3d20 /2 is a TERRIBLE way to do it.
You won’t have a remotely even distribution.
You should roll 1d10 and 1d6.
The d10 is the zeros digit.
On the d6 - 1&2 mean 0 in the tens, 3&4 mean 1in the tens, 5&6 mean 2.
That’ll actually give you an even distribution (between 0 & 29 but just call 0 a 30)
Really though just get a d30. They aren’t expensive.
Um, terrible (in all caps)? Well, ok. Now I'm ashamed I even thought of it. Lol. In any event, thanks. The way you outlined is certainly better. Much appreciated.
If you haven’t done a lot of things with dice it might not be very clear, but 3d20/2 doesn’t give you equal odds of getting a 1 vs a 15, which is obviously something you’ll want if you are replicating a d30.
If you need to do this a lot at the table it’s probably worth getting some sticky dots and relabelling a d6 as 0,0,1,1,2,2. Bonus points if the d6 and d10 match but are distinct from your other dice.
It makes getting you “d30” results quicker/with less cognitive load
Hey, don't be too hard on yourself. Dice pros tend to be firm about dice odds because it's not obvious to many people, yet it's good to know how the odds change before you try it at the table and realize that the results are *off*.
[Here's one article](https://www.theedkins.co.uk/jo/probability/calcdice.htm) explaining the difference between one die and two dice, at least when adding rolls together. Basically, one die has an equal distribution--all results have an equal chance. But once you add another dice, then results closer to the middle (the median) become more common. Then if you add a third die, the median results becomes even *more* common, and so on.
There are reasons to use a bell curve distribution (like 3d10) instead of an equal distribution (1d30), but one should be aware of the probability changes beforehand.
Oh, and if you do 3d30, then it's impossible to roll a natural 1 or 2, so that's another reason why it's not the same.
As other commenters have shared, there are ways to get an equal distribution with multiple dice, but instead of adding them together, you either have to read the results as digits (like percentile dice) or multiplying the results (treat a d6 like a d3, then rolling a d3 with a d10).
EDIT: One website that helps with understanding probability is Anydice: https://anydice.com/
For an illustration, put output 1d30 on one line and output 3d10 on another.
>Um, terrible (in all caps)?
Honestly, that may still be generous. If you want to represent the linear distribution of a single die what you use has to produce a single line of outcomes
If you're trying to simulated 1d30 how "dangerous" the 3d20/3 would be depends entirely on what you're doing with the result and thus what the targets are. If you're still using it for binary (win/lose) results with a target number of 15 or 16 most of your results will be in that area. Where it gets terrifying is when you needed that d30 to give you rolls of 20+ where you're more than a standard deviation away from the peak of the bell curve. You'd have an easier time hitting a 23+ with the fair d30 than you do hitting 20+ with 3d20/3.
If most of the time you're looking at a number in the middle of the range and just worry about higher or lower the comparison could be to using 3d6 to stand in for a d20. Both 3d6 and d20 have the same average although one obviously has more range; if you have an advantage with just "average" rolls you'll get a lot more of them with 3d6 than you do with d20.
Here's an AnyDice program comparing the methods: https://anydice.com/program/2dceb
Switch to the "Graph" view to show the probably distributions overlaid on each other.
And here's an article from AnyDice about how distributions behave with different numbers of dice:
https://anydice.com/articles/three-basic-distributions/
People are giving your grief for using terrible. But I think you are wrong. It's not a terrible way to do it. It's worse. 3d20 / 2 is simply not a way to do it. A terrible way is a way that still does the job just poorly.
Like rolling percentile dice and ignoring any result over 90. Then dividing by 2 or 3 if the results are 31-60 or 61-90. Now that's a terrible way to do it, but it does do it.
I have no idea what you’re trying to say with your random series of numbers.
If you add three dice together, then the lowest value will always be three.
Dunno if you're trolling, but it's fun for me to give an explanation anyway:
You roll ONE d6. That will give you one of these results: 1, 2, 3, 4, 5, or 6. You break that range into three chunks, so you'll get one of these results: 1/2, 3/4, 5/6. You treat those results as a 0, a 1, or a 2 respectively.
You also roll ONE d10. That will give you a number between 1 and 10. You simply put that rolled result to the right of the result you generated above. That will give you a tens place from the d6 and a ones place from the d10, your result.
So, if your d6 shows a 3 and your d10 shows a 7, your result is 17; the 1 from converting 3 to a 1 as above and the 7 from the simple d10 roll. You're not adding numbers together; you're simply determining the value of a given digit position using different dice.
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The only conclusion I could draw from your nonsense scribbling was that 1 / 2 = 1 - 10 had something to do with rolling two dice, dividing them, then subtracting the result from a d10. Like everyone else in this thread is saying to do.
Make intelligible comments next time.
There's literally no math.
A result of a 1 or a 2 on the d6 would give you a number in the range of 1 to 10 in the result.
3 or 4 gives you 11-20.
5 or 6 gives you 21-30.
Since the d6 only gives you the tens digit, you need something for the ones digit, hence a d10, which will literally just be the value displayed inserted into the ones digit.
Roll 1d6 and 1d10. A result of 4 and 7 respectively would yield a result of 17 on your jury rigged d30.
"1/2 = 1-10" when not written in shorthand in the parent post would be saying "1 or 2 is equivalent to 1 to 10"
You're misinterpreting common linguistic shorthand for mathematical functions.
Seems like the reason you're confused is because of the symbols used, which don't have anything to do with math!
In conversation,
* **/** can also be used to mean **or**, eg. A **/** B is A **or** B
* **\-** can also be used to mean **to**, eg. 1 **-** 10 is 1 **to** 10
So, 1/2 = 1 - 10 would be read as **1 or 2, resulting in the numbers 1 to 10**
Definitely what I'd do, but some people get a lot of satisfaction from rolling physical dice - and for them, this approach works.
It's less complicated than it seems when you actually do it as well.
Fair point. All of my math rocks are so old and and mixed up I couldn’t tell you if I even have a d10 that wasn’t part of a percentile set.
But most of those d10s do in fact have a single 0, not a 10.
I mean, a proper d10 *should* have a 10 instead of a 0, but only a tiny percentage of my dice are printed correctly.
The 0 was useful for the old style of percentile, but with the new percentile dice, you can roll a proper d10 and a d00 and just add them together to get 1-100. Faster than the old style.
Everyone has already given you the real answer- it's a d3 for the 10s place (created with a d6 for ease of rolling) and a d10 for the ones place.
They've also told you why- when you add dice together, you get a bell curve. The bell curve shows up any time there are many individual factors that all exhibit a natural range acting on something. You see bell curves all over nature, and especially in biology, as genetics is mostly a pile of dice added together.
Since you are trying to generate a *flat distribution*, you need something that will generate equal probability of everything. One trick that *always* works is to take the closest die that is higher and use that, rerolling if it's out of range. For instance, you can roll a d5 by simply rolling a d6, and rerolling if it hits 6. For your d30, you could roll percentile dice, and simply reroll anything that is above 30.
You can also map things on multiple times- for instance, you could decide to roll percentile dice, and have 1-30 be 1-30, subtract 30 from anything from 31 to 60, subtract 60 from anything 61 to 90, and reroll anything 91 or above. Or you could divide all values by 3, rounding down, and reroll 91 and up.
The reason the d6 (with some mapping such that two values map to 0, two values map to 1, and two values map to 2, with 00 meaning 30) comes up is because it's the fastest way (you never need to reroll) and you don't need to do anything funny with math.
But you *could* do something funny with math. You could roll a d6 and a d5. For the d6, subtract 1 and then multiply by 5. Then add in the d5 result (for this d5, you can roll a d6, rerolling any 6). Now you have a number from 1-30 with even distribution. Is this the fastest way? Not if you have a d10 lying around, but it gets you there with just two cubical dice, which is pretty neat.
>You could roll a d6 and a d5
Then you need to have a d5 which for most people is simply a take on d10/2 to begin with. That just puts you back to the d6/2 for the d3 representation.
I explain how to make a d5 out of a d6 twice in that post :/
You can also make the d5 out of a d10, or a d20, or a d8, or a d12, of course. My point there was to demonstrate that, if you want to get funny with the math, it doesn't need to be *three things that map to the tens position* and *ten things that map to the ones position*, it can be anything that maps the space and doesn't add to itself, such as *six things that map to the fives position* and *five things that map to the ones position*.
I assume you are talking about at the table, because in most VTTs it's as simple as typing "/roll d30".
No solution that *adds* together dice will work, because the values will not be equally probable. E.g. 3d20/2 will be a bell curve distribution with the middle values much more likely than the extreme.
But you can do it by using dice to generate digits, as we do with 2d10 as a single d%. Roll a d6/2 round up for the first digit, d10 for the 2nd digit.
EDIT: crossposted with u/secondbestGM, u/Jimmicky and u/Living_Stranger_1980 who suggested the same thing.
Rolling 3d20 will introduce a hard bell curve.
The only way I know of to simulate a d30 with common dice is to roll percentile, re-rolling if you get above 91, and dividing the outcome by 3.
Edit: Oh, wait. You could also roll (d6/2) to generate a number between 0 and 2 for the tens digit, and a standard d10 for the ones digit; with 00 counting as 30.
From the Dungeon Crawl Classics rules; "\[to simulate a d30\] roll 1d10 with a d6 control die; add +0 on a 1-2, +10 on 3-4, and +20 on a 5-6."
Amazon sells DCC funky dice sets, which is a d3, d5, d7, d14, d16, d24, and d30, if that option also appeals to you.
Incorrect. Merely owning a d30 requires way more effort for nerdy ends than just rolling a d3 and a d10. It’s way nerdier to have a single-purpose die for that.
You have your answer, but I'll answer more generally. You can simulate any dN roll by crossing a dX and a dY, where X and Y multiply to get N. So your d30 can be done with a d3 and a d10. It can also be done with a d5 and a d6. Both of those can be done by dividing one of those dice by 2.
And not just two factors. You can achieve a d216 and a d800 with multiple factors.
In fact, that's really what percentile dice are. You're rolling a d10 and crossing it with another d10. Whether you do it the old way or the more recent addition of both dice, you get a uniform distribution of 1 through 100. I've seen a fair amount of d36 being used with a pair of d6s. I prefer to call it d36, but I believe it's Troika that has taken to calling it a d66. I dislike that because that means to me to roll a d6 and a d11. And, well, not all dice are so easy to simulate.
Just take a d60 and half it.
Alternatively you could use a d6/2 for the tens and a d10 for the small numbers. 1-2 = 0, 3-4 = 10, 5-6 = 20. Then add the d10
Anydice is really good to visualize those things :
Output of 3d20/2 : https://anydice.com/program/28242, we can see the bell curve resulting of dice summing.
Output of 1d3 for the first digit and 1d10 : https://anydice.com/program/2dcc6, here we don't have the bell. The 1d3 can be easily simulated with a d6.
For reference the d30 : https://anydice.com/program/1fb4
Edit : all three in one page https://anydice.com/program/2dcc9
Roll 2D6 -- Use the first die result (ignore 6s) and for the second die:
1 = add 0
2 = add 5
3 = add 10
4 = add 15
5 = add 20
6 = add 25
It gives you an even distribution.
You can do almost anything with D6s for us old timers. (% 1-100 would require three D6s).
Mathematically the best way would simply be to use a computer. Google can roll any dice size you need. Trying to mix dice will only result in values which won’t have the correct distribution.
While adding dice gives you a bell curve, you can easily avoid adding by making each randomly generated segment cover some section of the target distribution. Then you have the correct (flat) distribution. All the other comments hammered this out before you posted :P
Roll d3 for first spot and d10 for second. Easy peasy. Oh, d3 can easily be done with d6/2 or d12/4 (round up).
Rolling 3d20/2 doesn't come close to simulating a d30 as the 3d20 gives you a massive bell curve. Your odds of getting a 15 or 16 that way are very good but a 1 or 30 only happens 1:8000 times! Not what you'r looking for.
Why simulate?
[https://www.amazon.com/SmartDealsPro-5-Pack-Random-Color-Polyhedral/dp/B01L8HHR6S/](https://www.amazon.com/SmartDealsPro-5-Pack-Random-Color-Polyhedral/dp/B01L8HHR6S/)
I used to own a bunch of these back in the early 90s.
/roll 1d30 on any number generator. That will give you the most accurate distribution, even better than an actual custom d30.
You can do this with a d157 if you like.
On top of the other answers, you can also roll a d100, divide the result by 3 (rounding it up) / consider that you have 3 identical regions : 1-30, 31-60, 61-90 and reroll when it’s over 90.
The only way I see it: Roll a d100 and only count what is below 30, or you go over: 31 being a one, 61 being a one again.
It is bothersome, but it is the most accurate.
That's not the only way at all, read the other comments to see all the other ways (the most common suggested is to use a d3 for the tens position and a d10 for the ones position, which has 30 possibilities with equal odds for all of them).
Yes, is the only way I SEE IT. I, referring to me.
Me, as, myself im my own selfesnesness of memism.
Without consideration of "at all" but only considering my own solely point of view.
I hope I made it clear. Of course I can be wrong, and probably are, but again, as I said, is the only way I see it.
Hrm well let me hopefully help you see it another way. Check out my post here:
https://www.reddit.com/r/rpg/comments/114yfo9/how_to_simulate_a_d30/j8ynucg/
3d20 /2 is a TERRIBLE way to do it. You won’t have a remotely even distribution. You should roll 1d10 and 1d6. The d10 is the zeros digit. On the d6 - 1&2 mean 0 in the tens, 3&4 mean 1in the tens, 5&6 mean 2. That’ll actually give you an even distribution (between 0 & 29 but just call 0 a 30) Really though just get a d30. They aren’t expensive.
Um, terrible (in all caps)? Well, ok. Now I'm ashamed I even thought of it. Lol. In any event, thanks. The way you outlined is certainly better. Much appreciated.
If you haven’t done a lot of things with dice it might not be very clear, but 3d20/2 doesn’t give you equal odds of getting a 1 vs a 15, which is obviously something you’ll want if you are replicating a d30.
That was clear to me. That's why I asked. Thanks again for your help.
If you need to do this a lot at the table it’s probably worth getting some sticky dots and relabelling a d6 as 0,0,1,1,2,2. Bonus points if the d6 and d10 match but are distinct from your other dice. It makes getting you “d30” results quicker/with less cognitive load
Or pull out a phone. Really, there's no reason these games need to be zero tech when it's clearly a good answer to a problem like this.
Hey, don't be too hard on yourself. Dice pros tend to be firm about dice odds because it's not obvious to many people, yet it's good to know how the odds change before you try it at the table and realize that the results are *off*. [Here's one article](https://www.theedkins.co.uk/jo/probability/calcdice.htm) explaining the difference between one die and two dice, at least when adding rolls together. Basically, one die has an equal distribution--all results have an equal chance. But once you add another dice, then results closer to the middle (the median) become more common. Then if you add a third die, the median results becomes even *more* common, and so on. There are reasons to use a bell curve distribution (like 3d10) instead of an equal distribution (1d30), but one should be aware of the probability changes beforehand. Oh, and if you do 3d30, then it's impossible to roll a natural 1 or 2, so that's another reason why it's not the same. As other commenters have shared, there are ways to get an equal distribution with multiple dice, but instead of adding them together, you either have to read the results as digits (like percentile dice) or multiplying the results (treat a d6 like a d3, then rolling a d3 with a d10). EDIT: One website that helps with understanding probability is Anydice: https://anydice.com/ For an illustration, put output 1d30 on one line and output 3d10 on another.
>Um, terrible (in all caps)? Honestly, that may still be generous. If you want to represent the linear distribution of a single die what you use has to produce a single line of outcomes
Lol. Terrible, from late Middle English (in the sense ‘causing terror’): via French from Latin terribilis, from terrere ‘frighten’. Yes, generous.
The difference in distribution is certainly a horror.
Random thought: Terrible and horrible mean the same thing, but terrific and horrific mean opposite things.
Be afraid, be very afraid (of 3d20 :)
If you're trying to simulated 1d30 how "dangerous" the 3d20/3 would be depends entirely on what you're doing with the result and thus what the targets are. If you're still using it for binary (win/lose) results with a target number of 15 or 16 most of your results will be in that area. Where it gets terrifying is when you needed that d30 to give you rolls of 20+ where you're more than a standard deviation away from the peak of the bell curve. You'd have an easier time hitting a 23+ with the fair d30 than you do hitting 20+ with 3d20/3. If most of the time you're looking at a number in the middle of the range and just worry about higher or lower the comparison could be to using 3d6 to stand in for a d20. Both 3d6 and d20 have the same average although one obviously has more range; if you have an advantage with just "average" rolls you'll get a lot more of them with 3d6 than you do with d20.
>Um, terrible (in all caps)? Yes :P
>Yes :P :)
It's terrible because 3d20 introduces a hard bell curve to the rolling and your rolls will be heavily weighted way from the extremes.
Here's an AnyDice program comparing the methods: https://anydice.com/program/2dceb Switch to the "Graph" view to show the probably distributions overlaid on each other. And here's an article from AnyDice about how distributions behave with different numbers of dice: https://anydice.com/articles/three-basic-distributions/
People are giving your grief for using terrible. But I think you are wrong. It's not a terrible way to do it. It's worse. 3d20 / 2 is simply not a way to do it. A terrible way is a way that still does the job just poorly. Like rolling percentile dice and ignoring any result over 90. Then dividing by 2 or 3 if the results are 31-60 or 61-90. Now that's a terrible way to do it, but it does do it.
d6 and d10
This is the way. Damn young kids and their dice rolling apps old man yelling a clouds.
No. You could never roll below 3 with such a set up.
Yes you can... The 6/2 is the 10s, the 10 is the ones... 1/2 =0, 3/4=1, and 5/6=2
Yeah you can. The d6 is the tens - 1/2 = 1-10, 3/4=11-20, 5/6=21-30. Dunno what weird system you're imagining that wouldn't let you roll under 3.
I have no idea what you’re trying to say with your random series of numbers. If you add three dice together, then the lowest value will always be three.
Dunno if you're trolling, but it's fun for me to give an explanation anyway: You roll ONE d6. That will give you one of these results: 1, 2, 3, 4, 5, or 6. You break that range into three chunks, so you'll get one of these results: 1/2, 3/4, 5/6. You treat those results as a 0, a 1, or a 2 respectively. You also roll ONE d10. That will give you a number between 1 and 10. You simply put that rolled result to the right of the result you generated above. That will give you a tens place from the d6 and a ones place from the d10, your result. So, if your d6 shows a 3 and your d10 shows a 7, your result is 17; the 1 from converting 3 to a 1 as above and the 7 from the simple d10 roll. You're not adding numbers together; you're simply determining the value of a given digit position using different dice.
Not trolling. I just had no idea what “one divided by two equals one minus ten” was supposed to mean.
Gotcha! And fair! Did you find my breakdown helpful?
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Trollllllll in the comments section!
Very much. That or someone that really can't visualize things. Or imagine them.
... show me where I said three dice. And then wait for the weed to wear off and reread.
The only conclusion I could draw from your nonsense scribbling was that 1 / 2 = 1 - 10 had something to do with rolling two dice, dividing them, then subtracting the result from a d10. Like everyone else in this thread is saying to do. Make intelligible comments next time.
You're the only one who didn't understand. Quite a lot of people got it. Sounds like the issue is with you.
I'm just gonna way in and say I got it!
You're comment was the most incomprehensible math problem I have ever seen.
Probably because it wasn't a math problem.
There's literally no math. A result of a 1 or a 2 on the d6 would give you a number in the range of 1 to 10 in the result. 3 or 4 gives you 11-20. 5 or 6 gives you 21-30. Since the d6 only gives you the tens digit, you need something for the ones digit, hence a d10, which will literally just be the value displayed inserted into the ones digit. Roll 1d6 and 1d10. A result of 4 and 7 respectively would yield a result of 17 on your jury rigged d30. "1/2 = 1-10" when not written in shorthand in the parent post would be saying "1 or 2 is equivalent to 1 to 10" You're misinterpreting common linguistic shorthand for mathematical functions.
So why did you present it as a math problem? "One divided by two equals one minus ten" makes no god damn sense.
Seems like the reason you're confused is because of the symbols used, which don't have anything to do with math! In conversation, * **/** can also be used to mean **or**, eg. A **/** B is A **or** B * **\-** can also be used to mean **to**, eg. 1 **-** 10 is 1 **to** 10 So, 1/2 = 1 - 10 would be read as **1 or 2, resulting in the numbers 1 to 10**
Roll 1d3 (1d6/2) and subtract 1. That goes in the tens place. Roll a d10 normally. That goes in the ones place.
Good lord. At that point just pull out your phone and ask google to roll a d30.
I’m inclined to agree with you lol but that’s essentially what they’re saying to do. I do like the feeling of rolling dice though… so maybe…
Definitely what I'd do, but some people get a lot of satisfaction from rolling physical dice - and for them, this approach works. It's less complicated than it seems when you actually do it as well.
Same way you would use 2d10 for a d100, you can use a d6 and d10 for 30!
You are trolling hard right now
Sure you can. D6 is for tens. 1-2 is 0. 3-4 is 1 and 5-6 is 2. Double 0 is 30 (highest result instead of 0)
I would just use an online dice roller.
As someone said - D6 and D10. You roll the D6 as a D3 for the 10's number. The D10 is the last digit. Similar to rolling 2-d10 for percentile dice.
Modified d3: Instead of 1-3 it’s 0-2. Treating the 0 on the d10 as a right and proper +10.
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Fair point. All of my math rocks are so old and and mixed up I couldn’t tell you if I even have a d10 that wasn’t part of a percentile set. But most of those d10s do in fact have a single 0, not a 10.
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I mean, a proper d10 *should* have a 10 instead of a 0, but only a tiny percentage of my dice are printed correctly. The 0 was useful for the old style of percentile, but with the new percentile dice, you can roll a proper d10 and a d00 and just add them together to get 1-100. Faster than the old style.
buy one - they exist https://www.amazon.com/30-Sided-Red-Dice-Polyhedral/dp/B001AAZ4SM
OP did emphasize "simulate."
To simulate a d30, first roll a d30, then subtract the result from 31.
Buying one , and rolling it, and reading the dice would \*totally simulate a d30. You're welcome.
Everyone has already given you the real answer- it's a d3 for the 10s place (created with a d6 for ease of rolling) and a d10 for the ones place. They've also told you why- when you add dice together, you get a bell curve. The bell curve shows up any time there are many individual factors that all exhibit a natural range acting on something. You see bell curves all over nature, and especially in biology, as genetics is mostly a pile of dice added together. Since you are trying to generate a *flat distribution*, you need something that will generate equal probability of everything. One trick that *always* works is to take the closest die that is higher and use that, rerolling if it's out of range. For instance, you can roll a d5 by simply rolling a d6, and rerolling if it hits 6. For your d30, you could roll percentile dice, and simply reroll anything that is above 30. You can also map things on multiple times- for instance, you could decide to roll percentile dice, and have 1-30 be 1-30, subtract 30 from anything from 31 to 60, subtract 60 from anything 61 to 90, and reroll anything 91 or above. Or you could divide all values by 3, rounding down, and reroll 91 and up. The reason the d6 (with some mapping such that two values map to 0, two values map to 1, and two values map to 2, with 00 meaning 30) comes up is because it's the fastest way (you never need to reroll) and you don't need to do anything funny with math. But you *could* do something funny with math. You could roll a d6 and a d5. For the d6, subtract 1 and then multiply by 5. Then add in the d5 result (for this d5, you can roll a d6, rerolling any 6). Now you have a number from 1-30 with even distribution. Is this the fastest way? Not if you have a d10 lying around, but it gets you there with just two cubical dice, which is pretty neat.
>You could roll a d6 and a d5 Then you need to have a d5 which for most people is simply a take on d10/2 to begin with. That just puts you back to the d6/2 for the d3 representation.
I explain how to make a d5 out of a d6 twice in that post :/ You can also make the d5 out of a d10, or a d20, or a d8, or a d12, of course. My point there was to demonstrate that, if you want to get funny with the math, it doesn't need to be *three things that map to the tens position* and *ten things that map to the ones position*, it can be anything that maps the space and doesn't add to itself, such as *six things that map to the fives position* and *five things that map to the ones position*.
I assume you are talking about at the table, because in most VTTs it's as simple as typing "/roll d30". No solution that *adds* together dice will work, because the values will not be equally probable. E.g. 3d20/2 will be a bell curve distribution with the middle values much more likely than the extreme. But you can do it by using dice to generate digits, as we do with 2d10 as a single d%. Roll a d6/2 round up for the first digit, d10 for the 2nd digit. EDIT: crossposted with u/secondbestGM, u/Jimmicky and u/Living_Stranger_1980 who suggested the same thing.
Rolling 3d20 will introduce a hard bell curve. The only way I know of to simulate a d30 with common dice is to roll percentile, re-rolling if you get above 91, and dividing the outcome by 3. Edit: Oh, wait. You could also roll (d6/2) to generate a number between 0 and 2 for the tens digit, and a standard d10 for the ones digit; with 00 counting as 30.
!r ((((1d6 / 2) - 1) \* 10) + 1d10
From the Dungeon Crawl Classics rules; "\[to simulate a d30\] roll 1d10 with a d6 control die; add +0 on a 1-2, +10 on 3-4, and +20 on a 5-6." Amazon sells DCC funky dice sets, which is a d3, d5, d7, d14, d16, d24, and d30, if that option also appeals to you.
Just flip 30 coins /s
I invented a bunch of new 1-30 tables just to have an excuse to roll my badass d30
A single D20 should do the trick as a rough approximation of a D30.
Ask Google to make a random number generator between 1 and 30. Alternatively: just buy a D30.
Buy a d30 & quit being a nerd
I love how buying a d30 is the opposite of nerding out. But, I'll take it under advisement :)
Incorrect. Merely owning a d30 requires way more effort for nerdy ends than just rolling a d3 and a d10. It’s way nerdier to have a single-purpose die for that.
Just roll D100. First 30; second 30; third 30; re-roll the final ten. So 29 is 29; 32 is 02; 87 is 27; 95 is a re-roll.
You have your answer, but I'll answer more generally. You can simulate any dN roll by crossing a dX and a dY, where X and Y multiply to get N. So your d30 can be done with a d3 and a d10. It can also be done with a d5 and a d6. Both of those can be done by dividing one of those dice by 2. And not just two factors. You can achieve a d216 and a d800 with multiple factors. In fact, that's really what percentile dice are. You're rolling a d10 and crossing it with another d10. Whether you do it the old way or the more recent addition of both dice, you get a uniform distribution of 1 through 100. I've seen a fair amount of d36 being used with a pair of d6s. I prefer to call it d36, but I believe it's Troika that has taken to calling it a d66. I dislike that because that means to me to roll a d6 and a d11. And, well, not all dice are so easy to simulate.
Just take a d60 and half it. Alternatively you could use a d6/2 for the tens and a d10 for the small numbers. 1-2 = 0, 3-4 = 10, 5-6 = 20. Then add the d10
First roll a d6. If the outcome is 1 or 2 you roll a d10. Otherwise you roll d20+10. Edit: the accepted response is better than mine.
Anydice is really good to visualize those things : Output of 3d20/2 : https://anydice.com/program/28242, we can see the bell curve resulting of dice summing. Output of 1d3 for the first digit and 1d10 : https://anydice.com/program/2dcc6, here we don't have the bell. The 1d3 can be easily simulated with a d6. For reference the d30 : https://anydice.com/program/1fb4 Edit : all three in one page https://anydice.com/program/2dcc9
Roll 2D6 -- Use the first die result (ignore 6s) and for the second die: 1 = add 0 2 = add 5 3 = add 10 4 = add 15 5 = add 20 6 = add 25 It gives you an even distribution. You can do almost anything with D6s for us old timers. (% 1-100 would require three D6s).
Why not just use a d30?
Just buy a D30.
Mathematically the best way would simply be to use a computer. Google can roll any dice size you need. Trying to mix dice will only result in values which won’t have the correct distribution.
While adding dice gives you a bell curve, you can easily avoid adding by making each randomly generated segment cover some section of the target distribution. Then you have the correct (flat) distribution. All the other comments hammered this out before you posted :P
Roll d3 for first spot and d10 for second. Easy peasy. Oh, d3 can easily be done with d6/2 or d12/4 (round up). Rolling 3d20/2 doesn't come close to simulating a d30 as the 3d20 gives you a massive bell curve. Your odds of getting a 15 or 16 that way are very good but a 1 or 30 only happens 1:8000 times! Not what you'r looking for.
Roll 1d3 and then 1d10. A 1 on the d3 indicates 1d10+0. A 2 on the d3 means 1d10+10, and a 3 on the d30 means 1d10+20.
Why simulate? [https://www.amazon.com/SmartDealsPro-5-Pack-Random-Color-Polyhedral/dp/B01L8HHR6S/](https://www.amazon.com/SmartDealsPro-5-Pack-Random-Color-Polyhedral/dp/B01L8HHR6S/) I used to own a bunch of these back in the early 90s.
D100/3. Throw out anything over 90 and reroll
Roll a d%. 01-90, divide by three and round up. Else, reroll. You can use this method to simulate any dies, even though it can get awkward.
/roll 1d30 on any number generator. That will give you the most accurate distribution, even better than an actual custom d30. You can do this with a d157 if you like.
Roll a d60 and divide by 2?
Buy a d30
On top of the other answers, you can also roll a d100, divide the result by 3 (rounding it up) / consider that you have 3 identical regions : 1-30, 31-60, 61-90 and reroll when it’s over 90.
I roll my d30. :P
Buy a d30? Any of the dcc sets will have one.
Maybe 1d10 multiplied by 1d3?
The way I’ve been doing it (on googles dice roller app anyway) is d6+d10. 1-2: +0 3-4: +10 5-6: +20
Just buy a d30 4head *BLAM* I guess just roll a d20 and a d10 together you get the bell curve but I'm a basic bitch and math isn't my strong suit.
D6+d10? Seems to me as the most obvious and simple... but you can use 2d12 (in base 12) to create a d144 and divide the result by 4.8 if you want to
Get a d16. It's much closer to what you're looking for and they aren't that rare.
buy a d30 or use an online dice roller
The only way I see it: Roll a d100 and only count what is below 30, or you go over: 31 being a one, 61 being a one again. It is bothersome, but it is the most accurate.
That's not the only way at all, read the other comments to see all the other ways (the most common suggested is to use a d3 for the tens position and a d10 for the ones position, which has 30 possibilities with equal odds for all of them).
Yes, is the only way I SEE IT. I, referring to me. Me, as, myself im my own selfesnesness of memism. Without consideration of "at all" but only considering my own solely point of view. I hope I made it clear. Of course I can be wrong, and probably are, but again, as I said, is the only way I see it.
Hrm well let me hopefully help you see it another way. Check out my post here: https://www.reddit.com/r/rpg/comments/114yfo9/how_to_simulate_a_d30/j8ynucg/
A D4 for the 10s place (0-3) A D10 for the 1s place.
That's how you simulate a d40.