Database schema:

CREATE TABLE country (
	id INTEGER NOT NULL -- example: ['2', '1'],
	country_name TEXT DEFAULT NULL -- example: ['Australia', 'Argentina'],
	PRIMARY KEY (id)
)


CREATE TABLE ranking_criteria (
	id INTEGER NOT NULL -- example: ['2', '1'],
	ranking_system_id INTEGER DEFAULT NULL -- example: ['2', '1'],
	criteria_name TEXT DEFAULT NULL -- example: ['International', 'Teaching'],
	PRIMARY KEY (id), 
	CONSTRAINT fk_ranking_criteria_ranking_system FOREIGN KEY(ranking_system_id) REFERENCES ranking_system (id)
)


CREATE TABLE ranking_system (
	id INTEGER NOT NULL -- example: ['2', '1'],
	system_name TEXT DEFAULT NULL -- example: ['Times Higher Education World University Ranking', 'Shanghai Ranking'],
	PRIMARY KEY (id)
)


CREATE TABLE university (
	id INTEGER NOT NULL -- example: ['2', '1'],
	country_id INTEGER DEFAULT NULL -- example: ['72', '73'],
	university_name TEXT DEFAULT NULL -- example: ['Massachusetts Institute of Technology', 'Harvard University'],
	PRIMARY KEY (id), 
	CONSTRAINT fk_university_country FOREIGN KEY(country_id) REFERENCES country (id)
)


CREATE TABLE university_ranking_year (
	university_id INTEGER DEFAULT NULL -- example: ['5', '1'],
	ranking_criteria_id INTEGER DEFAULT NULL -- example: ['2', '1'],
	year INTEGER DEFAULT NULL -- example: ['2012', '2011'],
	score INTEGER DEFAULT NULL -- example: ['98', '100'],
	CONSTRAINT fk_university_ranking_year_ranking_criteria FOREIGN KEY(ranking_criteria_id) REFERENCES ranking_criteria (id),
	CONSTRAINT fk_university_ranking_year_university FOREIGN KEY(university_id) REFERENCES university (id)
)


CREATE TABLE university_year (
	university_id INTEGER DEFAULT NULL -- example: ['5', '1'],
	year INTEGER DEFAULT NULL -- example: ['2012', '2011'],
	num_students INTEGER DEFAULT NULL -- example: ['20152', '2243'],
	student_staff_ratio REAL DEFAULT NULL -- example: ['6.9', '8.9'],
	pct_international_students INTEGER DEFAULT NULL -- example: ['25', '27'],
	pct_female_students INTEGER DEFAULT NULL -- example: ['33', '37'],
	CONSTRAINT fk_university_year_university FOREIGN KEY(university_id) REFERENCES university (id)
)

Question: Calculate the average number of criterias among "Times Higher Education World University Ranking","Shanghai Ranking" and "Center for World University Rankings".
Evidence: average number of criterias refers to DIVIDE(SUM(id), 3); "Times Higher Education World University Ranking", "Shanghai Ranking" and "Center for World University Rankings" refers to system_name IN ('Times Higher Education World University Ranking', 'Shanghai Ranking', 'Center for World University Rankings');

SQL:
SELECT (SUM(CASE WHEN T1.system_name = 'Center for World University Rankings' THEN 1 ELSE 0 END) + SUM(CASE WHEN T1.system_name = 'Shanghai Ranking' THEN 1 ELSE 0 END) + SUM(CASE WHEN T1.system_name = 'Times Higher Education World University Ranking' THEN 1 ELSE 0 END)) / 3 FROM ranking_system AS T1 INNER JOIN ranking_criteria AS T2 ON T1.id = T2.ranking_system_id