[Svartalfiar]

If you examine a large enough list of random numbers drawn from the range from, let's say, 0 to 999,999 then I would expect that the least significant digit would be distributed evenly; that is, there would be the same amount of each of the ten digits.

Except these numbers aren't randomly pulled from a list, they count, and they count up. Therefore EVERY count will start with a '1' at some point, but not every count will end up reaching a '9' for the same power of ten you end the counting in. That's why this works, if you start counting and are stopped randomly (what voting is, basically) every power of ten of votes (10, 100s, 1000s, 10Ms, 100Ms, 1MMs, so on) will fill in every chance to start with a '1', and each higher digit will be less likely to have been reached, but once you hit the XX999 and give each number an even chance to have been the final starting digit, the next count will go back to starting with a 1 and make that the most likely starting digit.

EG for your random result to be, say, 328, you will have had the opportunity to stop at 111 '1's (1, 10-19, and 100-199), 111 '2's (2, 20-29, 200-299), but only 39 '3's (3, 30-39, 300-328), and 0 '4'-'9's. Any 'random' number you pick, will always fill in the lower numbers' possibilities, but not the higher ones.

[Svartalfiar]

and 0 '4'-'9's.

That's wrong, I meant to say 11 of each. Single digits and tens, but nothing from the 100s because we stopped counting in the 300s.

[William Tell]

"Except these numbers aren't randomly pulled from a list, they count, and they count up. "

Just for clarification, I was talking about the last two digits not the first digit. A random sample taken from the range 0 to 999,999 will have around 90% six digit numbers. A Benford analysis for the least significant digits will be different than for the first digit.