Three Texas Repukes get 18,181. What are the odds? A quick estimate.

[TruthIsAll]

KISS - keep it simple, stupid.

This is just a quick off-the-top exercise.

Lets assume that we would normally expect a variation of 1000 votes in three county elections. That is, assume they would each expect somewhere in the range 18,000 to 19,000 votes.

What is the probability that by coincidence, chance or fate, all three would each get the same number of votes.

This is an exercise in conditional probability.

Assume candidate A gets 18,181 votes. This is given.

The Probability B will also get 18,181 = 1 out of 1000
The Probability C will also get 18,181 = 1 out of 1000

The Probability (B and C will both get 18,181, given that A =18,181) = 1/1000*1/1000 or 1 out of 1 million.

Just a coincidence? Oh sure, and if you believe that, I've got a Diebold machine to sell you. One without a paper trail. This way, you can't prove that fraud occurred.

[JackSwift]

1. The races were tampered with pure and simple

.

[GAspnes]

2. Not this crap again

no text, because none is needed.

[TruthIsAll]

17. As usual, your comprehensive analysis makes one Gasp..

I really learned a lot from you in my prior threads, where your comments were just as incisive as this one...anyway, much thanks, your constructive efforts are appreciated..

[MisterP]

22. and it's repeated some time later

[mjb4]

3. translates to ahaha!!!!!

we are screwed. democracy is gone in the US and no one cares...

[ThoughtCriminal]

4. It looked odd to me too, but

Your calculations are not valid because the only sample is the three races in question. A more valid sample would be to look at all similar size elections that night - what are the odds the the vote totals would be the same in 3 of them?

[0rganism]

5. Odds aren't the best way to look at this

You can't really apply probability to specific vote counts in an election, except to correlate with margins of error in entry or exit polls which compare the same races to expected results. Unfortunately, you're comparing different races and the vote is non-random.

The real problem here is that 18181 is obviously a test pattern from a programmer. It is an alternating palendrome prime capable of exercising a 16-bit accumulator. I would be suspicious of a SINGLE vote count turning up that number. Here, we have three. It's either very sloppy of the tech to leave that number in the vote reporting buffer, or s/he was trying to send a message to the rest of the country that the system is rigged.

[Nomad559]

6. 18,181

I seem to recall reading that there was 5 RePUGlicans with a 18,181 vote count, 3 of them In the same county. I don't remember where the other two were from.

I'll do a Google search ... }(

Edit - I Found It :)

http://mathforum.org/epigone/sci.math.num-analysis/wixchaikrung

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!


Bart has a couple of links.

http://www.bartcop.com/0930.htm

[0rganism]

7. IIRC, we never got to see the individual precinct counts

The guy with the first post on the math forum is thinking in terms of appearance of numbers -- why does a certain number look special? And thus dismisses suspicion as a psychological phenomenon.

The thing is, the "specialness" of that number makes it a perfect accumulator test pattern. It's instantly recognizable, even to someone who's been beating on a computer for 8 hours. Mapping it to the alphabet to get "ahaha" is a neat trick, but also beside the point.

Speculation of odds, playing alphabet soup, and psychological pattern excuses are all distractions from the obvious. It's like you see three bloody footprints leading away from the axe murder, and instead of reaching an obvious conclusion, "we should damn well follow the footprints and see who's wearing the shoes," we get hung up on, "what are the odds that three bloody footprints would be made by different people," or, "the logo on the shoeprint looks more sinister when bloody," or even, "it's just a coincidence, lots of people are wearing blood on their shoes these days."

Point being, these results aren't coincidences, they're clues.

[tom_paine]

8. I agree Organism...not coincidences...clues

And we're running out of time to follow up on them.

[Fleshdancer]

14. creepy

I think you're right. They are clues. What do we do now?

[bpilgrim]

16. 'clues'

well said :toast:

especially when you consider the improbability of them happening by chance ;->

peace

[mumon]

13. Well, 5 is a different story....

beginning to believe you.

[gulliver]

12. It's weird that it's a palindrome and prime, but

... I think about the only way to determine whether or not it is a coincidence is to compare it to the frequency of other numbers. The mind plays funny tricks on you with probability. Just ask Bill Bennett.

For example, are there any clusters of 18,182 votes? How many clusters of 3 (or 5 or whatever) equal numbers of votes are there? Of those, how many are palindromes or primes?

Even if it is a coincidence (and I think it probably is), the very fact that Diebold or anyone would be able to get away with a voting machine that lacks a paper trail is a scandal. Only computer illiterates would buy a machine like that.

[mumon]

9. Closer than you think

How many voters? Let's say 25,000, in all districts, for simplicity.

(They're roughly equal.)

How many voters GIVEN the propensity to vote a certain way (number of Repugs in a district, e.g.)? Let's say it's roughly 18/25, or 72% just for argument sake. Again, because of Nazi gerrymandering, they're all equal. And for simplicty too.

So, the real question is: Given the above, what's the probablity that you get 18181, instead of the expected number, 18000?

And that number is pretty small, which means that it's not "significantly different" from 18,000. (You could use some inequalities from probability and statistics, such as Chebycheff's Inequality to get a good bound on the number, but I know I'm talking Sanskrit to you at this point.)

Bottom line: the odds, based on the above, of 2 such votes coming out the same is actually pretty close.

[TruthIsAll]

20. You and I are essentially saying the same thing, but...

After some thought, here is how I would modify the analysis:

1. Assume Comal county had N elections.
2. Using prior polls/election results, determine
V= total number of voters,
P= average Republican vote (%),
A= V*P = average Republican vote
R = H-L = Range = Highest Rep. vote total (H) - lowest Rep. total (L)
(assume R= 1000)

Calculate X = probability of the Republican vote = any given value
X = 1/R (assume all vote counts in the range are equally likely, in this case, assume it is 18,181)

Assume R= 1000 = 19,000 - 18,000

Use the Poisson Distribution:
Although the Normal (Gaussian) probability distribution is by far the most important one, there is one which has proven to be particularly useful. The Poisson Distribution is derived from, and is a special case of the Normal Distribution. The Poisson Distribution applies when the probability "P" for success in any one trial is very small, but the number of trials N is so large that the expected number of successes, pN, is a moderate sized quantity.


Assume we have N elections in a given County
Assume a Range R = High - Low repub vote count= 1000
p = probability that Repub Vote total = X =18,181
a= Expected Number of elections having 18,181 = pN
m = Actual Number of elections having X=18,181

The probability of m elections (out of a total of N elections) having a Repub vote = X, is: P(m) =a**m*exp(-a)/m!

Case 1: 15 elections
0.015 a=pN
m prob (exactly m) prob (at least m)
1 1.48E-02 1.49E-02
2 1.11E-04 1.11E-04
3 5.54E-07 5.56E-07 or 1 out of 1,797,883 <<<<<<
4 2.08E-09 2.08E-09 or 1 out of 479,795,793

Case 2: 30 elections
0.03 a=pN

1 2.91E-02 2.96E-02
2 4.37E-04 4.41E-04
3 4.37E-06 4.40E-06 or 1 out of 227,275 <<<<<
4 3.28E-08 3.30E-08 or 1 out of 30,348,978

Case 3: 50 elections
0.05 a=pN

1 4.76E-02 4.8771E-02
2 1.19E-03 1.2091E-03
3 1.98E-05 2.0067E-05 or 1 out of 49,831 <<<<
4 2.48E-07 2.5021E-07 or 1 out of 3,996,580

Case 4: 100 elections
0.1 a=pN
1 9.05E-02 9.5163E-02
2 4.52E-03 4.6788E-03
3 1.51E-04 1.5465E-04 or 1 out of 6,466 <<<<
4 3.77E-06 3.8468E-06 or 1 out of 259,954

[gristy]

10. Does this make sense?

I think the "odds" of getting the same number (18,181 in this case) in the three races you chose are 100%, since that was the criteria you used to choose your sample. So the analysis needs be done a little differently, though it's not necessarily more complicated.

I think you need to choose, say, all the congressional races in Texas. Say there are 30 of those. And, so as not to change too many things from your example, let's still assume that the expected number is between 18,000 and 19,000.

So the odds of any 2 of those 30 Texas races resulting in the same number as a third race is

(29/1000)*(28/1000)=812/1,000,000= 1 in 1231

And when you consider that there are "groups" of races all over the country where you could apply the same analysis, it becomes even more possible that somewhere 3 races in the same state would give the same result.

That's not to say there isn't fraud going down. I think there is. But the analysis needs to be solid. What I've done here is just another step towards accurate analysis, I think. It's not there yet.

[mumon]

11. Much closer than you think

You have to take into account the facts that:

1. The districts are roughly equal in population & number of voters along a certain way (gerrymandering).

2. The real question is how much the votes would deviate from an EXPECTED result.

When you do this, the difference is actually quite close.

[ISUGRADIA]

15. The equipment used

according to a link I found this county used computerized precinct tabulators with paper ballots, not touch screen machines(and ES&S not Diebold as the vendor):

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

I'd venture this is more a strange anomoly rather than conspiracy or vote fraud.

Questions

1) How would someone be able to hack into the memory cards at 22 precincts and be sure to come up with this result? There is no linkage during the day so when election officials print out results at the end of the night and upload results for the races there is no way that machine knows what another machine across town is reporting. Multiply this by 22 times. Of course if this county is a central count one the process would be more open to tampering.

2)Are the results out of the ordinary for the party for that election or when compared to previous results? Here are some vote totals for countywide Republican candidates:

18,045; 18,031; 17,778; 17,894; 18,051; 18,259; 18,558; 19,601. Three election results that were the same totals would be odd, but not totally unbelievable given the nature of this county's voting.

3) What kind of safeguards are there in Texas law for elections? There should likely be a form accounting for the ballots used at each precinct, the number of voters at each, etc.

4) Who programed their tabulator memory cards? A vendor or was it done on site?

Some of these questions are rhetorical, but are intended to show that tampering to get such a result may have been harder than some here think or again, just one of those bizarre occurances in elections. You need a lot more proof than just numbers to prove tampering.

[zeemike]

21. I know little about computers

But if it was me I would put a little worm or virus in the program that would simply give the highest number of votes to the party of my choice. When cheating the KISS principle works very well.
For instance in Florida in 00 they never cleaned out the punched chads from the machines, and so only one person controlled the election. The one person that stored and distributed the machines to the precincts only needed to send the oldest ones to the democratic precincts to insure a high reject rate for that precinct.

[TruthIsAll]

18. What do you mean by "closer than you think"? n/t

...

[GAspnes]

19. not this crap again

TIA, give it up. You can't prove anything.

[TruthIsAll]

25. Gasp, I love your approach to solving the problem..

Your analytical expertise is amazing. I am sure your results are right on the money.

Once again, thanks for your counsel, advise and encouragement. You are one helluva mentor. You are my hero.

[Name removed]

27. Deleted message

Message removed by moderator. Click here to review the message board rules.

[ShaneGR]

23. TRUTH, tell us what races

KISS... what races, and when

[TruthIsAll]

30. Not sure what you mean. We are talking about localTexas elections here...

Is 18,181 just an odd coincidence to occur in 3 races in a single county of Texas. Check other posts in this thread for the specifics.

[Ivory_Tower]

24. Before I commit to conspiracy theory...

what I'd like to know is: out of all of the races across the country that day, how many winners had the same vote totals? Were there five races nationwide where all five winners had 12,834 votes, for example? Or are these the only cases where the winners had the same vote totals?

Mainly curious, but also concerned about completeness of the investigation.

[GAspnes]

26. don't bother with facts or reason

TIA is immune to them.

[TruthIsAll]

28. Have you ever contributed a single original post, or just sit there on

your crapper making an idiot out of yourself knocking down others who do?

Gasp on this, meathead..

[GAspnes]

29. dear heart

I have (along with many others) disputed your argument with facts and logic. You seem impervious. Now my sole goal in life is to continue refuting this illogical discussion with sarcasm.

[TruthIsAll]

31. That may be your goal, but you have achieved the result of just

making an ass of yourself. Give it up. Go back to school. But this time, stay there long enough to graduate.

Go ahead. Get your next inanity in. Dig yourself deeper into the shit you wallow in.

As far as I am concerned, you no longer exist. You are not worthy of another keystroke.

[GAspnes]

32. thanks for paying attention

you don't know probabilities, but you argue mathematics.

You don't know computers, but you somehow know what computers can do.

You have this idea in your head. Wrong, but an idea. You Believe.

Nothing wrong with that. But passing your faint wishes as fact goes too far.

I don't exist? Well, such is life.

[TruthIsAll]

35. Don't know computers; don't know math?

I have programmed computers since 1965 (using Fortran on IBM mainframes in the defense industry) to Visual Basic and C++ on Wall Street.

I have developed commercial financial valuation and acquisition software which I have marketed to to major corporations as an independent software developer.

I have 3 degrees in Mathematics (two Masters degrees in Applied Mathematics and Operations Research).

What the fuck is your experience?

[Forkboy]

37. GAspnes,I'm a dummy when it comes to math

can you explain why this isn't a big deal,and whats wrong with TIA's computations?

[TruthIsAll]

38. I don't believe that question was posed to you, was it?.

You are in rare form tonight.

What's up? See your boy Shrub going down?

You can't fool me. You give your politics away with your Repuke-stream of hateful nonsense.

Once again. Did you ever contribute ANYTHING to DU except for your bile? If so, point us to it.

[absolutezero]

33. i think

thats the same number you see in a lot of elections...i seem to remember seeing it on blackboxvoting.org after they mentioned that tree districts in georgia won by the same number (18181)

[GAspnes]

34. it's not significant

unless you can prove some sort of underlying method.

[Philosophy]

36. Come back when you understand probability

First of all, your initial assumption is baseless. Second, you can't use that assumption in your calculation the way you did. Basically all you've done here is numerology.

[TruthIsAll]

39. I'm back. Now would you tell me what you would do to solve it?

What is your solution? Tear apart my math in post #20. I know its not perfect. I am sure it can be fine-tuned. Now just go ahead. Show us what you can do. We already know that you, like Gasp, can criticize.

Gasp hasn't shown me shit yet.

Now its your turn. Put up or shut up.

[JackDragna]

40. There's two things being ignored here..

1) For all the votes in a particular country for a particular election to be the same isn't beyond reason, assuming close to the same number of people voted for each of the candidates in question.

2) An anomaly like this might very well happen if you hold enough elections. The tendency is for one to think something's fishy, which, to be fair, is probably a good instinct. There are going to be situations, however, where this sort of thing will happen given enough repetitions. In all of voting history, there have likely been times when an overly religious candidate has won by 666 votes or two candidates with the same last name won elections in different states by the same number of votes. This all boils down to the tendency to see patterns in individual events, such as election results, that may not actually be significant outside of our perceptions.

[0rganism]

41. Not really, TIA is right on top of this issue

While I disagree with his application of probability to the outcome, I do understand why he's choosing the numbers he does.

1) TIA initially gave a probability of 1/1000 to the second number being the same, which already accounts for the fact that similar numbers of republicans would be voting for their various candidates. If we were assuming a linear distribution, we'd be looking at a variance from 17681 to 18681, or about +/- 3%. A gaussian would allow for a wider range centered at 18181. It doesn't matter, really, his assumption is generous.

This isn't a product of straight-ticket voting either. According to the county elections office, the three candidates received different quantities of votes in each precinct. Only if the variation balances out exactly will they receive the same number of votes.

2) Qualitatively quite different. Consider that we have the context of vote counting automation and republican election theft to deal with. The FIVE candidates receiving these totals of votes in 2002 were republicans. The people who own the companies building the vote counters in question are republicans. This is no accident.

While I don't think it makes sense to apply statistics to the individual elections as if they were random outcomes, I also don't think we should ignore all patterns of unusual occurances simply because we're used to picking out patterns of unusual occurances. We have a gift for spotting the unnatural and contrived, and ignore it at our own peril.

[JackDragna]

42. Depends on what you consider an accident

As to the assumption of the likelihood of several candidates getting the same vote, again, if one holds enough elections, the probabilities of getting a result that looks anomalous increases. Candidates might receive dissimilar quantities of votes, but even in this example the totals fall within a similar range.

As to all of the winners being Republicans, this is not out of the question either. Texas votes heavily Republican, so if three people are going to get the same amount of votes in a particular county, then it's likely they're going to be Republicans. Similarly, Republicans did fairly well in 2002, so even with Texas nonwithstanding, the chance of five candidates sharing the same anomalous result and the same party affiliation increases.