Question: Suppose the winning combination consists of $7$ digits, each digit randomly ranging from $0$ to $9$. So the probability of $1111111$, $3141592$ and $8174249$ are the same. But $1111111$ seems (to me) far less likely to be the lucky number than $8174249$. Is my intuition simply wrong or is it correct in some sense?

Answer: 
The reason $1111111$ seems less likely is because it is part of an easily identifiable pattern, and the pattern itself is less common than the patterns that you see in $8174249$.

For example $8174249$ belongs to the set of numbers between $0000000$ and $9999999$ which have no repeated digit. That set is quite large, it has $10 \* 9^6 = 5314410$ numbers in it.

$5314410/10^7 = .531441$

So you have a greater than $50$% chance of the number having no repeated digits.

Whereas there are only $10$ numbers which are a single digit repeated $7$ times, so you have a very small ($10/10^7 = 0.000001$) chance of getting a number in that set.

So at this point, it seems like picking a number with no repeated digits is a much better choice, but the size of the set is so huge that you will end up with the exact same probability of winning. This is no coincidence.

If you pick a number with no repeated digits, you have a $.531441$ chance that the result will be in the same set, but there are 5314410 numbers in that set, so the odds of winning the lottery, given that the chosen number will have no repeated digits, are still $1$ in $5314410$. The probability of both events happening (You picking the right number out of the 5314410 choice, and the lottery system picking a number with no repeated digits) is exactly what you would expect your odds of winning to be: $.531441 \* 1/5314410 = 0.0000001$

The odds of winning given that the lottery system picks a number with all repeating digits are quite high for a lottery, $1$ in $10$, you only have `10` choices to choose from! But the probability of that pattern being picked are so low, that the probability of both happening are exactly the same as the probability of $8174249$ being picked: $0.000001 \* 1/10 = 0.0000001$

That is true for any pattern. The larger the set of numbers that fit the description of a pattern, the more likely it is that the pattern will be picked, but as it becomes more likely for that pattern to be picked it becomes less likely for you to pick the correct number in the pattern. It balances out perfectly like you might expect, and the odds of winning are the same no matter which number you pick.
Another point to consider is how is the winning number selected. Here the lottery numbers is decided through a powerball mechanism (<http://en.wikipedia.org/wiki/Powerball>). So given that there is only a fixed amount of numbered balls the winning number is drawn from, this skews the probability of the winning number being 111111 versus another random number.

For example normally, if the numbers chosen are truly random, the odds of 1111111 being the winning number is 1 / 10 000 000.

But lets say in the powerball bin, there is 7 '1' balls, 7 '2' balls and so forth, then the odds that first drawn number will be '1' is 7 / 63. That ball is then removed from the bin. The odds that second number will be '1' is now 6 / 62 etc.

That brings the odds of 1111111 to 7\*6\*5\*4\*3\*2\*1 / 63\*62\*61\*60\*59\*58\*57 = 2 / 1 000 000 000.

Quite a bit less... The effect will vary depending on how many balls of each number there is in the bin, and how balls are replaced once selected. Some powerball lotteries has a preselection draw which determines which balls are in the final draw, but the same argument is still valid.