Skiltao's Data, Part 1

I saw Skiltao talking about odds on his blog and asked myself why that would happen. It's been a while since I did much with Battletech and had to wrestle with numbers like this (because it's always Battletech). I'd suggest reading and understanding his original post before continuing because this thing runs long without me recapping anything. Besides, Skiltao's a pretty good guy and you should read his(?) blog.

The first question is why the odds of rolling a tie on 1D6 versus 2D6 was equal to the odds of rolling a '7' on 3D6. Well, to make a tie, 1D6 has to equal 2D6. That means that odds are equal to the odds of rolling 2-6 on 1D6 and 2-6 on 2D6. Instead of actually dealing with the fractions, I'm going to express these sets of probabilities as (2-6) and [2-6] respectively. I'll do this for all of my sets of probabilities for the rest of this article, forgoing any subsequent use of parenthesis for the sake of clarity (mostly). I'm sure guys who've been through probabilities classes more recently than I have can express this more professionally and/or better than I am, but I'm what you've got.

Okay, the next step is to find the odds of getting a '7' on 3D6. One way to work with 3D6 is to present it as 1D6 + 2D6. What combinations of 1D6 and 2D6 make '7'? (1,6], (2,5], (3,4], (4,3], and (5,2]. The odds are: (1-5) [2-6]. You might notice they're almost like the sets for getting a tie, but not quite.

The next step is realizing the bell curve in 2D6. If you've hashed the probabilities for Battletech at all, this is pretty fundamental for you, but let me review for people who don't have obsessive tendencies related to wonky board/roleplaying hybrid games with systems older than the average human being. You can express probabilities like so:

2D6       1D6 #1        1D6 #2
[ 2]      (1)           (1)
[ 3]     (1,2)         (2,1)
[ 4]    (1,2,3)       (3,2,1)
[ 5]   (1,2,3,4)     (4,3,2,1)
[ 6]  (1,2,3,4,5)   (5,4,3,2,1)
[ 7] (1,2,3,4,5,6) (6,5,4,3,2,1)
[ 8]  (2,3,4,5,6)   (6,5,4,3,2)
[ 9]   (3,4,5,6)     (6,5,4,3)
[10]    (4,5,6)       (6,5,4)
[11]     (5,6)         (6,5)
[12]      (6)           (6)
I could have expressed these as (1-2), (2-3), etc., but I wanted to demonstrate the correlation between physical size and probability as well as each individual combination of 2D6 rolls.

Here's the crazy thing: because of the shape of this symmetrical probability diamond, the odds from the top and bottom are the same. That sentence isn't clear, so let me explain:
[1] = [12]
[2] = [11]
[1,2] = [12,11]

Now, I wouldn't be blowing your mind if I told you that the 1D6 odds from the tie set, (2-6) are equal to the 1D6 requirement for the 3D6 '7' set (1-5), but I needed to establish that that the relationship (Minimum + X) = (Maximum – X) holds for all these dice combinations.

Because here's where step two comes in; because there's always a "decreasing" 1D6 set that mirrors the "increasing" 1D6, whenever you add it to the "increasing" 2D6 set, there's always a single sum that's the result of that.
For example: 4D6 and 5D6
Odds of a tie result:
(5-24) [5-24]

We know:
( 5) = (23)
( 6) = (22)
...
(24) = ( 4)

Therefore:
(5-24) = (4-23)
(4-23) [5-24] = (5-24) [5-24]

Finally:
4 + 24 = 28
5 + 23 = 28
6 + 22 = 28
...
23 + 5 = 28

Which checks with Skiltao's formula: odds of a tie between ND6 and (N+1)D6 = odds of rolling N*7 on (N+N+1)D6.

Indeed, it works with any two D6-based summation rolls.
For example: 243D6 and 587D6
Odds of a tie result:
(587-1458) [587-1458]

We know:
(587-1458) = (243-1114)
(243-1114) [587-1458] = (587-1458) [587-1458]

Finally:
243 + 1458 = 1701
...
1114 + 587 = 1701

So the odds of tying 243D6 and 587D6 is equal to the odds of rolling 1701 on 730D6.
Alright, next week I'm going to cover the "less than" bit because I'm seriously over words this week.

And I swear to the gods, popping the Enterprise's registry number was just a spooky coincidence.