My brute force script says that of the 1 million possible numbers, 55252 are 'lucky.'

There are 1000 possible 3-digit numbers. The sum of 3 digits range from 0 to 27. A few moments in Excel gives us the following distribution:

0 Count 1

1 Count 3

2 Count 6

3 Count 10

4 Count 15

5 Count 21

6 Count 28

7 Count 36

8 Count 45

9 Count 55

10 Count 63

11 Count 69

12 Count 73

13 Count 75

14 Count 75

15 Count 73

16 Count 69

17 Count 63

18 Count 55

19 Count 45

20 Count 36

21 Count 28

22 Count 21

23 Count 15

24 Count 10

25 Count 6

26 Count 3

27 Count 1

The odds of getting a lucky '0' are (1/1000)². The odds of getting a lucky '1' are (3/1000)². Sum these odds together, and we get 5.5252%, which is what the brute force method suggested.

There are 1000 possible 3-digit numbers. The sum of 3 digits range from 0 to 27. A few moments in Excel gives us the following distribution:

0 Count 1

1 Count 3

2 Count 6

3 Count 10

4 Count 15

5 Count 21

6 Count 28

7 Count 36

8 Count 45

9 Count 55

10 Count 63

11 Count 69

12 Count 73

13 Count 75

14 Count 75

15 Count 73

16 Count 69

17 Count 63

18 Count 55

19 Count 45

20 Count 36

21 Count 28

22 Count 21

23 Count 15

24 Count 10

25 Count 6

26 Count 3

27 Count 1

The odds of getting a lucky '0' are (1/1000)². The odds of getting a lucky '1' are (3/1000)². Sum these odds together, and we get 5.5252%, which is what the brute force method suggested.