First-seeded Dorothy Bundy defeated Dorothy Stevenson 6–3, 6–2, in the final to win the women's singles tennis title at the 1938 Australian Championships.^[1]

The seeded players are listed below. Dorothy Bundy is the champion; others show the round in which they were eliminated.

1. Dorothy Bundy (champion)
2. Nancye Wynne (semifinals)
3. Thelma Coyne (quarterfinals)
4. Joan Hartigan (quarterfinals)
5. Nell Hopman (semifinals)
6. May Hardcastle (quarterfinals)
7. Dorothy Stevenson (finalist)
8. Do Workman (quarterfinals)

• Q = Qualifier
• WC = Wild card
• LL = Lucky loser
• r = Retired

	Quarterfinals							Semifinals							Finals
	1	Dorothy Bundy	6	6
	6	May Hardcastle	3	4
								1	Dorothy Bundy	6	6
								5	Nell Hopman	2	3
	3	Thelma Coyne	4	2
	5	Nell Hopman	6	6
															1	Dorothy Bundy	6	6
															7	Dorothy Stevenson	3	2
	2	Nancye Wynne	6	6
	8	Dorothy Workman	4	3
								2	Nancye Wynne	3	3
								7	Dorothy Stevenson	6	6
	4	Joan Hartigan	3	6	1
	7	Dorothy Stevenson	6	1	6

	First round							Second round							Quarterfinals							Semifinals
								1	Bundy	6	8
									Walters	3	6
															1	Bundy	6	6
															6	Hardcastle	3	4
								6	Hardcastle
									Blick	w/o
																						1	Bundy	6	6
	3	Coyne	6	6																		5	Hopman	2	3
		Woods	2	0				3	Coyne	6	6
		O'Halloran							O'Halloran	3	4
		Cox	w/o												3	Coyne	4	2
	5	Hopman	6	6											5	Hopman	6	6
		Stern	1	2				5	Hopman	6	6
		Mauger	6	0	6				Mauger	2	1
		Coate	1	6	4

	First round							Second round							Quarterfinals							Semifinals
	2	Wynne	6	6
		Weston	2	1				2	Wynne	6	6
		Vickery	6	6					Vickery	4	3
		Bond	2	0											2	Wynne	6	6
	8	Workman	6	6											8	Workman	4	3
		Stevenson	1	3				8	Workman	6	6
		Wilson	9	6					Wilson	2	1
		Skinner	7	3																		2	Wynne	3	3
																						7	Stevenson	6	6
								4	Hartigan	7	6
									Chitty	5	2
															4	Hartigan	3	6	1
															7	Stevenson	6	1	6
								7	Stevenson	6	4	6
									Westacott	4	6	3

• [1] Source for the draw at ausopen.com