Rolling a d20 with six-sided dice

[waysoftheearth]

We all love our polyhedral dice -- most of us have huge piles of them to prove it. Some of us love our chit sheets too. But let's face it, the six shooter will always, always be King!

Now it's pretty obvious that you can get a roll of d2 or d3 out of a six sider, and with a little imagination you can figure out how to make a roll of d36 or d66 with a pair of them.

But did you know you can also roll a fair d4, d8, d10, d12 and even a d20 with nothing more than your good ol' chance cube? As an added bonus you can roll a fair d9 and d18 too!


Genuine time waster looming...


This is all academic if you're the kind of guy who carries a d20 on your person at all times "just in case". But if you're not, and you find yourself in a house full of d6s with an urgent need to roll a d20 then this might just save you



Rolling a d4 with Six-Sided Dice

Option One:
Roll one regular six-sided die. Reroll any result of 5 or 6.

Option Two:
Take two six-sided dice.
Mark the faces of the first die {0, 0, 0, 2, 2, 2}.
Mark the faces of the second die {1, 1, 1, 2, 2, 2}.
Roll and sum.



Rolling a d5 with Six-Sided Dice

Roll one regular six-sided die. Reroll any result of 6.



Rolling a d8 with Six-Sided Dice

Take three six-sided dice.
Mark the faces of the first die {0, 0, 0, 2, 2, 2}.
Mark the faces of the second die {1, 1, 1, 2, 2, 2}.
Mark the faces of the third die {0, 0, 0, 4, 4, 4}.
Roll and sum.



Rolling a d9 with Six-Sided Dice

Take two six-sided dice.
Mark the faces of one die {0, 0, 3, 3, 6, 6}.
Mark the faces of the second die {1, 1, 2, 2, 3, 3}.
Roll and sum.



Rolling a d10 with Six-Sided Dice

Take two six-sided dice.
Mark the faces of one die {0, 0, 0, 5, 5, 5}.
Leave the second die as is (1 to 6).
Roll and sum, re-rolling any individual die result of 6.



Rolling a d12 with Six-Sided Dice

Take two six-sided dice.
Mark the faces of one die {0, 0, 0, 6, 6, 6}.
Leave the second die as is (1 to 6).
Roll and sum.



Rolling a d18 with Six-Sided Dice

Take two six-sided dice.
Mark the faces of one die {0, 0, 6, 6, 12, 12}.
Leave the second die as is (1 to 6).
Roll and sum.


And the "big one"...

Rolling a d20 with Six-Sided Dice

Take three six-sided dice.
Mark the faces of the first die {0, 0, 0, 10, 10, 10}.
Mark the faces of the second die {0, 0, 0, 5, 5, 5}.
Leave the third die as is (1 to 6).
Roll and sum, re-rolling any individual die result of 6.



Rolling a d36 with Six-Sided Dice

Take two six-sided dice.
Mark the faces of one die {0, 6, 12, 18, 24, 30}.
Leave the second die as is (1 to 6).
Roll and sum.


So long as there are d6s you'll never lack for your freaky gamer dice

[gloriousbattle]

Rolling a D20 with a phonebook and a calculator (or just your head, if you love math):

1. Open phone book.

2. Put finger down on random phone number.

3. Divide random phone number by 20, stop when you get a remainder less than 20, or if the number divides evenly.

4. The remainder will always be a number between 1 and 19. This is your d20 roll. If there is no remainder, your roll is 20.

Note that this can be done to simulate any type of die roll.

[talysman]

You can simplify the calculations for the phonebook method: just use the last two digits of the phone number. This gives you a number from 00 to 99 (d100,) which you can divide by 20 and find the remainder. Easier to do in your head than a 7-digit number (with the first three digits not being evenly distributed...)

My own suggestion is the crazy-looking Perverse Polymorpher dice map I whipped up. There's a couple ways to get a d20 out of that, but probably the easiest is to drop a penny and a nickel on the printed version of that dice map and read the location of the penny as a d3 (using the innermost d3 ring) and use it as an index number, combined with the nickel's location in the d5 ring, to get the d20. I still have to do the full document explaining all the tricks.

[isomage]

d20 with two six-sided dice: using one die as a d5 (re-roll sixes) and the other as a d4 (re-roll fives and sixes), roll (d5-1)*4 + d4.

d20 with a nickel, a dime, and a d6: heads on the nickel is worth 5; heads on the dime is worth 10; re-roll sixes on the die and add (this is equivalent to the method of the first post, using coins instead of marked dice)

[thorswulf]

Why n ot just use a bell curve for D18?

[isomage]

Aug 16, 2011 17:12:21 GMT -6 thorswulf said:

Why not just use a bell curve for D18?

A die is usually assumed to have a uniform distribution, so a d18 would be expected to have all outcomes equally probable. Also, 3d6 has sixteen outcomes, not eighteen (lacking 1 and 2).

[waysoftheearth]

Aug 16, 2011 16:08:37 GMT -6 isomage said:

d20 with two six-sided dice: using one die as a d5 (re-roll sixes) and the other as a d4 (re-roll fives and sixes), roll (d5-1)*4 + d4.

Yes, I thought of that too; but trying it out it requires too many re-rolls. 16 chances in 36 of requiring at least one re-roll. And quite frequently requiring several re-rolls to get the one result.

[waysoftheearth]

Aug 16, 2011 17:12:21 GMT -6 thorswulf said:

Why n ot just use a bell curve for D18?

True, you could. But they are two different things. 3d6 results in a normal distribution of results (the bell curve), while 1d18 results in a flat distribution of results -- an equal probability of any one result.

In any case, it's all just for fun... to see what you could do if you cared. I'm not saying anyone would actually bother

[kent]

Neat, I like it. Could do with some blank d6 in fact.

[talysman]

It looks like the technique in the OP would require 10 blank dice, each specially numbered. I tried to find some dice that could serve double-duty, but I don't think labeling one die {0, 0, 0, 5/6, 5/6, 5/6} is such a great idea.

That could be a problem. One of the good features of using the standard polyhedrals is that they're easy to find -- and I don't mean in the stores. They're easy to find in a pile of dice on your table, just by shape. The specially-numbered d6s, in contrast, will require looking at several faces of the die to make sure you have the right one.

[kesher]

Aug 16, 2011 7:35:32 GMT -6 gloriousbattle said:

Rolling a D20 with a phonebook and a calculator (or just your head, if you love math):

1. Open phone book.

2. Put finger down on random phone number.

3. Divide random phone number by 20, stop when you get a remainder less than 20, or if the number divides evenly.

4. The remainder will always be a number between 1 and 19. This is your d20 roll. If there is no remainder, your roll is 20.

Note that this can be done to simulate any type of die roll.

What's a phone book?

Also, you're all crazy, but in a good, Gygaxian way. All I need now is for someone to tell me how to roll a d20 using a d6 and a polearm, and I can die happy...

[waysoftheearth]

Aug 16, 2011 19:11:28 GMT -6 talysman said:

The specially-numbered d6s, in contrast, will require looking at several faces of the die to make sure you have the right one.

This is true.

I was thinking of either colour or size coding them into groups (i.e., the three dice you need to roll a d20 are either all red, or all the same size -- different to all the others).

Another option I was thinking of was that each face could have a big number on it (the result of the roll), and also a tiny little "d8" (or whatever) in the lower left of each face.

I might put a few together and see what is easier ;D

[talysman]

Aug 16, 2011 21:13:12 GMT -6 waysoftheearth said:

Aug 16, 2011 19:11:28 GMT -6 talysman said:

The specially-numbered d6s, in contrast, will require looking at several faces of the die to make sure you have the right one.

This is true.

I was thinking of either colour or size coding them into groups (i.e., the three dice you need to roll a d20 are either all red, or all the same size -- different to all the others).

Another option I was thinking of was that each face could have a big number on it (the result of the roll), and also a tiny little "d8" (or whatever) in the lower left of each face.

I might put a few together and see what is easier ;D

Hmm.

You could use a square around the numeral for both dice used in the d4; one would be a blue outline around a blue number, the other would be a solid blue square with the numeral as the same color as the die (I picked "blue" as a mnemonic for "two", the high number on both of those dice.)

The {0, 0, 0, 4, 4, 4} die used for the d8 could be a solid red "stop sign" (octagon) with the numeral in the inverse color.

The d3 would be a green triangle outline; the other die used for the d9 could be a solid green triangle.

The {0, 0, 0, 5, 5, 5} die used for d10 rolls would be a black pentagon. The {0, 0, 0, 6, 6, 6} would be a black hexagon.

It may seem weird, but I'd put the numerals for the {0, 0, 6, 6, 12, 12} die inside a green hexagram, to indicate that it is based on the d3, but with multiples of 6.

That leaves the {0, 0, 0, 10, 10, 10} and the {0, 6, 12, 18, 24, 30}, which I'm unsure of. Maybe a blue pentagram for the first one.

[waysoftheearth]

I reckon a nice option would be to have a shape on each face that looks like the polyhedron the die is simulating...

How about a set of these for Christmas?

d4


d6


d8


d10


d12


d20



Now I just need to find someone to print them for me ;D

[Finarvyn]

Aug 16, 2011 7:25:24 GMT -6 waysoftheearth said:

And the "big one"...

Rolling a d20 with Six-Sided Dice

Take three six-sided dice.
Mark the faces of the first die {0, 0, 0, 10, 10, 10}.
Mark the faces of the second die {0, 0, 0, 5, 5, 5}.
Leave the third die as is (1 to 6).
Roll and sum, re-rolling any individual die result of 6.

Very cool. I'm pretty sure I could have solved this puzzle myself, but it's neat to see how you did it.

I love "time wasters" like this!