help me with a math problem

procco

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!

FinneousPJ

No, it's not equivalent to rolling a d20. In a d20 there is an even distribution of values. In your case, you can roll a 10 in many different ways, for example. Hence, it will be weighed towards the middle.

procco

That's what I thought...how does one show that mathematically?

FinneousPJ

What do you need to show? That the distribution is different? Or what the distributions are? It's easy to show that rolling a 1 in your scenario is way unlikelier than in d20. Rolling a 1 means rolling a 1 in each separate roll.

P(1) = 1/4 * 1/8 * 1/10 = 1/320

Whereas in d20

P(1) = 1/20

[Deleted User]

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AstroBryGuy

Roll d100*, and divide by 5 (round up).

1-5 = 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)

semiticgoddess

And Microsoft Excel has a random number generator built in. If you put
=RANDBETWEEN(1,20)
into a cell and click delete in an empty cell, it will give you a random number between 1 and 20.

Fardragon

procco said:

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!

No, it's not the same. A d20 has exactly the same probability of getting each number: 5%. Using your suggested method, the probability of getting a 1 or 20 is 1/(4x8x10), which works out at about 0.3%, while the probability of getting 10 or 11 is much higher (to early in the day for me to work out exactly what).

BillyYank

semiticgod said:

And Microsoft Excel has a random number generator built in. If you put
=RANDBETWEEN(1,20)
into a cell and click delete in an empty cell, it will give you a random number between 1 and 20.

There's a button on the Formula tab called "Calculate Sheet" that will do the same as a delete. If you're going to do any extensive die rolling in Excel, I would recommend setting calculation options to "Manual". That way you won't accidently reroll the dice and lose a roll you wanted to keep. I use Excel for all my character creation die rolls. I like it because you can set up complicated rolls like "roll 4d6 and drop 1".

Edit: 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

procco

Thanks, fellas. I didn't actually lose my d20, it was just something I was curious about for no good reason. I understand that the numbers towards the middle of the scale are going to have a higher chance of coming up since there are more ways of them to come up...I'm going to try to work out on my own how to figure out what the chance of each number is, then check back to see if anyone has a solution for that...

Teflon

Then throwing three d6s at once vs. d6 three times in a row is different?
Anyway i think cannot replace d20 with two d10s, alas I cannot prove mathematically but just intuition.

AstroBryGuy

Teflon said:

Then throwing three d6s at once vs. d6 three times in a row is different?
Anyway i think cannot replace d20 with two d10s, alas I cannot prove mathematically but just intuition.

You can, but don't add them. Use them as percentile dice (one d10 for the tens place, one d10 for the ones place).

01-05 = 1
06-10 = 2
11-15 = 3
etc..
96-00 = 20

FinneousPJ

@procco It's simple. It's the number of combinations that result in a roll over the total number of combinations:

P(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

Teflon

: What school of magic are you?
AstroBryGuy : Math.

BillyYank

The 1d4 + 1d8 + 1d10 - 2 method comes out as a standard bell curve:

http://anydice.com/program/a8fc

AstroBryGuy

procco said:

Thanks, fellas. I didn't actually lose my d20, it was just something I was curious about for no good reason. I understand that the numbers towards the middle of the scale are going to have a higher chance of coming up since there are more ways of them to come up...I'm going to try to work out on my own how to figure out what the chance of each number is, then check back to see if anyone has a solution for that...

The key is calculating the number of permutation (different dice rolls) that can result in a sum x.

Brute 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).