Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[JoshuaFH]

So I'm not a college educated person, and it's been forever since I've last WANTED to think about math and numbers, so please bear with me here while I try to explain.

I've known since forever that rolling any two D6's has the most likely outcome being a 7.

In my mind, that was that, I just never bothered thinking about it again, until I saw a graph showing each possible outcome shown in a graph, and how it was simply the highest outcome (6) plus the smallest outcome (1) that held true no matter how large or small the dice got.

What I'm trying to figure out is how you'd figure out what the most likely outcome would be if you were throwing 3 d6's instead. I'm trying to imagine a 3-dimensional graph since that seems like the next logical step, but I'm not sure how it would look.

But I'm also curious about if you just kept adding more D6's, how would you go about determining what the most likely outcome is? I'm drawing a blank here, but I was wondering if anyone could steer me in the right direction.

[RedKing]

It'd still be a bell curve, no matter how many d6's you add. Now, if you start adding dice of different sizes (and different probability ranges), then the graph starts getting wonky looking.

The math for it is here, along with some graphs showing 1d6, 2d6, 3d6, 4d6 and 5d6. Basically, 2d6 is a triangular graph and as you add more dice you flatten out the graph more and more into a traditional bell curve.

[Mech#4]

This is all probability isn't it? 6+6+6 = 18 which is the highest you can role on 3d6 but that would be as likely as 1+2+3 = 6 or 3+4+6= 13 in my book, baring the kind of pickiness like a one has less dents so it's the heavier side or whatever.

Though from what I can vaguely remember at 3:00am, there's more combinations of dice that make up a number such as 13 then there is that make up 18, thus it's more likely. You could have 6+6+1 or 5+4+4 or 3+4+6 on any of the dice but you need to role three 6s to get 18. Though as I said, to me those would be all as likely as each other, as the outcome of one dice doesn't affect the outcome of the others, so each dice is running off a 1/6 chance to come up whatever.

[Virex]

The odds are independent, so the expected total for n rolls can be calculated by n * M, where M is the mean of a single roll. For a 6-sided die, the mean is (6+1)/2 = 3.5, so for n 6-sided dice, the expected total for n rolls is 3.5*n.

[RedKing]

For 6-sided dice, a quick and dirty way to find the most common result is to add the lowest possible result (all 1s) and the highest possible result (all 6s) and divide by two. If you get a fractional result, that means you have two equally common midpoints, since dice produce integers.

2d6: 2-12 (which sum to 14. Divide by 2, you get 7.)
3d6: 3-18 (which sum to 21. Divide by 2, you get 10.5. So 10 and 11 are your most common results.)
4d6: 4-24 (which sum to 28. Divide by 2, you get 14.)

and so on.

[Leafsnail]

This is all probability isn't it? 6+6+6 = 18 which is the highest you can role on 3d6 but that would be as likely as 1+2+3 = 6 or 3+4+6= 13 in my book, baring the kind of pickiness like a one has less dents so it's the heavier side or whatever.

Though from what I can vaguely remember at 3:00am, there's more combinations of dice that make up a number such as 13 then there is that make up 18, thus it's more likely. You could have 6+6+1 or 5+4+4 or 3+4+6 on any of the dice but you need to role three 6s to get 18. Though as I said, to me those would be all as likely as each other, as the outcome of one dice doesn't affect the outcome of the others, so each dice is running off a 1/6 chance to come up whatever.

This is kindof true, but bear in mind it rarely matters which dice gets which number.  So the chances of rolling three sixes are 1/216, while the chance of rolling a 1, a 2 and a 3 are 1/36 because there are 6 different ways of getting a 1, a 2 and a 3.

[JoshuaFH]

For 6-sided dice, a quick and dirty way to find the most common result is to add the lowest possible result (all 1s) and the highest possible result (all 6s) and divide by two. If you get a fractional result, that means you have two equally common midpoints, since dice produce integers.

2d6: 2-12 (which sum to 14. Divide by 2, you get 7.)
3d6: 3-18 (which sum to 21. Divide by 2, you get 10.5. So 10 and 11 are your most common results.)
4d6: 4-24 (which sum to 28. Divide by 2, you get 14.)

and so on.

The odds are independent, so the expected total for n rolls can be calculated by n * M, where M is the mean of a single roll. For a 6-sided die, the mean is (6+1)/2 = 3.5, so for n 6-sided dice, the expected total for n rolls is 3.5*n.

Ah, I should've deduced something to that nature, but it didn't occur to me. For die of different sizes being rolled together, could you basically just add the mean of each individual die? So, like for a d6, d10, and d20, the most likely outcome would be 19.5 (3.5 + 5.5 + 10.5)?

[Nadaka]

The average roll of any die is (min + max) / 2.

For a standard d6, that is 3.5.

When rolling multiple dice, the most likely result is the sum of the average (or the nearest whole numbers to that sum).

3d6 most likely is 10 or 11 (10.5 average)

4d6 most likely is 14.

2d8+5d20 most likely is 4.5+4.5+10.5+10.5+10.5+10.5+10.5 = 61.5 or 61 and 62.

[Darvi]

It gets tricker when you say "Roll mdx, pick the largest/smallest n"

[Another]

"Roll n times and discard the m least results" makes the bell curve asymmetrical with no easy way to find the maximum.

On the other hand stuff like 1d100+1d6 will not have a single maximum at 50.5+3.5 but a long flat top of equally probable numbers from 7 to 101.

[Darvi]

Well yeah, but on average you're still getting 54.