I think it is true. We head off discussion here by referring to the naive approach, but what you're doing is past naive coupon collector's and past the technical scope of the context.

There are several steps where your method could (will) fail. Since you don't introduce any machinery for the hand, you imply an O(k^2) checking algorithm. You also imply an unsorted hand. If you generate the hand in random order, then you fall to an O(k*lg(k)) sort. "Sorted" is an implicit requirement that we initially omit, but revisit later, because introducing it early makes the discussion harder to follow.

If you try to introduce machinery like a hashtable, the next thing you might suggest, you violate an implicit O(1) additional space requirement, which I've grudgingly reintroduced to the more-formal description ("why is this here?"). This conversation keeps going, but other failure paths we leave unexamined.