A box modeling the die has six tickets labeled 1, 2, ..., 6. Let this box be F
F # F contains 6 * 6 = 36 tickets. They are labeled (1,1), (1,2), ..., (1,6); (2,1, 2,2), ..., (2,6); ..., (6,1), (6,2), ..., (6,6).
F + F is computed by relabeling these tickets by the sums of their components. The table below shows the relabeling. Columns of the table correspond to values on the first die and rows correspond to values on the second die.
| 1 | 2 | 3 | 4 | 5 | 6 | |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
The outcomes are the sums 2, 3, ..., 12.
To compute the probabilities, note that there is one ticket with a 2, two tickets with a 3, ..., six tickets with a 7, five tickets with an 8, ..., 2 tickets with an 11, and one ticket with a 12, for a total of 36. The probabilities are therefore:
2: 1/36
3: 2/36
4: 3/36
5: 4/36
6: 5/36
7: 6/36
8: 5/36
9: 4/36
10: 3/36
11: 2/36
12: 1/36
Clearly the probabilities are not all equal. (If you draw a histogram, it will have a symmetrical triangular shape with the peak in the middle at the value 7.)