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Imagine that all and each line are marked by the first 13 letters of English alphabet:
1 2 3 4 5 6 7 8 9 10 11 12 13
A B C D E F G H I J K L M
Then the space of all possible events is the set of all 6-letter words comprising of these letters.
Repetitions of letters in these words are allowed.
It is easy to calculate the number of all such 6-letter words.
Any of 13 letter can stay in the 1-st position. This gives 13 opportunities.
Any of 13 letter can stay in the 2-nd position. This gives 13 opportunities.
And so on . . .
In all, there are such words.
Correspondingly, there are elements in the space of events, in all.
Now, the winning events are those 6-letter words what have no repetitions.
The number of such words is exactly 13*12*11*10*9*8 = 1235520.
Therefore, the probability under the question is equal to
= = 0.256 = 25.6% (approximately).
Solved.