise.

Soooo. Suddenly I had to vomit. So I went and worshipped the porcelain altar. Or I thoroughly reconsidered my breakfast, whichever you prefer.

Came back, read the thread again, et voila, I GET IT. I iz happy Puck. Not that I have anything meaningful to add, but there you have it.

[FreakyCheeseMan]

I have a semi-general solution that's less computationally intensive, though requires more die rolls.

So, you can always "Cast" a die to a lower number, by saying any higher number is a reroll- so if you roll a 1-6 on a d20, you take the result, and if you roll anything higher, you roll again until you get 1-6. If this takes too long for you, you can mix and match by going with multiples- reroll on a 19 or 20, and assign 3 numbers to each, 1-6.

You can use this to make lower dice function as perfect divisors of higher dice. Example- trying to make a d20 out of d6s. Roll a d6 cast to a d4- so reroll on a 5 or 6. let this determine the range of answers- 1-5, 6-10, 11-15,  16-20. Then roll again, this time casting the d6 to a d5, to narrow it down.

The advantage to this system is that it requires no complicated arithmetic, besides factoring the higher die.

[Vertigon]

The one thing that bothers me here (sensitiveedit: this thread, not anyone's post) :

Singular: die

Plural: dice

LERN IT.