Why are there no combinations for certain hands when the dealer doesn't qualify for Ultimate Texas Holdem?

Question

I am trying to understand why the probabilities are 0 for the "Four of a kind" and "Full house" hands when the dealer doesn't qualify for the game described at https://wizardofodds.com/games/ultimate-texas-hold-em/ .

I don't see anything in the rules mentioning these hands specifically and the probabilities don't seem to be 0 (the dealer could have a less than a pair hand and the player could have a "Four of a kind" hand I believe).

Game rules:

1. The game is played with a single ordinary 52-card deck.

2. The player must make an equal bet on both the Ante and Blind, and can also make an optional Trips bet.

3. Two cards are dealt face down to the player and dealer. The player may look at his own cards.

4. The player can check or make a Play bet equal to three or four times the Ante.

5. The dealer turns over three community cards.

6. If the player previously checked, then he may make a Play bet equal to two times his Ante. If the player already made a Play bet, then he may not bet further.

7. Two final community cards are turned over.

8. If the player previously checked twice, then he must either make a Play bet equal to exactly his Ante, or fold, losing both his Ante and Blind bets. If the player already raised he may not bet further.

9. The player and dealer will both make the best possible hand using any combination of their own two cards and the five community cards.*

10. The dealer will need at least a pair to open.

11. The following table shows how the Blind, Ante, and Play bets are scored, according to who wins, and whether the dealer opens:

WINNER	DEALER OPENS	BLIND	ANTE	PLAY
Player	Yes	Win	Win	Win
Player	No	Win	Push	Win
Dealer	Yes	Lose	Lose	Lose
Dealer	No	Lose	Push	Lose
Tie	Yes or No	Push	Push	Push

[....]

The returns table lists probabilities of 0 for only those two hands when the dealer doesn't open and I don't understand why those probabilities are 0.

A full house hand is 3 matching cards of one rank and two matching cards of another rank and a Four of a Kind hand has all 4 cards of one rank plus a card of any other rank.

From the returns table:

DEALER QUALIFIES	HAND	PROBABILITY
No	Flush	0.001829
No	Full house	0.000000
No	Four of a kind	0.000000

Answer 1

(the dealer could have a less than a pair hand and the player could have a "Four of a kind" hand I believe)

No, that's impossible. In order for the player to have four of a kind (or a full house for that matter) there needs to be a pair on the board. Thus the dealer always has a pair or better and qualifies.