Here its mention that we would roll the dice for another time if we got face value of 1, 2 and 3.
So, let us consider a case where the face value be : [1 , 1 , 1 , 1 , 1 , 1] (for upto min 6 times the dice shall roll)
Similarly for : [2 , 2 , 2] (3 times the dice rolled) , [3 , 3] (2 times the dice rolled) and [1, 2, 3] (1 time the dice will roll.
Above 4 cases satisfies the condition the face value is atleast 6.
Now, considering another favoured conditions where the face value can also be atleast 6:
[1, 5] , [1, 6]
[2, 4] , [2, 5] , [2, 6]
[3, 3](already counted above so don’t repeat) , [3, 4] , [ 3, 5] , [3, 6]
[4] , [5] , [6]
So, now favourable conditions are : $\frac{(4 + 11)}{36}$ = $\frac{15}{36}$ = $\frac{5}{12}$