How to calculate the percentage of remaining election ballots that your candidate needs to win in order to catch up

I’m a math nerd and a political nerd, so I came up with the following equation tonight for other nerds. Regardless of who you want to win the presidency, this equation will tell you how to calculate the percentage of remaining ballots that your candidate needs to win in order to catch up to the other cat. Anything more than that percentage, and your candidate wins! 🙂
0.5 * (Ballots Remaining – Deficit) + Deficit
                     Ballots Remaining
You just need to know the number of remaining ballots to be counted and the deficit.
For example:
As of tonight, Thursday, 11/5/2020, in Arizona, Trump is behind Biden by about 47,000 votes with about 204,000 ballots remaining.
For simplicity, let’s just get rid of the thousands (000’s). The calculation will be the same.
   0.5 (204-47) + 47
= ——————–
       78.5 + 47
= ——————–
= 0.615
So, Trump needs to win 61.5% of the remaining ballots to pull even. More than that, and he wins.
Conversely, Biden needs to win more than 38.5% of the remaining ballots to win (1.0 – .615 = 0.385)
Math class is over. Drink a beer. Or a beverage of your choice.

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Martin Gardner: Mathematical Games : A Cipher I broke that Defeated Poe

Peter Bakke : I solved this cipher in 1987 while working on my master’s degree in systems science at SUNY Binghamton (University of Binghamton, now).

“Ge Jeasgdxv,
Zij gl mw, laam. xzy zmlwhfzek
ejlvdxw kwke tx lbr atgh lbmx aanu
bai Vsmukkss pwn vlwk agh gnumk
wdlnzweg jnbxw oaeg enwb zwmgy
mo mlw wnbx mw al pnfdcfpkh wzkex
hssf xkiyahul. Mk num yexdm wbxy
sbc hv wyx Phwkgnamcuk?”

In 1839, in a regular column Edgar Allan Poe contributed to a Philadelphia periodical, Alexander’s Weekly Messenger, Poe challenged readers to send him {cryptograms (monoalphabetic substitution ciphers), asserting that he would solve them all “forthwith.” One G. W. Kulp submitted a ciphertext in longhand. It was printed as shown above in the issue of February 26, 1840. Poe “proved ” in a subsequent column that the cipher as a hoax—”a jargon of random characters having no meaning whatsoever.” In 1975 Brian J.Winkel, a mathematician at Albion College, and Mark Lyster, a chemistry major in Winkel’s cryptology class, cracked Kulp’s cipher. It is not a simple substitution — Poe was right — but neither is it nonsense. Poe can hardly be blamed for his opinion. In addition to a major error by Kulp there are 15 minor errors, probably printer’s mistakes in reading the longhand. Winkel is an editor of a new quarterly, Cryptologia, available from Albion College, Albion, Mich. 49224, at $16 per year. The magazine stresses the mathematical and computational aspects of cryptology. The first issue (January, 1977) tells the story of Kulp’s cipher and gives it as a challenge to readers. So far only three readers have broken it.


The width of the Veil Nebula is 3 fingers

So how does one calculate the width (in degrees) of any distant object – from say the Veil Nebula 2100 light years away to the width (in degrees) of that mountain just 21 miles away? Simple. But you will need a scientific calculator. Here’s one:

English: A wide-field image of the Veil Nebula...
English: A wide-field image of the Veil Nebula, made as a colour composite from individual exposures from the Digitized Sky Survey 2. The field of view is 4.2 x 4.4 degrees. (Photo credit: Wikipedia)

Using the information given to you, say in an article about the Veil Nebula , calculate a ratio of distance over width.

In this case, we are told the Veil Nebula is 2100 light years away and it is 110 light years in width. It is very convenient that they gave us distance and width in the same units (light years), else we’d have to do that ourselves.

So, 2100 / 110 = 19.1

Using your scientific calculator (see above), enter 19.1 and hit the COT function (COTANGENT).

The answer of 5.9 (degrees), or about the width of your three fingers extended at arms length towards the fabulous night sky – in this case in the direction of the Swan Constellation.

Happiness is Astronomy !

(Conversely, of course, if you estimate the distance of an object and also estimate its width (in degrees), then you can determine the approximate width of the object {in feet, miles, light years, etc.}).

The Veil Nebula. What’s a few trillion miles between friends?


Article Source:

I believe someone’s math is incorrect in this article. I’m not spoiling the fun, just correcting the record. The image is truly breathtaking and wondrous! But Numeracy counts. :~)

63,000 just didn’t seem big enough, and I was right. Somebody’s calculation re: the width of the Veil Nebula is off by a factor of 100.

A light year is approx 6 trillion miles, or 6,000,000,000,000 miles.

At a width of 110 light years, the Veil Nebula is therefore about 660,000,000,000,000 miles across.

The sun is about 93 million miles away, or 93,000,000.
If you do the math (660 trillion / 93 million), 110 light years across is not about 63,000 times the distance between earth and the sun as claimed in the article, but instead is about 7,000,000 (7 million) times! An even more incomprehensible number than 63,000, I know.

But, what’s a few trillion miles between friends?