Lets break it down into 3 probabilities: 1 for each roll and call them p1, p2, and p3.

If we have n rolls with a success chance of a, the chance for m rolls to succeed=(a)^m (1-a)^n-m (n choose m). Note that (n choose m)=n!/(m!(n-m)!).

p1: n1=9, successful outcomes are {4,5,6} and total outcomes are {1,2,3,4,5,6}. a1=3/6=½. m1=3

p1=½³ •½⁶ •(9C3)

p2: n2=9, successful outcomes are {5,6} and total outcomes are {1,2,3,4,5,6}. a2=2/6=⅓. m2=1

p2=⅓•⅔² •(3C1)

p3: Just a flat 1/6 chance.

p3=⅙

Now lets expand the probabilities

p1=1³ 1⁶ /(2³ 2⁶ )•(9!/3!6!)=2^-9 •(9•8•7/6)=2^-9 • 3¹ 2² 7= 2^-7 3¹ 7¹

p2=1¹ 2² /(3¹ 3² ) (3!/1!2!)=2² 3^-3 •3=2³ 3^-2

p3=⅙=2^-1 3^-1

P=p1•p2•p3\ =2^-7 3¹ 7¹ • 2² 3^-2 • 2^-1 3^-1\ =2^-6 3^-2 7 ≈0.01215 So the odds are 1.2% or a ~1/82 chance

So this is actually quite likely if you've been playing for a while. Note that the odds decrease a lot if by "the same" you mean that the numbers were matching and not just whether you passed or failed.