Expert: Scott A Wilson - 11/1/2009

Question
A six digit number (abcdef)each digit can vary from 0to9. A number is said to be a lucky number if sum of first 3 digits is equal to the sum of last 3 digits.what is the probability that the 6-digit number is a lucky number?

Answer
Take a foud sided die.
Flip it once, it would be
1: 1
2: 1
3: 1 and
4: 1.

If you flipped it again, you would add 1 to the numbers four numbers above the number rolled.
Zero out the new total first.
For the old total being 1, add one to 2, 3, 4, and 5.
For the old total being 2, add one to 3, 4, 5, and 6.
For the old total being 3, add one to 4, 5, 6, and 7.
For the old total being 4, add one to 5, 6, 7, and 8.
This would give
2: 1
3: 2
4: 3
5: 4
6: 3
7: 2 and
8: 1.

Make these the old totals.  Zero out the new totals.  For each of the old totals, add the number there to each of the four numbers above it.
For 2, add 1 to 3, 4, 5, and 6.
For 3, add 2 to 4, 5, 6, and 7.
For 4, add 3 to 5, 6, 7, and 8.
For 5, add 4 to 6, 7, 8, and 9.
For 6, add 3 to 7, 8, 9, and 10.
For 7, add 2 to 8, 9, 10, and 11.
For 8, add 1 to 9, 10, 11, and 12.

This would give
3: 1
4: 1+2=3
5: 1+2+3=6
6: 1+2+3+4=10
7: 2+3+4+3=12
8: 3+4+3+2=12
9: 4+3+2+1=10
10: 3+2+1=6
11: 2+1=3
12+ 1

This means the number of times for each number is
3: 1
4: 3
5: 6
6: 10
7: 12
8: 12
9: 10
10: 6
11: 3
12: 1


Now that you have that down, I'll do it for 10 choices for each number, 0,1,2,3,4,5,6,7,8, or 9.
Once
0: 1
1: 1
2: 1
3: 1
4: 1
5: 1
6: 1
7: 1
8: 1
9: 1

Twice
0: 1
1: 2
2: 3
3: 4
4: 5
5: 6
6: 7
7: 8
8: 9
9: 10
10: 9
11: 8
12: 7
13: 6
14: 5
15: 4
16: 3
17: 2
18: 1

Three:
0   :   1
1   :   3
2   :   6
3   :   10
4   :   15
5   :   21
6   :   28
7   :   36
8   :   45
9   :   55
10   :   63
11   :   69
12   :   73
13   :   75
14   :   75
15   :   73
16   :   69
17   :   63
18   :   55
19   :   45
20   :   36
21   :   28
22   :   21
23   :   15
24   :   10
25   :   6
26   :   3
27   :   1

Now the possibility for each of these is the number/1000,
since there are 1000 ways to make a three digit number with leading 0's.

That means to find the number that are the same, it has to have the same sum it had last time.
To find, for example, a 23, there are 15 ways each time.  That means there 15*15 ways to do it both times.  In other words, square the column of numbers and add them up.
Realize that that new total is 1,000,000 choices (1000 * 1000).

The number, ways of getting it, and the square is
0   1   1
1   3   9
2   6   36
3   10   100
4   15   225
5   21   441
6   28   784
7   36   1296
8   45   2025
9   55   3025
10   63   3969
11   69   4761
12   73   5329
13   75   5625
14   75   5625
15   73   5329
16   69   4761
17   63   3969
18   55   3025
19   45   2025
20   36   1296
21   28   784
22   21   441
23   15   225
24   10   100
25   6   36
26   3   9
27   1   1

That is a total of 55,252 out of a million, are a little better than 5%.
That the probablity of having the same sum for the first three and the last three.