Over and Under: The Electoral College in 2016

Come the evening of Tuesday, November 8, nothing on Earth will be as important as the Electoral College. Yeah, some folks are going to get out their quadrennial placards bitching about the college and then quickly stow them before they get any splinters.

Except not this year because about 60% of the electorate is going to be preemptively cheering a Hillary Clinton win and the other 40% will find that like most collective bubbles of ignorance, their collective bubble of ignorance will be popped by The Very Serious and Adult World of Voting in America, a world created by two of the greatest enemies of willful ignorance: math and law.

Actually, cancer edges out law, but law's still third.

Skiltao's Dice: Return of the Comebackening

Some weeks ago, I went over some numbers that Skiltao mentioned on his blog (Part 1 & Part 2). I didn't do any rigorous mathematical analysis on them—what I did was more akin to recognizing the geometric qualities of dice probabilities—but I did manage to illuminate the qualities of the dice rolls and why certain relationships existed between them. Whenever I was done, Skiltao asked about alternative dice relationships, for example, if different-sized dice would have similar relationships or how dice which have repeated numbers would be affected.

The first question is straightforward. The odds of a tie between, say 1D10 and 3D4 being expressed as a certain value on a 1D10 + 3D4 roll is easy to test.

1D10 ranges from 1 to 10, while a 3D4 ranges from 3 to 12. The odds of a tie are the compounded probabilities of a set of values on the table below:

1D10      1D4 #1        1D4 #2        1D4 #3
[ 3]      (1)           (1)           (1)
[ 4]      (2)           (1)           (1)
[ 5]     (2,3)         (2,1)          (1)
[ 6]    (2,3,4)        (2,1)         (1,2)
[ 7]     (3,4)        (3,2,1)        (1,2)
[ 8]     (3,4)         (3,2)         (1,2)
[ 9]     (3,4)         (3,4)        (3,2,1)
[10]      (4)          (3,4)         (2,3)

Skiltao's Data, Part 2

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. 

This is part two. Part one is here, but they're mostly unrelated. The second part of Skiltao's post was how the odds of 1D6 beating 2D6 were equal to the odds of rolling a 6 or less on 3D6. In order for that to happen, you've got to roll a three or greater on the 1D6 and anywhere from two to five on the 2D6. I'm expressing these probabilities as (3-6) and [2-5], respectively. 

Really though, you need pairs of these odds:
1D6  2D6  1D6  2D6  1D6  2D6  1D6  2D6
(3)  [2]  (4)  [2]  (5)  [2]  (6)  [2] 
          (4)  [3]  (5)  [3]  (6)  [3]
                    (5)  [4]  (6)  [4]
                              (6)  [5]

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.