Poll of the Day > d20's are the d6's of the tabletop gaming world

(+) Yeah! (+)

As mentioned before, rolling 4d6 and making 21,22,23 = 1, 2 and 3 respectively and 24 get thrown out, does not create the same chance for each number, you are not simulating a 1d20 because to roll a 4, you need to roll 4 1's, which will only happen 1 way, so out of the possible combinations of rolls (which is 6^4), you have only 1 combination that is equal to rolling a 4 on a 20 sided dice.

My method has exactly a 5% chance of rolling a 1, a 5% chance of rolling a 10, and a 5% chance of rolling a 20.

Meanwhile with 4 dice, you can roll a 10 MANY different ways (even if you don't consider each dice different, and just count the number combinations).

So basically using that method, you end up with a WAY larger chance to roll numbers in the middle of the range, then numbers in the extreme ranges.

My method is fast, easy, can be done for any dice range pretty fast in your head, and gives you the exact same chance for any number as rolling a dice with that number of sides would do. It's easy enough that anyone with basic math skills can use it. Even if you could only count, someone could teach you how to use it as well. I could teach 1st graders to use it easily (considering I taught them how to play 21 (not the card game, the adding game), which is actually more complicated.
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wolfy42 posted...

Nope there are not, because you literally are forcing there to be only 20 results in every case, other then when you re-roll 6's (making them not count.....even though it's technically a result).

"Reroll on #" causes a theoretically-infinite number of results, thus such rerolls should be discarded when working out the statistics.

Questionmarktarius posted...

wolfy42 posted...

Nope there are not, because you literally are forcing there to be only 20 results in every case, other then when you re-roll 6's (making them not count.....even though it's technically a result).

"Reroll on #" causes a theoretically-infinite number of results, thus such rerolls should be discarded when working out the statistics.

Not really, there are an infinite number of possible rolls, but only a set amount of results. We are simply trying to get a 5% chance for every results (Between 1 and 20) and my method delivers that perfectly. Yes, because you are rerolling 6's you technically could roll an infinite number of times, but in reality, you will only need to reroll at all 1/6th of the time (about a 16.6% chance) and need to reroll a second time about 1/36 times (less then a 3% chance).

If you simply have 2 dice of different colors, you could roll them both the first time (to determine 1-10 and 1-5) and then roll them a second time (with the colors determining order), if the first color is a 6, you use the second colors number. If both are 6's, you reroll them again etc. You would need to re-roll a second time extremely rarely, and probably wouldn't need to re-roll a 3rd time even once.

Meanwhile every single time you roll, you have the same 5% chance for every possible number 1-20. This simulates a d20 exactly, as far as the chance to roll each number at least. In fact, it's slightly better on average then a d20 because quite often then are not balanced completely etc, which means you are more likely to roll/land on a certain number then others.

The d6 method both removes the weighted dice effect (or reduces it), especially if you use multiple dice.

So you might actually get a more random, truer, 1-20 randomization from the d6 method then an actual 20 sided dice.
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Btw, 1-100 would be pretty easy as well.

First roll odd/even for 1-50 and 51-100

Second roll reroll 6's for a 1d5.....determining the 1-10 range above (1-10, 11-20 etc).

Third is odd even again (1d5 within the 10 range or 6-10)

And last roll is again 1d5 (reroll 6's).

So 4d6 can easily be used to get a accurate 1d100 roll as well. As long as they are different colors you can quickly get the result.

Blue 1-50 or 51-100
Red 1-10, 11-20, 21-30, 31-40,41-50 (or the same for 51-100 based on blue roll)
Green 1-5 (or 6-10)
White reroll 6's for the actual number 1-5 or 6-10.

Do that enough times and again, you'll read it automatically.

You could also just do the roll for 1d10 twice (like rolling two d10s) if you wanted and that is too complicated...it's the same thing, just melded together.
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If you really want to nerd up a d6->d20 mapping:

Roll, multiply by 4, reroll 6
Roll again, subtract from the first result, treating 4 as 0 and rerolling 5 and 6.

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

31 results becomes 20, with 11 discards. Not particularly efficient, but workable, and with 5% probability between each useful value.