Playing Snap - probability of no matches after 4 cards

Two people are playing Snap with exactly half a deck of cards each (total 52).

What is the probability of there being no matches after four cards (in total) have been played?

My thinking is as follows (but it's wrong):

52/52 * 48/51 * ((44/50 * 46/49) + (3/50 * 47/49) + (3/50 * 47/49))

Explanation:
[FONT=&quot]52 ways to draw the first card, 48 non-matches for card 2.

[/FONT]Then the 3rd and 4th cards could be either:
- an unseen card rank is drawn for card 3 (44/50) plus a non-match (46/49) OR
- card 3 is the same as card 1 (3/50) plus a non-match (47/49) OR
- card 3 is the same as card 2 (3/50) plus a non-match (47/49)

This equals approx 0.88, however the correct answer is approx 0.83.

Can anyone help?

Rules for Snap:

1 standard deck of cards, shuffled.

Player 1 gets half, Player 2 gets half. Cards are face down.

ROUND 1: Player 1 turns over their top card, Player 2 turns over their top card. If they match (i.e. same rank) it is a 'snap'.

ROUND 2: Player 1 turns over next card, Player 2 turns over next card. If these two match, it is a snap.

etc.

This question is asking the probability of no matches after two rounds (i.e. four cards in total played. Cards 1 and 2 don't match, and Cards 3 and 4 don't match).