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Dicebending: One Weird Trick to Use D6 for Other Dice Results |
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Players of popular mainstream tabletop RPGs are often collectors of miniatures and dice. Some games use only 6-sided dice (or d6 for short), which is also the case for a lot of board games out there. Others use d20s or d10s up to a full set of polyhedral dice, as is used for Dungeons & Dragons. As such, it can be a point of pride to have this set of clear or frosted plastic polyhedral dice or perhaps a Dwarven rune-themed set made of metal. Even so, at times a gamer might find themselves in need of a polyhedral set of dice, but without their gaming supplies. What to do?
The first thought most people would have is to try to use smaller dice that simply add up to the larger, absent die. For example, if you needed a d12, just roll 2d6 and add them together. In fact, however, this won't give you a fair balance of values found on a d12. In fact, if you think about it, there's no way to get a "1" by adding 2d6. The values will be between 2 and 12. Furthermore, the possible outcomes aren't equally weighted, because math. Just to provide a couple of comparisons that demonstrate the unbalanced nature, there is only one possible roll of 2d6 that will result in a 2. Likewise, there is only one roll that will result in a 12. By contrast, six different combinations will result in a 7. So, that's not anywhere close to accurately providing the same random probability as a 1d12. |
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What can be done, instead, is to take the range of values and divide it into smaller, evenly-sized sectors, such that the number of sectors is equal to or less than the range of possibilities on the available randomizer (in this case, the 1d6 in your hand). This sounds complex, I'm sure, but this is essentially similar to the method employed when using 2d10 as percentile dice. The "tens" die is basically chunking the 100-number range into ten evenly-sized ranges of ten values, then the second d10 selects one value in that range. This is easy to do and comes naturally for people who use a base 10 number system, since we basically treat numbers that way all the time, even off of the dice.
With my suggested method, the above percentile dice using 2d10, you would not treat "00" as 100, but would just roll the dice, treat them as the displayed digit and add 1. That means the lowest value that would display on the dice would be "00" and would evaluate to a 1 and the highest value displayed on the dice would be a "99" which, after adding 1, would give you 100. |
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If, however, you only had d6 available, you would need to chunk the 100 number range into ranges smaller than 10... something that you can do on a d6. Since you can't divide 100 into 6 even integer ranges, you would need to use something smaller. In this example, I would recommend dividing the 100 number range into five sectors of twenty. You can choose which sector you're in by rolling 1d6 and rerolling sixes since you're shooting for a value between 1 and 5. Once you have randomly selected which of the 5 available 20-number sectors you're going to be in, you can determine what value this sector contributes to your final result. The first twenty values will be at most twenty, and we will be simulating a d20 next to determine that value, so being in the first sector contributes nothing to the total. The highest possible value on a d100 (that we're simulating, here) would be 100, which would be the 20 that is the highest possible on our next simulated roll (of a d20), plus 80. So, 80 is the value contributed by being in the fifth sector. The basic formula we get from that is that you subtract one from the sector you roll and multiply that value times the number range of each sector. So, in this case, 1d5*-1 x 20.
The next step is to simulate rolling a d20 with a d6 to complete the d100 simulation. Again, as in the d100, a d20 can't be evenly divided by 6, but it can be divided evenly by 5, so we roll 1d6, rerolling any sixes. Then, we get the contributed value from this roll by using the same formula, but modified to suit the subset range for the d20 simulated roll. If we're dividing the 20-number range into five evenly-sized ranges, each sector has four possibilities. That makes our formula: 1d5*-1 x 4. Add this value to our amount.
Finally, we now just need to simulate 1d4 and add the result to our total to get our final value. Simulating a d4 is just as easy as simulating a d5 on a d6 (or any die with more sides, for that matter) ...simply roll the d6 and reroll on invalid values. So, to simulate 1d4, you can roll 1d6, rerolling on any 5s or 6s. |
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So, if you want to use this trick to obtain fair and balance random rolls of dice other than a d6, using a d6, you can understand the above, or just use the formulas below:
Anything LOWER than 6:
1d6, reroll if the value is larger than the highest allowed value.
Eight-Sided Die (1d8):
(1d4-1 x 2) + (1d2)
Ten-Sided Die (1d10):
(1d5-1 x 2) + (1d2)
Twelve-Sided Die (1d12):
(1d6-1 x 2) + (1d2)
Twenty-Sided Die (1d20):
(1d5-1 x 4) + (1d4)
One Hundred-Sided Die (1d100):
(1d5-1 x 20) + (1d5-1 x 4) + (1d4)
I hope this comes in handy. You could, perhaps, keep this written on a card in your wallet for that time you end up at a rental vacation cabin with games, but not all the needed dice, for example.
Anyway, happy gaming! |
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-Geck0, GameVortex Communications AKA Robert Perkins |
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