What's so special about 18,181

[scottxyz]

This is not a blockbuster revelation, just a little daydreaming...

I'm a computer programmer and a sort of amateur mathematician, so I'm always adding and subtracting and multiplying and dividing numbers to pass the time.

Especially when I'm sitting on the stationary bike at the gym, hoping and praying that the time will go by quicker. If the timer is set for 10 minutes, I start counting down the fractions, thinking: at 1:00 I'm 10% done, at 2:00 I'm 20% done, etc. When I'm 20% done, that's one fifth, but even better, I can tell myself I just have to do what I did *four* more times. "One-fifth" doesn't sound as good as "four more times".

What's the point of all this? The point is, when you're starting from a round number (a multiple of ten, say, like 10 or 100 or 1000), and you chop off a tenth or you add on a tenth, you end up with math problems involving numbers one away from 10 - ninths or elevenths. When you start off with a nice round number and you chop off or tack on a fraction, you no longer have that nice round number to work with anymore - the original nice round number is bigger or smaller by that fraction, and all your previous fractions are out of whack now.

A typical situation is where you bump up a number by 10%, like when you go from 100 to 110. Now if you've used up 10 units of that, you're looking at 1/11, not 1/10.

OK, so what's your point? My point is that ninths and elevenths come up a lot when you're starting with a multiple of ten and bumping it up or down by 10% - something we do in a lot of situations.

When you write out 1/11 as a decimal, you get 0.09090909... which looks really cool. The fraction 2/11 looks like 0.1818181818... which also looks pretty cool.

Which brings us to the mysterious electoral number 18,181 which crops up in so many suspicious elections in Texas. Some people have ventured an explanation by noticing that a=1 and h=8 in the alphabet, giving "ahaha".

http://www.scoop.co.nz/mason/stories/HL0307/S00064.htm

I think there's might be a more mathematical way of looking at this. 18,181 is 200,000/11 (if you truncate it, instead of rounding it. Rounding would give 18,181.81 = 18,182 - but let's just assume Republican hackers are too stupid to know that rule about rounding up when it's over a half, and they just truncate all the time.)

That, in my opinion, is what's so special about 18,181. It's not a weird number because it's a palindrome or because it maps to ahaha - it's an important number because it results when you're dividing a nice round number by 11 - a situation that often happens when you're incrementing or decrementing some starting number by 10%.

[silverweb]

1. Thank you!

This episode in our election fraud history has not been getting nearly enough attention. Your post is very, very relevant and telling.

Now, if only we could get the media to pick up on the story....

[NJCher]

2. very insightful

I just knew something like this would come along and 'splain that.


Cher

[DS1]

3. 10%

Is very easy to test and debug too, lazy programmers.

Oh yeah, did you mention it was a prime number?

[Aaron]

4. Good observation!

BTW I do the same thing with regard to workouts and timing. Although I don't think I ever would've thought of this 18181 rounding thing. That's sharp work.

[Trajan]

5. I am NOT a mathematician ....

But, having dabbled with puters and programming over the years, I recognize 18181 as a 'marching' binary bit pattern, where only the first and last bit in each byte toggles ....

1 = 0001
8 = 1000
1 = 0001
8 = 1000
1 = 0001

WHY election counts in separate races would simultaneously result in 18181 CAN be simply the natural result of honestly adding the votes, and getting 'lucky' ....

It could also indicate a common process operating on different data by a common method, ... in this case a 'blunt force' replacement of REAL count data with a predefined value: .... I say replacement, because integer operators, if used against different integer values, would bring different integer results .... the fact that the SAME result is present in three places would indicate EITHER a REAL count of a common value, or REAL counts replaced by a common value by an identical process ...

If one were to look at OTHER races in different locations, can one find ANY three races ANYWHERE that match precisely ? ...

18181, could also result if a mechanical counter were somehow disabled by blocking sensing paths for bit 2 and 3 in a discrete 4 bit (single byte) path .... I dont know the nature of the balloting and tabulation mechanisms in those cases, so I dont know if such devices are used ....

A GOOD test would be: .... "did ALL three races occur on IDENTICAL voting and counting equipment ?"

IF they did NOT, then we could discount the 'common process' notion, since it would be unlikely that the SAME idiosyncratic process would be programmed on differing platforms ...

IF they are identical voting systems, then it is possible that the simultaneous 18181 results could be EITHER the result of value replacement through a common process, or a stroke of luck ....

Ever read "The Report on the Barnhouse Effect" by Kurt Vonnegut? ...

Where Professor Barnhouse learned how to use "Dynamopsychism" to roll TEN sevens in a row in a crap game ? ....

An excerpt: ... http://www.geocities.com/ResearchTriangle/Node/4729/barnhouse.html

"From time to time Private Barnhouse was invited to take part in games of chance by his barrack mates. He knew nothing about the games, and usually begged off. But one evening, out of social grace, he agreed to shoot craps. It was terrible or wonderful that he played, depending upon whether or not you like the world as it now is.
"Shoot sevens, Pop," someone said. So "Pop" shot sevens -ten in a row to bankrupt the barracks. He retired to his bunk and, as a mathematical exercise, calculated the odds against his feat on the back of a laundry slip. His chances of doing it, he found, were one in almost ten million! Bewildered, he borrowed a pair of dice from the man in the bunk next to his. He tried to roll sevens again, but got only the usual assortment of numbers. He lay back for a moment, then resumed his toying with the dice. He rolled ten more sevens in a row.
He might have dismissed the phenomenon with a low whistle. But the professor instead mulled over the circumstances surrounding his two lucky streaks. There was one single factor in common: on both occasions, the same thought train had flashed through his mind just before he threw the dice. It was that thought train which aligned the professor's brain cells into what has since become the most powerful weapon on earth.
The soldier in the next bunk gave dynamopsychism its first token of respect. In an understatement certain to bring wry smiles to the faces of the world's dejected demagogues, the soldier said, "You're hotter'n a two-dollar pistol, Pop."

-snip-

Yeah: .... three 18181 ballot counts in one day in one state is "hotter'n a two-dollar pistol, Pop" .......

[Merlin]

7. Good God! Don't we know that yet?

"did ALL three races occur on IDENTICAL equipment ?"

I mean you just asket the MOST critical question. With all the manpower on this board, and all the hoopla over the probable gaming of this election, it's hard to believe nobody has found this out yet.

Yet, I'm sure if they had, we'd know about it by now.

Interesting about the 'marching' binary bit pattern.

[Sick of Bullshit]

14. Since all these results are from Comal County, Texas,

I would assume they occurred on identical equipment

http://www.co.comal.tx.us/election_results2002.htm

[sandnsea]

46. Optec III Eagles

http://pub103.ezboard.com/fsoldiervoicefrm49.showMessage?topicID=12.topic

A scanner machine with penciled in ballots, a paper trail. The whole county went Republican with votes like 18000 to 5000 so I hardly think there was any tampering swinging those counties Republican. It would be completely unnecessary, unless somebody were just trying to find a county nobody would notice to get ready for 2004. I don't know, just trying to swing something out there.

[Must_B_Free]

49. There were two others in the nation that also had 18181

I recall seeing a laundry list and it it little recognized because it is so unbelievable, but there was the occurrence of the three 18181s in one race in texas, AND there were two other 18181 numbers each in a different state, if I recall correctly.
oh, here

TWO MORE 18,181 vote counts
--------------------------------------------------------------------------------
Yep! Both republican. The fix is in!

miboecfr.nicusa.com/elect...10000.html
Candice Miller (R) Lapeer Co, MI, 10th District Congressionsl Rep
www.mdarchives.state.md.u...02del.html
Michael D. Smigiel, Sr. (R) Dist 36 Maryland House of Delegates

[Mairead]

9. Just a bit of pedantry

A byte is 8 bits. 4 bits is a nybble.

And 18181 is not a repeating bit pattern, it's 100011100000101.


Sorry for the cold water, and I apologise if I misunderstood your point.

[Trajan]

27. My bad ....

Of course: ... this is conjecture, so I take criticism in this issue 'lightly' ....

But, these patterns DO exist in binary computing systems, and are peculiar and recognizable because of this .....

Your right: .... 4 bits is a nibble, 8 bits is a byte ....

Unfortunately: ... I am old enough to remember when 8 bits was a word ..... and 4 bits was a byte .... That is my understanding at any rate, right or wrong: ...

But YOU are right: 8 bits is a byte .....

Next: I was looking at EACH digit separately, NOT the entire number as a singular integer entity ....

Computing systems are arranged with data paths that are 4,8,16,32,64 and 128 bits wide ... arranged in groups of 4 bits each .....

Binary coded decimals and binary coded hexadecimal are the STANDARD data types in computer equipment, and certainly these systems can and do convert between binary, hexadecimal, decimal and BCD types when needed ... so one can switch from one numeric type to another quite easily .....

Cripes: ... I have entered data in the 4, 8, 16, and 32 bit hexadecimal digit , and 3 bit Octal digit formats DIRECTLY into the old style computer front panels, to initiate boot programs and to operate within executable programs, entered the operands and operators by hand .....

When I see '18181' .. I see '0001 1000 0001 1000 0001' as binary coded digits in hexadecimal format, where EACH group of 4 bits represents a single digit ...

From the compiler level: 18181 = 100011100000101 .... this is a direct decimal to binary conversion as a whole integer .... which is what you expressed ....

from the MACHINE level: .. 18181 = 0001 1000 0001 1000 0001 in binary coded hexadecimal ....

DEEP in the bowels of a CPU, in the Arithmetic Logic Unit (ALU) ..... lies rows of symbolic logic devices (logic gates) in configurations that operate on data mechanically .... old forms like "half adders" and "full adders" are STILL used within these sections, to perform basic addition functions, and THEY operate on binary digits in 4 bit groups ....

Addition is the PRIMARY FUNCTION of a CPU: .... Addition, Subtraction, Multiplication and Division are ALL performed using the Addition logic in the ALU, using bit level logic executed is parallel groups of 4 bits each ....

Its been a long while since I have even considered the architecture of the CPU and ALU, but it is my understanding that these mechanical structures STILL form the heart of EVERY modern CPU ....

They do so in binary coded hexadecimal digits ..... where 18181 = 0001 1000 0001 1000 0001 .....

AGain: ... this is stil conjecture: ... but these patterns are fascinating and peculiar to computing systems, which makes these results stand out ......

[retread]

73. More pedantry

Don't we use 16 bits in this case? If so there should be 16 digits. If I recall the distant past doesn't it start with 0. We then have 2^0 to 2^15. For example 0000000000000101 is 2^2 + 2^0 or 5.

[papau]

19. Excellent - how to get 18181 in binary sounds like the path used

:-)

But I like the C programing approach. The playing with bits is a PDP-1 trick that GOP folks are not up to.

My new programing guru god - scottxyz - is the title of RogueTrooper's post where he gives the C program - and that I like a lot.

[ElsewheresDaughter]

31. 18181 in binary looks like this

0011000100111000001100010011100000110001

here is a fun tool...binary it's digitalicioushttp://nickciske.com/tools/binary.php

democraticunderground.com looks like this 01100100011001010110110101101111011000110111001001100001011101000110100101100011011101010110111001100100011001010111001001100111011100100110111101110101011011100110010000101110011000110110111101101101

[Merlin]

6. Still don't get it. Could you translate further, please.

Although I'm a programmer, too, I'm no math expert. I understand the numbers you're presenting. I just don't grasp how this might apply to a hacked election.

Are you suggesting that the typical hack might want to bump the numbers for the Rethug by 10%? Ok. Makes sense, I guess. But then how does this relate to the Texas type results?

Or maybe I'm just missing the obvious.

[foo_bar]

11. it's astrology for engineers

It says more about the salience of repeating sequences than their distribution. If three gerrymandered Texas districts yield the same result it isn't because 18,181 human integers had to pass 20 Arbys on the way to 11 polling stations. But it's fun to speculate at their expense.

[DUreader]

15. I'm glad you are keeping this thread at the top, but...

what are you talking about?

Are you making a joke?

I don't get the astrology reference here.

[nolabels]

8. I will more than admit that I am stupid about some (or many) things

But I try not take apart or fix things that I couldn't figure out how they work. The statement "Rounding would give 18,181.81 = 18,182 - but let's just assume Republican hackers are too stupid to know that rule about rounding up when it's over a half, and they just truncate all the time.)". This brings me to the conclusion that is does have a trunk


http://www.funny--pictures.com/elephant_legs.html

[izzie]

10. Gee I have always counter things but only items.

So many flags out, so many TV dishes, so many red cars, so many yellow birds, and now may be I should start div them into some thing? Do you think we are crazy? I have been told it is to keep your self calm.

[DoYouEverWonder]

12. Here's a list of the 5 candidates

Who won by 18181 votes in 2002 election.

DANNY SCHEEL (Texas): 18,181 votes (Comol County)
CARTER CASTEEL (Texas): 18,181 votes (Comol Country)
JEFF WENTWORTH (Texas): 18,181 votes (Comol County)
CANDICE MILLER (Michigan): 18,181 Votes (Lapeer County)
MICHAEL SMIGIEL (Maryland): 18,181 Votes (St. Anne's)

In addition, all five won and all five were Republicans. Even odder 3 of the 5 came from not just the same state but the same county!


edit: added more county info

[foo_bar]

13. what's odd about it?

I think that part of Texas has the vote machines with the lever on the left and one on the right. Assuming right-handed folk have an 9:20 torque ratio to the Democratic lever, well you know how it goes.

all five won

If they didn't they'd be candidates who lost by 18181 votes.

Even odder 3 of the 5 came from not just the same state but the same county!

If 18181 Rush-bots voted the same party line, this is an extremely accurate count.

[DUreader]

16. Another kinda confusing reply...

Is this another joke?

[Trajan]

28. Blah Blah .....

Funny thing is: those who do NOT believe tampering has occured COULD just pick up and move on to other climes more suitable ....

Yet they instead hang out, making their pitifully small and insulting pronouncments .... as if only to hear themselves being insulting .....

Blah blah ....

[RogueTrooper]

17. scottxyz

if you do this in C ( c does not round up numbers when you cast )

float f_value;
int i_value;

f_value = 200,000 / 10;
// this would make f_value = 18,181.81 ( and on... )

now, if you cast that value into an integer...

i_value = ( int )f_value;

wound make l_value 18,181 not(!) 18,182

then there is no rounding. C casts downward, taking no account for the mathematical correctness.

VB was orginally written in C. As was Access.

[DUreader]

18. Another indicator of a possible programmer with a lack of accounting

acounting knowledge where the standards for rounding are clear.

Except I learned that in 8th grade science/math where it is also standard.

[RogueTrooper]

20. Indeed

it looked like his did a straight casting, rather than something mathemaatically correct.

[Cronus Protagonist]

29. 200,000/10 = 20,000 not 18181.81

I think you meant 200,000/11



http://kucinichforpresident.com - Kucinich Is The One
http://cronus.com/prayer - One of Kucinich's speeches

http://brainbuttons.com/home.asp?stashid=13 - cute little buttons
http://bushspeaks.com - sardonic political toons
http://cronus.com - enlightening and educational liberal fun

Conceptual Guerilla

[Skittles]

21. hello scottxyz

someone mentioned in another thread that you are an impressive poster. I have to agree! :D

[myrna minx]

22. Thank you so much for your insight.

As a keeper of books, I feel your info is spot on. :hi:

[qb]

23. Could be a screwed up patch

Check out Bev Harris's report on the "patch" fiasco in Georgia just prior to the 2002 elections.

http://www.talion.com/lies.htm

It's long, but worth the read. In short, Diebold's software was so screwed up, they had people running all over the place frantically installing patches in voting machines. There was no regard for security or verification that the machines worked properly after installing the patches.

[THUNDER HANDS]

24. this kinda stuff

makes me which I would have paid attention in math class.

:)

[Eloriel]

33. That kinda stuff

makes ME understand all over again why I didn't. :evilgrin:

Eloriel

[laylah]

68. And for me

it is one of the reasons I became a Social Worker......no advanced math required! :silly:

jenn

[homelandpunk]

51. and perhaps in english class??

haha...i'm buzzed so I am just trying to rib you in fun. I coouldn't resist when I saw "which" for "wish" ...lol...oh come on...i'm trashed...it is kinda funny if you think about it and know i'm not being a prick, just a friend. Hey, I'm almost readty to say: (in garbled voice) "Iiii love yewww guys." hahahaha

[Ivory_Tower]

25. I dunno...I still don't get it

Not to throw cold water on this, but I'm not making a full connection here. Let me see if I understand:

Take a number, increase it by 10%, take the difference between the new value and the original number, and divide it by the new value, then multiply by 2. You always get 0.18181818....

That much I get. But that's a ratio. I don't see how you can go from a ratio back to straight difference. In other words:

y=0.1*x
z=x+y
2*(y/z)=0.18181818... (for any value of x)

But this doesn't guarantee that z-x=18,181.

There's still a disconnect as to how a ratio can be transformed back to a difference between the original value and a modified value.

Even if it was a bug in a nefarious patch, why would it only trigger in five districts?

Some additional questions I'd have before I claimed there was something definitely wrong here:

What were the vote totals in those five suspicious districts?
Were they all identical?
What were vote differences like in previous elections?
Has nobody ever won by 18,181 votes before?
Have there never before been multiple elections where the winners won by the same margin?

I don't know the answers to those questions, and while the 18,181 number looks intriguing, I don't see a smoking gun yet. Not writing it off, but I'm not convinced.

Sorry.

[nolabels]

26. "Sorry" was a word that echoed through the barracks

The trainee would say "Sorry Drill Sergeant" and the Drill Sergeant would say "Well, you certainly are a sorry thing". Not in the literal sense (I hope) but just in a motivational sense. I never got to learn that much about math either, but it's so interesting some times. But since I am a simple layman and have pulled a couple of slot machines (before I found how rigged they were), the possibility of the number 18181 coming up 5 times in a row has to be astronomical. But like I said am no math wiz, but it does seem like common sense

[Ivory_Tower]

34. Ummm....

I guess that was an insult, but I don't understand it, so I suppose I'm okay with it. :)

Yeah, if 18181 came up five times in a row, that would be suspicious, but remember there were literally hundreds of elections. I wouldn't be surprised if there were five elections where the winners all had 19,328 votes.

One correction to my post above -- the five candidates in question had vote totals of 18,181 and I was saying that they won by 18181. Oops. (Sorry?)

The fact that three of the candidates are all in the same county certainly makes me raise an eyebrow, but I still don't see how 18181 votes can relate back to a 2/11 ratio. For the three races in question:

County Judge
Rep: 18181
Dem: 5547
Total votes: 23728

State Senate District 25
Rep: 18181
Dem: 4988
Lib: 723
Total votes: 23892

State Rep District 73
Rep: 18181
Dem: 5303
Total votes: 23484

I got these numbers from the website provided in another post on this thread. btw, there were a total of 36 races listed on that page, if I counted correctly.

Looking at these numbers, I still don't see a formula that would result in these three candidates getting exactly 18181 votes. If anything illegal occurred in those races, I don't think it was automatically programmed. Given Bev Harris' revelations about the ease with which vote totals can be manually changed, however, I wouldn't be horribly surprised if those numbers were manually entered.

But barring proof of that....
(Edit: Argh, hit "Post" instead of "Preview". Just wanted to add something to the effect of I'll leave it to other people to find a connection here, because I just don't see it. Still leaving open the possibility of post-election tampering, just not in-line code tampering -- at least not for these cases.)

[ret5hd]

40. BevHarris, if i had any money, and i was a gamblin' man...

i would be grepping (ask one of your programmers) all 40,000 files for a variation of just such an equation.

then, i would wonder how those files w/ nothing but a number in it (253 or 57 or whatevr they were)

then i would look for vote results that corresponded (differences or totals) to .181818 * 253 or 57 (or whatever they were)

but then, if i knew what i was talkin' about, i would either know that what i was askin' was too "variable" to find or i would do it myself for you.

keep it up.

[Eloriel]

30. Just a little

:kick:

Eloriel

[DrBB]

32. COOL POST man. I always thought it had an octal

...thing to it, but of course you wouldn't GET the number "8" in octal, just 0-7. So I knew that wasn't it, but I figured if it was something generated by the system--noise, or some kind of flaw because of insufficiently-sophisticated tampering--that it must have a mathematical explanation of some kind. You're just pointing out that some idiot was trying to decide a nice easy way to add votes to his guy or subtract 'em from the other guy, so he just did a simple little 10% fiddle. Sure, 10%, why pick something hard?

Brilliant, sez the English Major. Has Bev or the Black Box Hack crew seen this?

[BevHarris]

35. Correcton on the news reports -- they won WITH 18181 not BY 18181

Total registered voters in Comal County about 60,000
Turnout was about 40%
Three candidates won with exactly 18181 votes each

Go HERE and see for yourself -- something else I noticed about the results was that there was significant fluctation in the number of votes cast in the various races.

http://www.co.comal.tx.us/election_results2002.htm

[Philosophy]

52. But the races weren't even close

If the races were close it might have been a little suspicious. But all those guys won by huge margins. I still don't see why this couldn't just have been a coincidence. Improbable things happen all the time.

[punpirate]

36. One question on this...

... and other posts of a similar nature. Votes, by definition, are whole integers. A single vote, a block of votes, or an aggregate vote total still have to be natural numbers, that is, whole non-negative numbers.

My limited understanding is that an arithmetical unit can get into trouble (such as the rounding errors in some Pentium chips a while ago) when it has to split a decimal number into a natural number, store that result, then multiply the decimal portion to make a natural number, do whatever calculation is required, then divide again by the same amount and sum the result with the stored natural number.

That would be necessary, say, to provide a percentage of registered voters actually voting, or the percentage of votes won by a candidate. But, those rounding and casting errors would not come into play when doing simple addition of votes, unless the resulting number were too high to be stored in a single memory address (certainly not the case with the numbers cited in these elections).

So, at what point in the calculation by the arithmetical unit might it be required to do division or multiplication when doing simple addition of natural numbers? Anyone know?

I guess what I'm getting at is that adding up vote totals should be a very simple process to program and for the arithmetical unit to compute, since votes cannot be anything other than natural numbers. Any complication in that process using mathematical operations other than addition has to be the result of either sloppy programming, a chip-level hardware error, or intentional mischief.

Cheers.

[Pobeka]

37. This is similar to same birthdates in a group of 30 people

I don't remember exact numbers here, but if you randomly select 30 people whose birthdays are randomly distributed between 1 and 365, what are the odds at least 2 of those selected have the same birthdate?

At furst blush you'd think it's near impossible. But when you break down the problem, it turns out the odds are pretty good that 2 people will share the same birthday.

The twist on this one, is now you have *many* more people (i.e precincts), and *many* more birthdays (i.e. number of votes cast for a candidate).

Let me get out my calculator and do the numbers on this (actually, it will take a small C program). I'll get back later with the answer (and code for others to review).

p.s. Can someone tell me the total number of precincts we have to consider?

[Pobeka]

39. A link explaining the birthday problem

http://mathforum.org/dr.math/faq/faq.birthdayprob.html

[jimmynochad]

42. precinct numbers in US approx 192000 nt

[Pobeka]

48. Some numbers

Say you could have up to 2 million votes for a candidate in a precinct.

Here are the probabilities that 2 precincts will have exactly the same number of votes, based on the total number of precincts:

# precincts odds
2000 0.63
4000 0.98
6000 1.00
...

Now our problem is a little different here, but I'll bet the chances of multiple precincts with exactly 18,181 are going to be within the range of reason.

The reason I asked about other duplicate numbers, is because we should see other duplicates showing up -- if we don't then something may well be fishy.

--

[lynndew2]

66. I am sorry but, The #s are all from one district

So it seems if they voted on a party line it wouldnt be that wierd.

Total registered voters in Comal County about 60,000
Turnout was about 40%
Three candidates won with exactly 18181 votes each

[ParanoidPat]

67. You're missing the fact....

....that the districts these people ran for were in the same county but the voter pools that voted for each candidate were different!

[ConsAreLiars]

43. Calculating the odds

"This is similar to same birthdates in a group of 30 people"

I have just been looking at the page with those numbers and had the same hunch. I've forgotten the math, but the variables can be defined in several ways. This is tricky because vote-totals are less random than birthdates. But anyway, it looks to me that:

These results combine several precincts and machines into distict-wide totals. So any wierdness is at the level of the county-wide counting process.

If it's actually one congressional district then we have to remember that there were 535(?) chances to turn up a threesome. But given the size it is probably only a part of one district.

If we look at the size of the sample, about 25000 voters, and figure there were 50,000,000 votes nationwide, then this means there are 2000 more cases where such a coincidence might occur. This means that if the odds of 3-matching numbers in one sub-district are 2000:1 against, that this is just exactly what coincidence theory would predict.

As to the numbers, that's trickier. We see that in contested countywide races R's got around 17,000-19,500, so there are around 2500 possible "birthdays" in the "year." (The difficulty is not knowing if this range is typical of the range of votes in other districts or subdistricts.) But whatever the best guess, it is certain that throwing lower and higher numbers into the calculations would be making the obviously false assumption that vote-totals of 0 or 1 or 2 or 3... are just as commonly found as 19000,19001,....

Anyway, that's as far as I got. The important assumptions are about the range of numbers that replace 365, and the the number of iterations to replace 30 before the likelihood exceeds 50%. Trying it with 2500 and 2000 should be interesting. (Unless I've gotten it all wrong.)

A complicated puzzle, but an interesting and instructive one.

[Pobeka]

45. For the nerds in the group, source code

For the birthday problem WHICH IS NOT THE SAME PROBLEM WE HAVE!!

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

double paired_odds(long p, long n)
{
long i ;
double odds_different = 1.0 ;

for(i = 1 ; i < n ; i++)
odds_different *= (double)(p-i) / (double)p ;

return 1.0 - odds_different ;
}

int main(int argc, char **argv)
{
long pop_size ;
long n_selections ;
double p ;


/* uncomment this to check against the birthday problem, should get .507 */
/* printf("%lf\n", paired_odds(365,23)) ; */

for(pop_size = 10 ; pop_size < 1000000 ; pop_size *= 10) {

for(p = 0.05 ; p < 0.35 ; p += 0.05) {
n_selections = p * (double)(pop_size) + 0.5 ;

printf("pop_size:%10ld n_selections:%10ld odds:%lf\n", pop_size, n_selections, paired_odds(pop_size, n_selections)) ;
}
}
}

[CHIMO]

47. Hint

The probabilities for birthdays occuring on the same date is approached by looking at it from the reverse angle of what is the probabilty that there is not a common birthday in the population.

Putting letters to numbers. If base 0 is taken then "1" is not "a" it would be "b".

It does appear the 18181.... is a prime. Interesting. How many instances of winners having a prime number of votes that are the same as other winners that are the same prime. Maybe could be approached by normalizing the numer of voters?

Liked the method of arriving at the 18181, although if one were so far ahead then why would one have to call in a function? An unecessary back-up?

[Pobeka]

50. Numbers using 535 congressional districts

(pop_size is the range of possible votes, in bumbler's example he
suggests 2500 might be the right number, even at 20,000 odds are
very good for a single duplicate)

pop_size: 2000 n_selections: 535 odds:1.000000
pop_size: 20000 n_selections: 535 odds:0.999258
pop_size: 200000 n_selections: 535 odds:0.510740
pop_size: 2000000 n_selections: 535 odds:0.068937

--
Edit - typo

[ConsAreLiars]

53. I recall enough

to remember that this was the point at which I was convinced that math was not my natural talent. Next we need the odds of a threeway match.

As you mention, this is not exactly the birthday math since we will expect numbers at the middle of the "year" to be most likely, but even assuming a flat distribution across a limited range shouldn't be too misleading.

If I'm following this, however, the pop_size would be 535 (Districts) or 2000 (groups of 25,000 voters). And the n_selections would be 2500 (or 2000 as you used). The terms are ambiguous, but the logic is that the probability of finding duplicates within a range of 2000 (or 2500), given 535 (or 2000) chances, are high but less than certain.

[CHIMO]

54. Backing Up to Birthdays

To calculate the probabilities for birthdays on the same day the approach is to calculate what is the probability that none in the population do not have the same birthday.

Would you please expand on "we will expect numbers at the middle of the "year" to be most likely", I do not understand the meaning?

I am getting a bit lost with the probabilities that are appearing. Are they bionomial, or a random population or biased from polls?

I am getting lost in the programming language.

A little help would be appreciated.

Thanks.

[Pobeka]

56. Some explanation

The code I wrote assumes we are drawing from a random distribution, which in this case is the possible vote total for a race. I think the statement "expect numbers in the middle of the year" from bumbler is based on the analogy back to the birthdates, most candidates are going to get between say 30% and 70% of the vote -- this corresponds to the "middle of the year" for the birthday analogy.

I would rather choose the conservative approach, and say any vote total is a possibility, from 0 up to the total number of voters.

[ConsAreLiars]

61. Sorry

Skipped a few words. Birthdays are more-or-less equally distributed across the full range of possible dates. Vote totals for each party within any district are clustered within a small range of numbers and concentrated around the middle. To further complicate things, the "year" for vote totals (the full range of possibly matching numbers (analogous to the number of birthdates in a year)) is the range of possible outcomes, but this range is not as definite as the number of days in a year.

More concretely, if you have 18,000 R's and 6,000 D's, the votes per race will fluctuate around those numbers, but not by a whole lot. And the further the vote numbers are from the average outcome, the less likely they are to occur. There is math to quantify this, but it is a bit more complicated than the case when the outcomes are all equally likely (like birthdates). Still, ignoring this difference between votes and birthdates only means that the model Pobeha is using is simplified, not that it is wrong.

(Sorry again. (Any explanation that involves nested sets of parentheses is probably not that helpful.))

[Pobeka]

55. Sorry for the confusion

Pop_size is the range of the final vote count. So, if I select from that population (n_selections), 535 times, then those were the odds I got.

Trying to wrap my head around the 3 way, 4 way, n-way match aspect now.

[CHIMO]

57. So Like Numbered Balls in a Bag

This presumes that there are no biases. Don't know if I am interpreting your point correctly?
1000 different numbers in a bag, then chance of any one is 1 in 1000.
But if one has 100 different bags each with numbers 1 to 1000 then what is/are the odds that 2 of the same are not picked from the bags. Or three, or four. Then one has to add up the probabilities that it would not happen and subtract from 1.

Only if things are approached from the correct logic can a proper deduction be made.

[Pobeka]

58. Yes, you've got it.

so if I have 10 balls in 2 bags, odds are 10% that I'll get a match.

10 balls, with 3 bags is 28%
10 balls, with 4 bags is 49.6%

and now say we have 30 bags (races), with 24000 balls (possible vote totals). The chance of a match is 1.7969%

Do you have a C compiler?

[CHIMO]

62. You are Getting a Bit Ahaead of Me

It's getting a bit late but trying to get back to basics. Maybe I am skipping a beat but here goes anayway.
3 bags with balls numbered 1 to 10.
Pick a ball from the first bag.
Chances that you will have 3 of the same is .1 x .1 = .01
Chances that bag one and two have the same pick is .1
Chances that bag one and three pick have the same is .1
Chances that bag two and three have the same pick is .1

Total is .31
I am not satisfied with my quick analysis here but it appears to me that any improvement would lead to a greater probabality?

I may be off on my logic at this late/early time or maybe have to go back to basics?


Don't have C on my machine. Have the program, back from 1995 but haven't used it since. Am running near my capacity on this 5 year old machine.

[CHIMO]

72. Correction

Agree probabilty two or more occurances in this scenario is 0.28

My previous post is out to lunch.

[punpirate]

69. Not a statistician...

... but it seems to me that the problem cannot be logically structured that way.

In reality, the structure of the problem is like this, as I conceive of it: County-wide, there are 60,000 registered voters. Of that number, only 40% actually come out to vote, which is 24,000 voters. County-wide, perhaps there are one hundred precincts (think of these as places where vote sub-totals are generated). With an even distribution, that would be 240 voters per precinct. But, the distribution is not uniform, because of geographical constraints. For geographic convenience, some outlying precincts may see only 50 voters, while precincts in heavily populated areas may see 1000. By the previous analogy, the number of balls in each precinct bag are different--perhaps by two orders of magnitude.

Additionally, there are poorly-defined variables related to geography--is a lightly-populated rural precinct composed of poor rural workers, or wealthy farmers? They tend to have different voting patterns (this is important when calculating odds around a narrow band of variance within say, the +/- 4% swing around a pair of candidates, because that swing is not replicated precinct to precinct, but rather in the county-wide totals).

Next, the problem cannot be expanded, for statistical purposes, to all precincts in all counties in all states, and then treat Comal County, Texas as a subset of that group. It's unique, because its ballot structure is unique. If the candidates and ballots were all the same, nationwide, then Comal County would be merely representative, and the rest could be included in the computation of odds. Because it's unique, however, all the rest of the country has to be excluded. Comal County is _the_ set. Further, in calculation of the odds, some consideration would have to made with regard to kind of voting machines used nationwide, because all have different rates of error.

Now, considering that Comal County is unique, and is its own unique set of numbers, there's a further consideration. Precincts tally votes, and then transmit or carry those vote totals to the county for county-wide toting. There are considerable variations in vote distribution from precinct to precinct because of the socio-economic variables described above, and there are considerable variations by precinct in the percentage of registered voters who actually vote. This can significantly skew the results. Affluent suburbs may enable a higher percentage of conservative registered voters to actually vote, while poor voters in other precincts may not be able to vote at all, because of job requirements, etc. It's another variable to be considered.

Now, with all these variables to consider, the _individual_ precinct totals, across 100 precincts, must then add up to exactly 18181 votes for a particular candidate, at the county level. In each of the races, 30 in all, the vote totals for one candidate or another at the precinct level (the ratio of votes in the bag), will be different, because of the variables described.

Then, the results are transmitted to the county level and when _added_ together (because that's the only way the votes can be toted), the result is that, in three races out of thirty, the candidate of one party receives a _total_ from the precincts of _exactly_ 18181 votes. No candidate of the opposing party (or any of myriad third parties) receives a similar identical number of votes.

The number of balls in each bag varies. The socio-economic reasons for voting in a particular way varies from bag to bag. And yet, when all the balls from all the bags are dumped into a common bin and counted, three candidates of one party receive exactly the same number of votes.

Finally, with a +/- 3% historical variation in the vote and 24,000 voters voting, with only two candidates in each race, there are about 1500 different combinations of vote totals possible at the county level, in each race, as derived from precinct totals. With three candidates on the ballot, the number of different combinations in vote totals grows to 12,000. With four, about 24,000 different combinations.

(On edit, I should also say that the above is predicated on a +/- 3% statistical variance about an equally split vote. Depending upon the historical trend in voting, the vote tally in this election could produce a much larger difference in vote ratio. For example, if the historical variance were +/- 3% around an even distribution of votes, then results in these elections of 18181 votes for and 6000-odd votes against would be wildly at odds with the historical variation, and would result in many, many more possible combinations of votes, because a 75/25 split would be well over 20 percentage points outside the statistical norm, further tending to the improbability of the same number of votes being cast in three of thirty races.)

That is, to my mind, the definition of the statistical problem, rather than the much simpler likelihood of vote distribution according to the formulae presented. {On further edit, it occurs to me that the birth date analogy might not be at all appropriate when looking at vote distribution. When looking at birth dates and a fixed group of 30 people, the odds are greatly different, for a couple of reasons. There is only one order of magnitude between the number of people and the possible number of dates available. Second, each date is fixed, and in computer terms, can be reduced to four bytes and cannot be modified by any other variable. This is an entirely different problem than that of determining odds in terms of voting ratios. In the case of birthdays, the maximum value for that variable is 365. With the number of possible vote combinations, the maximum value is much larger, and even more so when one considers that the vote combinations are derived from 100 different subsets.}

Cheers.

[punpirate]

70. Ummm. Forgot rough calculations of odds on the above....

If the possible vote combinations with only two candidates in a single race are 1500 to 1 (at the +/- 3% variance), then a rough calculation of the odds for three candidates of one party receiving the same number of votes in thirty different races would be 1 in 1500 * 3 in 30 * 1 in 2 or about 30,000 to 1. I wouldn't even begin to know how to calculate odds of 100 different precinct subtotals adding up to that exact amount in three different races, but I'd be willing to bet it would be, without considering socio-economic variables, about 3,000,000 to 1.

But, then, I'm not a statistician. Maybe it shows. *smile*

Cheers.

[Pobeka]

59. Calling it a night

Chimo PM me if you have more questions, I may not get back to this til tomorrow night...

[Pobeka]

38. One other question

How many other duplicate numbers occured, is 18181 the *only* duplicate?

[Pobeka]

41. Hello -- is anybody out there?

Just wondering, you guys usually respond pretty quick...

[punpirate]

44. Maybe they're all checking...

... this. But, with regard particularly to Comal County, I think those were the only duplications.

On edit, gotta get my counties straight....

Cheers.

[Pale_Rider]

60. Kinda like fixing a floating point error?

I dimly remember in the ancient past, having to type cast and do an (int *) to drop from a float value to an integer value in some thermal (IEEE-ASTM) engineering code. Don't remember an int() in the formulas. Is there any issues with using long to type-cast to an int in C++ (especially portable C)? Like int to long and vice-versa? Especially in regards to signed bits and perhaps bit wrap ...

? 1+1+1=3.2

[MoonGod]

63. Damn! I knew the number looked familiar...

... from back in the days when I spent a lot of time playing with baseball statistics. Early in the season, lots of guys (who get off to a slow start) will have a .182 (or .18181818181...) batting average.

[scottxyz]

64. I admit it's more numerology than mathematics...

Merlin (post 6) Ivory_Tower (post 25) have quite correctly reminded us that there's not much to my theory about 18,181 - in fact I can't make a theory at all, not even a conjecture. It really is little more than daydreaming.

All I can say is that this sort of number does come up a lot when you're dealing with elevenths, and elevenths do come up a lot when you've bumped something up by a tenth.

But, as has been pointed out, the thing you're bumping up by a tenth has to be a multiple of ten - and raw poll numbers usually aren't nice round multiples of ten.

So everything I said is really pure speculation, and I confess I'm as mystified as everyone else by this recurrent number 18,181.

I basically just wanted to get people off the 'ahaha' thing, which seems like it can't go anywhere because we're only dealing with numbers here, and it's unlikely that any tampering would have had any reason to translate the numbers into an alphabetic code.

I also doubt the binary notations (0001 and 1000) have much meaning, because although as we all know computers do their underlying integer computations in binary rather than decimal, any tampering would probably have also been done directly in decimal.

It's just that people were wondering if there's anything mathematically special about 18,181, and the way it's related to elevenths and the way elevenths come up so often when you're bumping up something by a tenth seemed like it could be some kind of a lead. A lead to where, I must admit I don't know.

I did try to color my language with suggestive terms such as 'increment' and 'bump up' because we all know that's what's going on when tampering happens. But, again, the problem is that if you're bumping up a Repuke's raw poll numbers, it's unlikely that what you're bumping up is a nice round multiple of ten.

I'm a little embarassed for engaging in such numerology, but it's a common temptation for some people who work with numbers all day.

[ParanoidPat]

65. I wonder if....

....there's any record of last minute software patches being loaded on the machines that gave those results? :shrug:

[OrdinaryTa]

71. Number Theory

I'm a little embarassed for engaging in such numerology

You shouldn't be. Number theory is fascinating, and number theorists are interesting people, especially the great ones.

A couple of interesting stories here. Erdos had very little practical sense, and he was often flummoxed by the ordinary details of life. But he never forgot a phone number! Every phone number has a little story associated with it, and Erdos would remember the story.

Hardy and Ramanujan were getting out of a taxi in London, and Hardy happened to notice the number on the taxi. "Not an interesting number," he sniffed. Ramanujan disagreed, and discoursed for the next 10 minutes on the interesting properties of the number, that it is the Nth prime in a sequence, etc. ...

I keep a pine cone on my desk to remind me of Fibonacci numbers - which, it turns out, are everywhere. Fibonacci himself was aware of the disdain for number theorists and other mathematicians who pursue problems of no immediate practical value. Fibonacci is a nickname meaning "jokester" or "unserious person".

http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Fibonacci.html

For a discussion of Fibonacci phyllotaxis, see:
http://www.math.smith.edu/~phyllo/About/math.html

[Nlighten1]

74. 18181 is a dihedral prime number

Dihedral Primes
Dihedral primes are defined by Mike Keith as integers n which displayed on a calulator have the property that n, n upside down, n in a mirror and n upside down and in a mirror are all primes. It's clear that the digits of a dihedral prime are restricted to 0, 1, 2, 5 and 8. The first dihedral primes are 2, 11, 101, 181, 1181, 1811, 18181, 108881, 110881, 118081, 120121, 121021, 121151, 150151, 151051, 151121, 180181, 180811, 181081, 188011, 188801, ... (Sloane's A038136). The number of dihedral primes consisting of 1, 2, ... digits is 1, 1, 2, 2, 1, 14, 40, 52, 228, 482, ...

See Dihedral Primes.

For what it is worth