Only the number seven left unchosen

Seven people each pick a number uniformly at random from $\{1,2,\dots,7\}$, independently. Given that no one picked $7$, what is the probability that $7$ is the only number in the set that no one picked?

$\tfrac{35}{648}$. Conditioned on avoiding $7$, all $6^7$ choices from $\{1,\dots,6\}$ are equally likely. Using all six values with seven people means one value is chosen twice: $\binom72$ ways to pick the pair, $6!$ ways to assign the six values, so $\tfrac{21\cdot720}{6^7}=\tfrac{35}{648}$.