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Say the d20 is missing from my set of dice, but I want to replicate the range of 1-20. Easy enough...roll a d4, d8, and d10 together to get a range of 3-22, then subtract 2 to get the 1-20 range. My question is, is that actually equivalent to rolling a d20, or will the chance of each number coming up be weighted towards the numbers in the middle, and how do you figure that out?

Thanks!

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## Comments

6,4569156,456P(1) = 1/4 * 1/8 * 1/10 = 1/320

Whereas in d20

P(1) = 1/20

03,4151-5 = 1

6-10 = 2

11-15 = 3

etc..

96-00 = 20

EDIT: *I mean 2d10 with one as a "tens" die - I keep forgetting that they actually sell real d100s now...

A variation on @Shandyr's method: d4 and d10.

d4 is 1 or 2:

Read the d10 as 1-10.

d4 is 3 or 4:

Read the d10 as 11-20.

(or do odds/evens on the d4 if you prefer)

14,833=RANDBETWEEN(1,20)

into a cell and click delete in an empty cell, it will give you a random number between 1 and 20.

4,5112,769Edit: There's a lot of free dice rolling apps for Android and, I assume, iOS. I played around with this one for a few minutes and it seems OK.

https://play.google.com/store/apps/details?id=fr.sevenpixels.dice&hl=en

915517Anyway i think cannot replace d20 with two d10s, alas I cannot prove mathematically but just intuition.

3,41501-05 = 1

06-10 = 2

11-15 = 3

etc..

96-00 = 20

6,456P(1) = 1 / 320, because there is only one way to get 3 on 1d4+1d8+1d10

P(2) = 3 / 320, because there are 3 ways to get 4 on 1d4+1d8+1d10

and so on. Of course, finding out the number of combinations may not be so simple

517AstroBryGuy : Math.

2,769http://anydice.com/program/a8fc

3,415Brute force (i.e., enumerating all the possibilities) is slow and painful.

As faster way uses recursion. Take a look at this spreadsheet:

https://docs.google.com/spreadsheets/d/1pSdm-IvFtbcslhsatAppx4a6Y8NtO9FFQhYX78tyOFg/edit?usp=sharing

The first column gives the possible values for the sum of the dice (I'm ignoring the -2 modifier and just doing d10+d8+d4 for ease of understanding). Note that I've started at -9 to make the formulas work easily.

The second column shows the number of permutations to calculate each sum with a single roll of a d10. There's a 1 next to each value from 1-10.

The third column shows the number of permutations to calculate each sum by rolling d10+d8. Using the d10 column as first roll, then for each possible sum, x, there are up to eight possible ways to get that sum (i.e., from rolling the d8):

- d10 shows x-1 and the d8 shows 1.

- d10 shows x-2 and the d8 shows 2.

- d10 shows x-3 and the d8 shows 3.

- d10 shows x-4 and the d8 shows 4.

- d10 shows x-5 and the d8 shows 5.

- d10 shows x-6 and the d8 shows 6.

- d10 shows x-7 and the d8 shows 7.

- d10 shows x-8 and the d8 shows 8.

So, I make this a formula, adding up the previous 8 rows in the d10 column. Some (or even all) of the terms may be 0, but that's okay. The formula is still valid. (This is why I have all the negative sums at the top, to create blank cells for the formula to reference).- For cell C11 (Sum=1), the formula is: "=B10+B9+B8+B7+B6+B5+B4+B3". The result in this case is, of course, 0 (d10+d8 can't equal a sum of 1).

- For cell C12 (Sum=2), the formula is: "=B11+B10+B9+B8+B7+B6+B5+B4". The result in this case is 1. There is one possible permutation of d10+d8 that gives a sum of 2, rolling 1 on both.

Once I enter the formula in cell C11, I use the spreadsheet's "Fill Down" function to automatically populate the rest of the column.For the d4 column, the formula only adds up the 4 previous rows in the d8 column (since the d4 has 4 possible values). Enter the formula for D11 and use Fill Down again.

I've also added a column to calculate probabilities. That's just the (number of permutations for each sum)/(total number of permutations).