# In common practice a d100 is effectively a 0-99 roll, with the stipulation that 0 be treated as 100

The game needs a way to roll 1 to 100 with equal chances, and no chance of getting zero.

Let's start by just looking at how we are set up to roll the results from 1 to 99, and then we'll get to the special case of getting 100.

In order to have a practical way to get the numbers 1-99, we roll a double-digit and a single-digit d10 together, and add them arithmetically. We add the two numbers we literally see, so a "90"and a "3" is 93; a "00" and a "3" is just 3; a "90" and a "0" is just 90, etc.

Two problems initially: we have no way to get 100, and we have the possibility of getting a total zero by rolling "00" and "0".

Solution to both: count the "00" and "0" combo as 100, instead of zero. With this simple adjustment, we have exactly what we want: a 1-in-100 chance of getting all possible results from 1 to 100, and never getting a final zero.

### Admittedly, it is a matter of convention

Understandably, you normally treat a "0" on a regular d10 as "10", and you could continue to do so even when using it in d100, such that if you rolled "80" and "0" and treat the "0" as 10, then you get 90. But game communities tend to converge on one common practice or another, and the one outlined above is the one that has taken hold. But technically, you could use either method, as long as everyone at the table understands and agrees.

### A final historical note

In the *old* days, few players had two d10s to enable rolling a d100 all in one go (maybe because dice sets were more expensive relative to income back in the day). We'd roll our single d10 for the tens-place, and then pick it up and roll it again for the ones-place. This made the zero-substitution rule very exciting and suspenseful! If my initial one-die roll was zero I'd be thinking, "Dang, I probably will end up with just a 1-9, but I've got a shot at a 100!". And everyone around the table would be thinking the same thing, and would watch with tense anticipation what the second roll was going to be.

This dynamic added some fun to the game (that you don't "feel" when rolling two d10s), and might help explain why the method "stuck."

## Emulating a d6 using a fixed number of d20 rolls is impossible.

Of course, as the other answer points out, it is possible to do so if we take into account more than just the rolled result, such as its orientation, but let's ignore that for a moment.

You rightly say that we can emulate a d6 with a d20 by mapping 1-18 to the d6 and rerolling on 19 or 20. This works, but this could in theory go on forever if we keep rolling above 18.

So, why is it impossible? As you said it has something to do with factorization. A d6 is impossible to create with a d20 not because it is the product of two primes, but rather because it has a prime factor which is not present in the prime factorization of 20. Firstly, note that the reason that we can't emulate a d6 with *a single* d20 is because 20 is not divisible by 6. Now, the prime factorization of 6 is \$2 \times 3\$, and that of 20 is \$2^2 \times 5\$. If we roll \$n\$ d20s, this gives a total of \$20^n\$ possibilities. Because multiplication is the same as adding the prime factors, \$20^n\$ will have \$n\$ times the factors:

$$ 20^n = 2^{2n} \times 5^n $$

That is, regardless of the amount of times we roll a d20, the total amount of possible results will never have a prime factor other than 2 or 5, and will therefore never be divisible by 3. Because every outcome is equally likely, there is therefore no way of distributing these outcomes over the faces of a d6 such that all 6 possibilities are equally likely.

More generally, a dX can emulate a dY if and only if X has all prime factors that Y has. Therefore, a d20 cannot emulate a d3, d6, d12 or d35, but it can emulate a d8, d10 or d4294967296.

## Example of emulating a d8 with a d20

If we wish to emulate a d8 with a d20, we first note that 20 is not divisible by 8. This is because 8 has three 2s in its prime factorization, and 20 has only two. However, \$20^2\$ has factorization \$2^4 \times 5^2\$, which encapsulates the prime factorization of 8, and therefore we need only two d20 rolls.

Now, all we needed from one of those rolls was a single factor 2, so we can use one of them as a coin flip. Therefore, we can roll the first d20 and if it lands on the top half (11-20), we take something from the top half of the d8 (5-8), and similarly for the bottom half (1-10) -> (1-4). Then we only have to divide the other d20 in four categories: (1-5) -> (1 or 5); (6-10) -> (2 or 6) and so forth.

Note that this is only an example, and there are plenty of ways to emulate a d8 with two d20s. All you have to do is ensure that out of the 400 possibilities with two sequential d20s, 50 of those possibilities are assigned to each of the 8 target numbers, which is equivalent to cutting the space equally in two a total of three times.

## Best Answer

The usual way of doing a conversion between systems is to not try to reproduce the mechanics, but to look at what the mechanics are trying to accomplish or model and then figure out the "native" way in the new system for how that would be done.

Since you're using D6 Fantasy, you're presumably using D6 Fantasy's combat system instead of trying to emulate d20's BAB system and armour classes – in the same way, make up a rule that "feels D6-ish" to handle this adventure mechanic. That's all the original authors did, after all.

But even given that, it's not hard to convert the mechanic fairly faithfully because you're

nottrying to convert a single roll, you're trying to convert a series of rolls and the probabilities involved there make the conversion easy.Consider that an accumulated total is actually going to be a bit of a bell curve regardless of how you roll the individual checks, because the average of a d20 roll is 10.5 and after a few rolls in the series, the total will be likely close to 10.5 times the number of rolls, because the addition of multiple rolls smooths out probabilities. Because of that, you can use 3d6 to generate 3–18 (average 10.5) and your accumulating average will end up being very similar after enough summed rolls.