How I can I roll a number of non-digital dice to get a random number between 1 and 150?

Question

This question is not actually related to RPGs but is more of a real-world dice rolling scenario I'm looking for help with, from dice rolling experts.

I'm currently memorizing the Book of Psalms, which is divided into 150 chapters.

I'm looking for an analog, elegant way of quizzing myself, by rolling some number of dice (or whatever means, really, just nothing digital) to get a random (equally likely) number from 1 to 150.

Not being very aware of the dice world I start googling, starting with "150 sided dice" ... which would obviously do the trick in one go.

Didn't take long for me to realize there are only so many 3D shapes that lend themselves to being dice with 120 sided dice being the largest (and pretty ridiculous looking) I could find.

So it'll be some combination of things. One solution I have thought of so far would be a d30 and a d5. I would subtract one from the d5 and multiply that number by 30, resulting in 0,30,60,90,120. I would then simply add the value of the d30.

Hopefully that example gives you an idea of what I'm trying to accomplish. Apologies in advance if I'm barking up the wrong tree and this question doesn't belong. If so I'll just edit my question "A Character in my RPG is memorizing the Book of Psalms..." :)

Answer 1

Use a "d50" and a "d3" - which are 2d10 and 1d6

You roll 2d10 to make up your percentile dice of 1-100.
Divide the result by two (round up) to make it 1-50. (If the dice are marked 0-9, then 00 = 100).

You roll the 1d6 to determine the offset: {1, 2} = 0, {3, 4} = +50, {5, 6} = +100.

Result is 1-50, 51-100, 101-150.

Per your comment, you need to buy dice. Buy two ten-sided dice, and use a single 6-sided die.

• I suggest each of the ten sided dice be a different color: say red and green. Red for tens, green for ones. Rolling a red 3 and a green 7 yields 37. Divide by two (round up) for a 19. (or 69 or 119).
• If the game shop (or other vendor) sells ten-sided dice that are labelled 10, 20, 30 ... as well as 1, 2, 3,... then the color coding is not needed. Thank you @MrZarq

Good luck.

Answer 2

The most straightforward method is to roll two dice: a die for the “ones” place and another to generate 0–15 for the “tens” place. This is quick to roll, easy to read, portable, and requires no mental calculations. You can do this with a pair of available dice: a 15-sided die and a ten-sided die (the latter usually numbered 0–9).

Mark the “15” on the d15 as zero: with a marker you can draw a zero over it, or black out white numerals, to remind you that this face means a zero in the “tens” place. (Or, if it’s easier, just always remember to read “15” as zero.)

Then roll and read them in order:

• d15 shows 7 and the d10 shows 1 gives you 71.
• d15 shows 4 and the d10 shows 0 gives you 40.
• d15 shows 0 (the blacked-out face) and the d10 shows 1 gives you 1.
• d15 shows 14 and the d10 shows 9 gives you 149.

This pattern holds except for a result of zero on both dice: read this as 150, which otherwise won’t show up literally if you’ve marked the d15. (This is akin to how a pair of d10s marked 0–9 are traditionally read to roll 1–100: they’re read literally for 00–99, but when two zeroes come up that’s read as 100, completing the range.)

Answer 3

Roll a D20 and D10, Reroll Results Over 150

The simplest solution for when you have no die of the right size is to use a higher die that does exist and drop results that are too high.

It is not the most elegant solution, but one way to generate very easy to interpret results for 1-150 without requiring particularly uncommon dice is to get a d20 and a d10. As very common rpg dice they should both be easy to find at most game shops (there's usually little bins of single dice of the various sizes). In my area they cost on the lower side of around a dollar each.

Roll d20 to determine the first two digits (hundreds and tens) and the d10 for the last digit. Count 20's as 0 (this is so that you can generate numbers below 10). This will generate results from 0-199. reroll results for which no psalm exists (which usually just means rerolling the d20, the d10 only if you end up with "psalm zero").

Once again not the most elegant solution, but it requires very little effort to interpret results, a $1.50-$3 or less investment, a trip to basically anyplace fine rpg materials are sold, and rerolling one die a quarter of the time.

Some d10s are marked 0-9 and others 00-90. Life will be easier if you buy the prior.

Of course, if you are more mathematically inclined, there are numerous other answers that are much more clever here and involve no rerolls.

Answer 4

Don't worry about those silly plastic things and use the internet*

You can use websites such as anydice.com in order to roll virtual dice, this will give you the same effect and then you won't have to worry about real dice.

Here is an example of how I setup anydice to allow you to roll 1d150s and generate a result

You mentioned breifly in your question about "real-world scenario" and I'm not sure if this meant you physical dice or not. Anydice (or similar websites) can be accessed from mobiles, desktop PCs and the likes.

---

*Title posted for comedic effect, I love real dice.

Answer 5

Roll from 1 to 600, then wrap around.

^{(...or from 0 to 599, or from 100 to 699, or from 101 to 700, or whatever. It makes no difference.)}

Probably the simplest way to do this is to roll a d6 for the first digit and a d100 (typically represented in RPG dice sets as a pair of d10s, one numbered 0–9 and the other in steps of 10 from 00 to 90) for the last two digits (counting a double zero on the d100 as 00, not as 100). Done straight, this gives you a number from 100 to 699, which is fine.

^{(You could also interpret 6 on the d6 as a zero, giving you a number from 0 to 599, which is also fine. Or you could additionally interpret a roll of 6,0,0 as 600 instead of 0, giving you a result from 1 to 600, which is also fine. Or you could count a double zero on the d100 as 100 and add it to the d6 × 100, giving you a result either from 101 to 700 or from 1 to 600, depending on whether you count 6 on the d6 as 0 or 600. Any of these methods is also fine, as long as you decide ahead of time what you're going to do and do it consistently. If you happen to have a d100 pair where the ones die is numbered 1–10 and prefer to just sum the tens and ones, that's just fine too.)}

Then reduce the result modulo 150. The naive way to do that would be to repeatedly subtract 150 from the result until it's 150 or less, but it's easier to make use of the fact that 2 × 150 = 300. So you only need to reduce the first digit by three until the result is at most 300, then subtract 150 once if the result is still more than 150.

^{(If the method of rolling you used happened to give you a zero, add 150 to it. Because we're not programmers, and prefer to count from 1 to 150 rather than from 0 to 149.)}

Ps. Mathematically, the reason why we have so much latitude in choosing how to roll the initial number is because any method that picks a uniformly distributed random number from a contiguous range whose length is a multiple of \$n\$, and then wraps that result around into a range of length \$n\$, will give a uniform result modulo \$n\$. So as long as there are 600 different possible dice rolls (which using a d6 and two d10s ensures), as long as each distinct roll yields a different number, and as long as the difference between the highest and the lowest possible number equals 600 − 1 = 599, then reducing the number thus rolled modulo 150 will give you a uniformly distributed random number from 1 to 150.

^{(Or from 0 to 149, if you're a programmer.)}

Answer 6

If you do not have RPG equipment and a utility belt full of weirdly shaped dice already, you can solve the problem with a standard six-sided die:

• Roll a number x from 1 to 5. To do so roll your die, but if you happen to roll a six, repeat until you don't.
• Roll another number y from 1 to 5.
• Roll a number z from 1 to 6

Note that this gives you \$5\times5\times6=150\$ different outcomes. To turn the outcomes to a number in the range 1-150, compute

$$ (x-1)\times30 + (y-1)\times6 + z $$

Answer 7

Similar to SevenSidedDie's answer to generate a number for the tens and the units. If you wanted to solve this without electricity or specialist dice, you could use a deck of cards.

To decide between 00-10-20-30-40...140, use the Ace to King of Spaces to represent 10-130 and Ace of Clubs to represent 00 and 4 of Clubs to represent 140.

To decide between 1-2-3-4...9 Use 1 to 10 of Hearts.

e.g.

7 of Spades and 4 of Hearts is 74.
4 of Clubs and 9 of Hearts is 149.
Ace of Clubs and 10 of Hearts is 150.

Shuffle and draw a card from each pile to get a number between 1 and 150.

Answer 8

A d30 marked 1-15 twice, and a d10 marked 0-9

I like SevenSidedDie's answer of a specialist d15 and a d10. It appears that specialist dice are available for every odd and even value from d2 to d20, and custom 3d-printed designs up to d30 and beyond.

However, a slightly cheaper alternative may be to mark a d30 with 1-15 twice, using stickers or the like. The advantage is that d30s are cheaper, more readily available, and you're more likely to already own one, it being perhaps the most common of non-standard dice. The disadvantage is your stickers can get dirty or peel off over time; they lack the durability inherent to dice.

You would, naturally, read this like percentile dice with the d10 as the second digit, giving a literal range of 10 to 159, with presumably numbers 151-159 read as 1-9. Since you know there are no actual Psalms numbered above 150 (ignoring apocryphal texts of course), this method ensures that you are unlikely to misinterpret rolls of 151-159, thus maintaining an even distribution.

It's possible for someone to 3d print an actual 150-sided die, but in my experience, the more sides you add past 1d20, the closer you approach a sphere. You need a very long table for it to stop rolling and it's hard to read the top precisely. Even a 60-sided die is impractical.

Answer 9

Since there exists a d10 die, and the last digit is uniformly distributed, we can ignore it entirely. Thus the problem becomes to generate numbers between 1-15 using only dice.

The problem here is obviously the 5 in the 3x5 factorization, since it only appears in d10 and d20.

You indicated in a comment that you don't want to do mental math or rerolls, so you might be against the relatively clean answer of:

• take 4*[d4] + [d4] - 4, re-roll on 16

which means you only roll 3 die (2d4 and d10) 93% of the time.

as well as the division-requiring answer, e.g.:

• take 10 * [d6] + [d10] and divide by 4 (rounding up) or take modulo 15 + 1

which never requires re-rolls.

Unfortunately any answer that involves direct addition of 1..n die is going to irrevocably break the uniformity requirement. However, if you are willing to add a little DIY arts and crafts (clear tape + paper to relabel), we can give you a simple addition version that satisfies your requirement.

• create a "5 * [d3] - 5" die using a d6 relabeled as [0,0,5,5,10,10]
• create a "[d5]" die using a d10 relabeled as [1,1,2,2,3,3,4,4,5,5]
• roll these two dice and add them together to get [1..15] uniform

This might require a few minutes to put together, but you would end up with 3 die in your pocket and only adding a small multiple of 5 once for the math.

Answer 10

Roll 3 different coloured d10s:

• One for the hundreds value
• One for the tens value
• One for the ones value

Reroll if the result is over 150.

Answer 11

Use a six sided dice to roll three digits of a base 6 number, assigning the value 0 to a rolled six. Three rolls give you three numbers, a, b and c, each between zero and five inclusive.

Now compute $$ m = a \times 36 + b \times 6 + c $$ m will be a number between 0 and 215 inclusive.

Then compute $$ n = \mathrm{int}\left(\frac{m \times 150}{ 216}\right) $$ Where int(x) rounds down to the largest integer less than or equal to x. This will give you a random integer uniformly distributed between 0 and 149. Add 1 to get 1 to 150.

Answer 12

Wow... some of these are pretty in-depth. Let's go with the least amount of dice necessary for this. A die 6 and a die 10. Roll the d6 looking for even or odd. Since 1 is odd, we'll make 0 "even". ie, you roll a 4. Even = 0. This will be your first digit. Since it's a zero, roll the d10 twice for the next 2 digits. ie, 4 and 8. Your number is 048. If you roll odd on your 1st d6, now your first digit is 1. Since you can't go over 50, roll the d6 again and subtract 1 for the 2nd digit and then the d10 for the third (obviously ignoring this last step if you rolled a "6", ie 5, for the 2nd digit).