"Best way to get a D16 from standard polyhedrals?" Topic

I know of a game that uses a D16 for some rolls. AFAIK, D16s are not available anywhere in the UK, so what is the best way to proxy a D16 using the standard polyhedral dice?

roll a d6 first then a d8, 1-3 on the d6 add nothing to a d8, 4-6 on the d6 add 8 to the d8

instant d16

regards

jim

Use a D20, and ignore or re-roll 17-20.

The other standard method is to roll a d20 and reroll 17-20.

Some people think one method doesn't "feel right," others vice versa.

Or you could just buy a d16:

link

Of course, if simplicity is not desired roll two d4s of different colors. Take the roll on one die (which must be determined before hand), subtract one, multiply by four, add the other die, you get an evenly distributed 1-16.

Andrew

Chessex makes them still…
link

Or…. I will risk a DH-ing here… I agree with what John says.

Sheez… I guess Andrew and I share the like idea… Just seconds apart…

:) Roll a d8. Add 8 if it points towards your side of the table. haha

I prefer the ignore and re-roll technique that the OFM stated. It works for everything (well, single die rolls) and is real easy to do in a pinch.

If you have to use it alot, then you might want to hunt for a d32 and subtract 16 from results larger than 16.

link

Or ignore games that have silly dice requirements.

Or ignore games that have silly dice requirements.

Why is it silly?
No more silly than any other polyhedral dice type.

Anybody that designs a game that requires more than the usual compliment of d4-d12 and d20 is being ridiculous.

Some folks would go so far as to say that anything other than d6 is silly. That's fine, I understand the utility of d6s, ease of access, etc.

The other main dice types are easily accessed as well.

The very point that a potential player has to go out of his way to track down a d16 proves the point.

Also, Chessex offers the d16 at three times the price of other, standard dice, illustrating that it's a low-demand die.

Roll 3d6, subtract 2. Same price, not perfectly distributed numbers, but hey, it's just a game.

Anybody that designs a game that requires more than the usual complement of d4-d12 and d20 is being ridiculous.

As I understand it, it was to do with the mathematical basis of the combat maechanics and the size of the units, so not silly at all.

Roll 3d6, subtract 2. Same price, not perfectly distributed numbers, but hey, it's just a game.

Exactly the point – rolling 3D6 gives a "bell curve" probability of results rather than the "straight line " probability of a single die. That doesn't work mathematically.

For D16 I just roll a D6 and a D8, add 8 if the D6 scores 4+. Easy, peasy.

Call or text people on your contact list and ask them to pick a number from 1 to 16.

Call or text complete strangers to eliminate any possible bias.

iPhone app….. no problem!

-W

Worst case scenario, if a computer is handy, use an Excel spreadsheet with the formula*:

=RANDBETWEEN(1,16)

Press F9 to "roll" again.

*Requires the Analysis ToolPak add-in for '03, should work fine with newer versions.

[warning, there is some silliness below]

You could do it with a d8+d10 – the D10 is your result with a 0 being a 10 unless the d8 shows up with a 6, 7, or 8, in which case you add 10 to the d10 result, unless it's a 7, 8, 9, or 0, in which case you roll again.

Answers involving Excel and iPhones are invalid since the OP specifically wanted to use other polyhedrals to make the roll. Ignoring games with silly dice is also a prohibited solution.

You could also a flip together a quarter, dime, nickel, and penny. Any tails counts as a zero, but a heads on the quarter adds 8, the dime adds 4, the nickel adds 2, and heads on the penny counts 1. This will give you an even distribution across

3d6-2 is INSANE as this gives you a NORMAL distribution instead of an EVEN distribution. Typing "it's just a game" would be a DH offense in a perfect world.

If you had nothing but d6s life is still easy. Roll a d6, if it's 1-3 you'll start with 0, if it's 4-6 you'll start with 1. Now roll the d6 again and if it's even add 8 to your total. Now roll the d6 and if it's odd add 2 to your total. Roll a fourth d6 (or the same one a fourth time) and add 2 to your total if the result is a prime other than two. Even distribution, there you go.

Wait, I got more. This dude has a marvelous web page that includes patterns for print-and-fold dice, including the requisite d16. It's not all that colorful but you can always edit the PNG and get what you like. So here's a *free* d16.

link

dicecollector.com/D16.PNG

Of course, there's big fat problems with this die. The extra weight of the seams between 12 and 4 and between 1 and 9 mean that 5, 13, 16, and 8 are slightly more likely results. The solution to this is easy, you need four different dice so that each of the sixteen results is more likely on exactly one die. Randomly select a die and roll and over time you'll get an even distribution.

But if you're actually going to play the game I'd spend the two bucks on a nice d16.

Andrew

You could also a flip together a quarter, dime, nickel, and penny. Any tails counts as a zero, but a heads on the quarter adds 8, the dime adds 4, the nickel adds 2, and heads on the penny counts 1.

Not polyhedral, so prohibited. Taking that idea though:

Take four six-sided dice.
Mark one with 3 lots of 1000 and 3 lots of 0000.
Mark one with 3 lots of 100 and 3 lots of 000.
Mark one with 3 lots of 10 and 3 lots of 00.
The last gets 3 lots of 1 and 3 lots of 0.

Roll the dice and add the results. The answer is from 1 to 16 in binary (with 0000 = 10000).

(If you don't know binary, you should!)

Or buy binary dice:
link
colour them so that they go from dark to light, to give the order for your binary answer.

Take four six-sided dice.
Mark one with 3 lots of 1000 and 3 lots of 0000.
Mark one with 3 lots of 100 and 3 lots of 000.
Mark one with 3 lots of 10 and 3 lots of 00.
The last gets 3 lots of 1 and 3 lots of 0.

Roll the dice and add the results. The answer is from 1 to 16 in binary (with 0000 = 10000).

You know, that's a really neat idea – I like it!!

Or – you could just ask Doug @ em4 Miniatures….

link

Always had great service from him.

As I understand it, it was to do with the mathematical basis of the combat maechanics and the size of the units, so not silly at all.

Naaah. What it really means is that the designer is incapable of using common dice, or does not have the imagination to do so.
What if your model "requires" D17? What do you do then?

Or – you could just ask Doug @ em4 Miniatures…

That's useful, didn't know Doug did them. Thanks.

As I understand it, it was to do with the mathematical basis of the combat maechanics and the size of the units, so not silly at all.

Naaah. What it really means is that the designer is incapable of using common dice, or does not have the imagination to do so.

I do not have the time or inclination to explain the maths here. Suffice to say that it is a perfectly logical mechanism. In some ways far more logical than forcing the game mechanisms into using the so called "common" dice just because they are "easier".

What if your model "requires" D17? What do you do then?

That would be an interesting challenge, but not insurmountable as posts here have shown.

"What it really means is that the designer is incapable of using common dice, or does not have the imagination to do so."

If you go for a target sucess number based on troop quality type system and then model different circumstances by increasing base dice size (say, cover, unit posture whatever) a D16 fits quite nicely into the sequence.

This is an elegantly simple solution to what is otherwise a bunch of tedious modifier lookups.

Personally I design games around D6s, but it does box you into a corner sometimes and you end up with boring charts or lists of modifiers.

Roll 4d6, each die of a different colour.

Rank the colours (say) red, green, blue, white, and note whether each coloured die scores odd or even.

If the score is odd, count it as 1, and if the score is even, count is as zero. Then compose these in colour order (red, green, blue, white) into a binary number.

This gives 16 different possibilities, and saves you buting stupid-sided dice.

Alternatively, roll 1d8 and 1d6, and add 8 to the d8 score of the d6 rolls odd, much as Martin suggested.

All the best,

John.

you could use 2d4, read them as base 4, with 4s as 0. So it would be like using 2d10 to get 1-100.

d17 = roll a d19 and re-roll the 18s and 19s.

Andrew

I bought a bunch of them from the UK off of Ebay. The seller TheDiceShopOnline has a bunch listed right now on there.