# help me with a math problem

This will probably turn out to have an easy answer, but math isn't my strong suit:

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!

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!

1

## Comments

6,4559156,455P(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)

14,829=RANDBETWEEN(1,20)

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

4,5112,768Edit: 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

915515Anyway 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

6,455P(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

515AstroBryGuy : Math.

2,768http://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).