"Easy To Use Online Dice Probability Tables-Do They Exist?" Topic

Ok, I'll admit it, I'm no math wiz by any stretch whatsoever, but when I'm working on new things for my homebrew games, I often need to work out various dice probabilities.

I've seen others here on TMP ask those who can easily work out this stuff for help, which is always gladly given. However, I dislike bothering anyone with my math impairment (seems to be getting worse the older I get) and since I'm always tinkering and tweaking some design, I'd have to frequently ask those kind souls here on TMP for help-which I don't want to bother anyone with.

So, that brings me to my question… is there any online source(s) that an aging brain dead guy can refer to that easily explains how to work out such things?

For example (and I know this will be elementary for some of you, but I just can't get my head around it anymore), I have 3D6 that each have a 1 or 2 inscribed on opposite sides. So, the odds of getting a 1 or 2 result on any one die is…? Then, the result of getting a 1 or 2 result on any two dice of the three would be…? What if I add another D6, how does that change the result? So, you see, in my feeble mind it makes sense as to the answer I'm looking for, but it's rather hard (awkward) to explain so that someone can understand what I'm after. Thus, if I had a source website I could go to to find these things out would be fantastic. Does any such site exist?

As always, thanks for your input.

My stats are a bit old, but let me give it a try….

If you roll a die, you have a 1 in 6 chance of any number coming up. When you roll multiple dice, your chance of a specific combination is equal to a 1 in (die type x die type). So rolling two ones on 2d6 is a 1 in 36 (6 * 6). Rolling a 1 and a 6 is also a 1 in 36 chance. Where things get confusing is that you could also roll a 6 and a 1 (if one die is red, the other blue, red could = 1 and blue =6 OR red = 6, blue = 1). Or you could roll a 5 and a 2, or a 2 and a 5 to also get a seven. Or a 4 and 3, or 3 and 4. So when you look at all the possibilities, there is actually 6 combinations out of the 36 (6 sided die * 6 sided die) that result in a seven. So while your chances of rolling a 2 on 2d6 is 1 in 36, your chances of rolling a 7 are 6 in 36 (simplified to 1 in 6).

I find a chart helps things flow.

Roll 2d6
2 1,1 = 1 in 36
3 1,2 and 2,1 = 2 in 36
4 1,3; 3,1; 2,2 = 3 in 36
5 1,4; 4,1; 2,3; 3,2 = 4 in 36
6 1,5; 5,1; 2,4; 4,2; 3,3 = 5 in 36
7 1,6; 6,1; 2,5; 5,2; 4,3; 3,4 = 6 in 36
8 2,6; 6,2; 3,5; 5,3; 4,4 = 5 in 36
9
10
11
12

Notice how the pattern forms? Each combination from 2 to 7 grows by 1 in 36. Each combination down decreases by 1 in 36.

So now, rolling 3d2 (if I read your post right, each side of the 6 sided die has only a 1 or 2 possible, with an equal chance of each, right?)

Since you are rolling 3d2, you have a total of 2*2*2 = 8 possibilites

3 1,1,1 1 in 8
4 1,1,2; 1,2,1; 2,1,1 3 in 8
5 1,2,2; 2,1,2; 2,2,1 3 in 8
6 2,2,2 1 in 8

Hope that's helpful, and not too confusing. Feel free to contact me if you need more help.

This any good?

link

Wow! Thanks for the quick replies, guys. Well, I hope I can clarify what I'm looking for:

Captain Oblivious said:
"So now, rolling 3d2 (if I read your post right, each side of the 6 sided die has only a 1 or 2 possible, with an equal chance of each, right?)"

Yep. What I did was take a blank D6 and inscribe a "1" on one face, then I inscribed a "1" on exact opposite face. I did the same with a "2." So, I made a D6 with three "1s" on exact opposite faces and three "2s" on exact opposite faces. This leads into Ditto's reply, which said:

"Captain Oblivious, I didn't understand what 000 is was talking about = if his dice all have 1s and 2s, there's exactly a 100% chance of rolling what he's asking."

Now, in my addled cranial cavity, if I roll my 1D6, I have an equal chance of getting a "1" or a "2," which to me means that I have a 50% chance of rolling either a 1 or a 2 on my 1D6. Correct?

(hope you guys don't hate me for this, I tried to warn you, but… ramping it up)

What if I inscribed each d6 (on opposing sides) with "1," "2," and "3?"

Now, if I roll 3D6 (all inscribed the same), I would get various probabilities of either a "1" or a "2" or a "3" coming up and because I'm rolling three dice, there should be the possibilities of:

(A) a "tie breaker" (i.e. rolling two "2s" and and a "3"), or

(B) all three dice resulting in the same number (i.e. "3," "3," "3.") or

(C) each die having a different result (i.e. Die 1 = "1," Die 2 = "2," and Die 3 = "3.")

So, with this in mind, what are the probabilities of rolling a "1," "2," or "3" using 3D6 as inscribed above?

(you can see why this is giving me a brain cramp and I truly appreciate all your help) :)

With what is essentially a D2 as above, there is a 50% chance that you will roll a "1". There is also a 50% chance that you will roll a "2". There is a 100% chance that you will roll a 1 OR a 2, because those are the only possible results. "OR" in this situation has a specific logical and mathematical meaning, and includes the probability of BOTH results.

Similarly, on a D6, the odds are 1/6 that you will roll any given number. The odds that you will roll a "1" *or* a "2" on a D6 is 1/3 (1/6 + 1/6 = 2/6 = 1/3).

If you are rolling 3 D3's (which once again is effectively what a D6 numbered 1-3 twice is) as a set, not counting order as different results, there are a total of 27 possible combinations (D to the Nth power, where D is the Die type and N is the number of dice. It's pretty easy to brute force the probabilities for this:

All the same = 3 combos (1,1,1 – 2,2,2 – 3,3,3)
All different = 6 combos (1,2,3 – 1,3,2 – 2,3,1 – 2,1,3 – 3,2,1 – 3,1,2)
Only two the same = 18 combos (27 minus {6+3}) because there are no other additional possible patterns when considered this way.

Dividing by 27 means 3/27 or one chance in 9 all three dice are same, 6/27 or 2/9 of all different, and 18/27 or 2/3 chance of two and only two dice the same.

Now if you ADD the score of the dice, that's a different problem, as CO demonstrated with his 2D6 example, although there are still D to the Nth power possible combinations, in that case 6 squared or 36 combos.

3D6 has 216 possible combinations (six cubed) – the method for rolling stats in classic D&D, hence a natural "18" score thus is only 1 combo out of 216, or less than 0.5%.

I'd better stop there.

Alright, let's see if I can answer this….

Each d6 is now actually a d3, so a 1 in 3 chance of each number coming up.

So, 3*3*3 = 27 possibilities

3 1,1,1 – 1 in 27
4 1,1,2; 1,2,1; 2,1,1 – 3 in 27
5 1,2,2; 2,1,2; 2,2,1; 3,1,1; 1,3,1; 1,1,3 – 6 in 27
6 1,2,3; 1,3,2; 3,2,1; 3,1,2; 2,3,1; 2,1,3; 2,2,2; – 7 in 27
7 2,2,3; 2,3,2; 3,2,2; 3,3,1; 1,3,3; 3,1,3 – 6 in 27
8 3,3,2; 3,2,3; 2,3,3 – 3 in 27
9 3,3,3 – 1 in 27

So using the above chart, you should be able to answer your questions. Attempting to answer them gives:

A) 2,2,3 has a 3 in 27 chance (1 in 9)
B) 1,1,1; 2,2,2; 3,3,3 all have an equal chance of 1 in 27
C) 1,2,3 has a 6 in 27 (2 in 9) chance of occuring

Hope that helps! I like math (it's the teacher in me coming out), so feel free to keep asking!

anydice.com

--
Martin Stephenson
blog.vexillia.me.uk
amazon.co.uk/shops/vexillia
twitter.com/vexillia

@vexillia

Wow! AWESOME! Thanks for sharing that.

Thanks for all the help everyone! I won't feel as bad now posting questions on these things because I know I'll get fast/friendly answers. :)

that anydice is awesome. already loving it for making rules