There’s an interesting article in the Washington Post today exploring one line of reasoning suggesting that the Iranian election is fraudulent. Basically, it comes down to this: the results aren’t random enough. In a fair election, you’d expect that each digit, from 0 to 9, would be the final digit the results in each region roughly ten percent of the time: you’d see a vote count like 12,43** 7** just as often as 12,43

**. But in fact certain digits come up more often:**

*5*The numbers look suspicious. We find too many 7s and not enough 5s in the last digit. We expect each digit (0, 1, 2, and so on) to appear at the end of 10 percent of the vote counts. But in Iran’s provincial results, the digit 7 appears 17 percent of the time, and only 4 percent of the results end in the number 5. Two such departures from the average — a spike of 17 percent or more in one digit and a drop to 4 percent or less in another — are extremely unlikely. Fewer than four in a hundred non-fraudulent elections would produce such numbers.

You can’t expect the *first* digits in a result to be random, because they represent tens of thousands of voters, and in any given region, one candidate probably *is* supported by more voters than the other candidates. But the final digits should be random in a fair election.

For some reason, people seem to pick numbers ending in “7″ as more “random” than other numbers. When we asked our readers to generate random numbers from 1 to 20, 7 and 17 were the most common answers, appearing almost three times as often as you’d expect if the numbers were truly randomly generated. Meanwhile numbers ending in 5 only came up about half as often as they should have. In fact, our results were quite similar to the Iran election results for those digits:

Beber and Scacco also found that the patterns in the last *two* digits of each number are not random. They calculate the chance of these two anomalous results in the elections occurring due to chance as less than 1 in 200.