The correct approach here seems to be to calculate all the possibilities that a hand contains 4 cards of the same rank (A) and all the possibilities that it contains exactly 3 cards of the same rank (B). Then the probability is equal to $\frac {A}{A+B}$. Where in this case: $A = 13 * 48$, $B = 13 *{48 \choose 2} * {4 \choose 3}$.
I understand this approach, but I am struggling to understand why it is (assumingly) wrong to transform the problem to the following: Three cards (all the same rank) have been removed from the deck, so the deck contains 49 cards and 2 will be dealt. There is only 1 card left in the deck that I need to get in order to have 4 of the same rank (including the 3 three that were removed in the beginning). What would be the correct probability in this situation, and why is it different (if at all) to the original problem?