I completely agree with Henning Makholm: the important difference between this problem and the classic Monty Hall problem is not whether the apples are chosen by the players or assigned to them — in fact, that makes *absolutely no difference*, since they have no information to base any meaningful choice on at the point where the apples are given to them.

Rather, the key difference is that, in the classic Monty Hall problem, the player knows that Monty will never open the door they choose. Similarly, if one of the players in this problem knew that they wouldn't be asked to bite into the first apple, they'd be better off switching apples with the other remaining player. But of course, if the apples are assigned randomly, it's impossible for more than one of the players to (correctly) possess such knowledge: if two of the players knew they'd never be chosen to go first, and the third one got the wormless apple, Monty would have no way to pick a player with a wormy apple to go first.

Anyway, you don't really have to believe my reasoning above; as with the classic Monty Hall problem, we can simply enumerate the possible outcomes.
Of course, I'm making here a few assumptions which weren't *quite* explicitly stated by the OP, but which seem like reasonable interpretations of the problem statement and match the classic Monty Hall problem:

- Each of the players is equally likely to get the wormless apple.
- Monty will
*always* choose a player with a wormy apple to go first.
- Of the two players with wormy apples, both are equally likely to be chosen to go first.
- All the players know all of the above things in advance.

Given these assumptions, there are six possible situations the might occur, with equal probability, at the point where the two remaining players are asked to switch:

- $A$ has the wormless apple, $B$ went first $\to$ $A$ and $C$ remain.
- $A$ has the wormless apple, $C$ went first $\to$ $A$ and $B$ remain.
- $B$ has the wormless apple, $C$ went first $\to$ $B$ and $A$ remain.
- $B$ has the wormless apple, $A$ went first $\to$ $B$ and $C$ remain.
- $C$ has the wormless apple, $A$ went first $\to$ $C$ and $B$ remain.
- $C$ has the wormless apple, $B$ went first $\to$ $C$ and $A$ remain.

From the list above, you can easily count that, for each player, there are four scenarios where they remain, and in two of those they have the wormless apple. So it makes no difference whether they switch or not.

But what if one player, say $A$, knew that they'd never be chosen to go first? Then, if $B$ or $C$ got the wormless apple, Monty would have to choose the other one of them to go first. Thus, scenarios 4 and 5 above become impossible, while 3 and 6 *become twice as likely*. Thus, if, say, $A$ and $B$ remain, they know that they have to be in scenarios 2 or 3 — and of those, the one in which $B$ has the wormless apple (3) is now twice as likely as the one in which $A$ has it (2), so $A$ should want to switch but $B$ should not.