According to the FIDE handbook (C.04.1 Basic rules for Swiss Systems):
For each player the difference between the number of black and the number of white games shall not be greater than 2 or less than –2.
Each system may have exceptions to this rule in the last round of a tournament.
No player shall receive the same colour three times in a row.
Each system may have exceptions to this rule in the last round of a tournament.
And according to the FIDE handbook (C.04.3 FIDE (Dutch) System):
non-topscorers (see A.7) with the same absolute colour preference (see A6.a) shall not meet (see C.04.1.f and C.04.1.g).
An absolute colour preference occurs when a player’s colour difference is greater than +1 or less than -1, or when a player had the same colour in the two latest rounds he played. The preference is white when the colour difference is less than -1 or when the last two games were played with black. The preference is black when the colour difference is greater than +1, or when the last two games were played with white.
So how can you legally get a better ratio than what you propose? Longer tournaments are out; the longer you go the more times you'll have to play White in order to keep it within 2. The obvious way would be a 3 round tournament, but that seems awfully short and on the edge of your "unnatural" 1 or 2 round tournaments.
But a 3 round tournament isn't the only way you could play 3 games. You could have a player in a 4 round Swiss with a first-round bye. The colors could then go XBBB. This would seem to be somewhat improbable, however, since any players who had played all the games would have a weaker color preference in the last round. You'd need multiple people at XBB going into the last round that have to play each other.