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 3-letter words comprising of these letters.

Repetitions of letters in these words are allowed.


It is easy to calculate the number of all such 3-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.
Any of 13 letter can stay in the 3-rd position. This gives 13 opportunities.


In all, there are  such words.
Correspondingly, there are  elements in the space of events, in all.


Now, the favorable events are those 3-letter words what have no repetitions.

The number of such words is exactly  13*12*11 = 1716.



Therefore, the probability under the question is equal to  

 = 0.7811 = 78.11% (approximately).     ANSWER