Looking at this diagram and comment:
With three stones, there are some problems. An attempt by white to hane-tsugi may be met by a throw-in (a liberty stealing tesuji) where white expects to play . The ko that results if white does not connect with does not help white. Therefore at is often correct (a case of One-Two-Three).
I don't understand the remark "the ko... does not help white". It looks to me like black has a lot to lose in such a ko, so it's a picnic for white. Instead, black can play at a, then white can't play at , right?
Here's my 3k guess:
I assume that this is your reasoning? (" and get captured because white cannot play at . In fact, at a is gote.")
What happens if = tenuki? Well, Black can certainly play at b, and white can tenuki again. But compare this with the following diagram:
This seems slightly better than the result above... although I can't give an exact number in miai counting terms (Bill Spight, are you there? ;-) If is played elsewhere, white can play at for an even bigger reduction.
Karl Knechtel: Black takes the ko first, can extend backwards on the first line for several moves if desired, or just connect at a later... in which case White will eventually have to spend two moves capturing Black's throw-in stone before she can fill (i.e. Black implicitly gets more virtual threats too).