Restoring a six-digit house number after two digits fall

A house displays the six digits $1,2,2,5,7,7$ in that order, i.e. the number $122577$. A storm knocks off two of the six digits, chosen uniformly at random among all pairs of positions. The owner, who has forgotten the number, puts the two fallen digits back into the two empty positions in a random order. What is the probability that the house again reads $122577$?

$\tfrac{17}{30}$. Of the $\binom62=15$ equally likely pairs, two (the two $2$s or the two $7$s) consist of equal digits and are always restored correctly; for the other $13$ pairs the owner is right half the time. So the probability is $\tfrac{2}{15}+\tfrac{13}{15}\cdot\tfrac12=\tfrac{17}{30}$.