This is actually pretty poor statistics. They are basing their entire argument on ~100 provincial results. The deviations they are talking about with this few number of samples isn't likely to be significant.
They sum it up nicely but it's too easy to miss:
"The probability that a fair election would produce [these number distribution disparities] is less than .005. In other words, a bet that the numbers are clean is a one in two-hundred long shot."
Key thing to take away from the article if you're going to perform election fraud, use a computer to produce your random numbers.
Careful with that reasoning. Suppose there are four candidates, and the last digits of their total vote counts are 8,7,2,8. Fewer than 1 in 9999 fair elections would produce a vote count with exactly those digits!
Does this mean that the vote was rigged?
Post-factum searching for (perceived) oddities and calculating probabilities of their occurences in uniform distributions shows some misunderstanding of statistics.
How did they calculate this probability. I can see what they did in other spots, but this number seems to come out of nowhere to me.
This is all academic. Even the Guardian Council has admitted that there were more voters in 50 areas than eligible to vote.
http://www.nytimes.com/2009/06/23/world/middleeast/23iran.ht...
Also note: "But many districts where the excess votes were recorded are small, remote places rarely visited by business travelers or tourists, analysts said"
And so on, problem after problem.
I dare say, the next election there will be "handled" better... :-)
In isolation you can look at the odds for 17+ and 4-, but two <5, or two > 16 would also look odd. As would a single >20 or <4. And, they started looking at 2 digits but the pattern is only in the last digit so there are looking at 3 groups.
PS: Also how did you calculate the odds? A single number of 17 or more increases the odds of a number of 4 or less. I am not really sure how you would calculate the odds outside of a simulation.
> "The probability that a fair election would produce [these number distribution disparities] is less than .005. In other words, a bet that the numbers are clean is a one in two-hundred long shot."
Man, reading that makes me roll my eyes and wonder how the average person reading this site has such poor statistical intuition as to mod this up.
It's like rolling a die six times, and arguing that getting the exact numbers I got has an exceptionally low chance of occurring, so the die must be rigged.
You have two options: either you trust your (unproven) assumption that the coin is fair and that you just witnessed a very unlikely event and say "the past outcomes do not affect the probability of the next throw, it's 0.5."
The other option, on the other hand, is to conclude that your assumption that the coin is fair is false, at which point you estimate the next throw based on the previous sequence and say "1.0".
None of these are right or wrong, it just depends on what prior you trust.