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.
This is a case of the devil being in the details.\*

> 
> 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.
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With a small stretch of English grammar, these two statements are true, but it is the difference between the statements that is the key to understanding your confusion.

You are conflating two quite different things as if they were one.

1. The possibility of X being the winning number.
2. The possibility of the winning number being X.

On the one hand, all numbers in the range have an equal possibility of being the winner. 1111111 is one number out of a million and has once chance out of a milion—the same one chance out of million that 111112 has, as does 923652.

On the ther hand, the chance of the winning number being a specific number (or specific pattern) is bounded by the number of patterns in your set. Assuming zeros are allowed, there are 10 sets of repeated digits. In other words the winning number has a 1 in 100k chance of being a set of repeated digits.

The chance of the winning number being any specific number pattern does not in any way change the chance of a specific number pattern being the winning number.

\* For the sake of simplicity, I've glossed over complexities in the math to show the general idea at stake.