Conditional probability in SQL - sql

Question

I think I have end up in a bit of a dead end.
Let's say I have a dataset, which is fairly easy -
person_id and book_id. Which is pretty much factual table that says person X bought books A, B and C.
I know how to find out how many persons have bought Book X and Book Y together.
This is
select a.book_id as B1, b.book_id as B2, count(b.person_id) as
Bought_Together
from dbo.data a
cross join dbo.data b
where a.book_id != b.book_id and a.person_id = b.person_id
group by a.book_id, b.book_id
Yet again this is where my brain decided to shut down. I know that I would probably need to do it so that
count(b.person_id) / all the people that bought book A * 100
but im not entirely sure.
I hope I was clear enough.
EDIT1: I'm using SQL Server 2017 currently, so i think the correct answer is T-SQL?.
In the end the format should be something similliar to this. Also there is no cases where person A could have bought three copies of book X.
Book1 Book2 HowManyPeopleBoughtBook2
1 2 50%
1 3 7%
2 3 15%
2 1 40%
3 1 60%
3 2 20%
EDIT2: Let it be said there is hundreds of thousands of rows in the database. Yes this is bit related to a data science course i am taking - hence huge amounts of data.

Answer 1

You can extend your logic to do this:
select a.book_id as B1, b.book_id as B2,
count(b.book_id) as bought_second_book,
count(b.book_id) * 1.0 / book_cnt as ratio_Bought_Together
from (select a.*, count(*) over (partition by a.book_id) as book_cnt
from dbo.data a
) a left join
dbo.data b
on a.person_id = b.person_id and a.book_id <> b.book_id
group by a.book_id, b.book_id, a.book_cnt;
This assumes that people buy a book only once. If there are duplicates, then count(distinct) would adjust for that.

Answer 2

If you would like to generate all possible combinations of the pairs of books bought together along with the percentage of the persons who bought that combination the following can help
create table data1(book_id int, person_id int)
insert into data1
select *
from (values(1,300)
,(2,300)
,(2,301)
,(1,301)
,(3,301)
)t(book_id,person_id)
with books
as (select distinct book_id
from data1 a
)
,tot_persons
as (select count(distinct person_id) as tot_cnt
from data1
)
,pairs
as (
select a.book_id as col1 /* This block generates all possible pair combinations of books*/
,b.book_id as col2
from books a
join books b
on a.book_id<b.book_id
)
select a.col1,a.col2
,count(b.person_id)*100/(select tot_cnt from tot_persons) as percent_of_persons_buying_both
from pairs a
join data1 b
on a.col1=b.book_id
where exists(select 1
from data1 b1
where b.person_id=b1.person_id
and a.col2=b1.book_id)
group by a.col1,a.col2

Answer 3

On my phone, apologies for typo's
SELECT
SUM(bought_b) * 100.0 / COUNT(*)
FROM
(
SELECT
person_id,
MAX(CASE WHEN book_id = 'A' THEN 1 END) AS bought_a,
MAX(CASE WHEN book_id = 'B' THEN 1 END) AS bought_b
FROM
data
WHERE
book_id IN ('A', 'B')
GROUP BY
person_id
)
person_stats
WHERE
bought_a = 1
On my phone, apologies for typo's
EDIT : just saw that you want all combinations, just just one set combination.
WITH
book AS
(
SELECT DISTINCT book_id FROM data
)
SELECT
book_a_id,
book_b_id,
bought_b * 100.0 / bought_b
FROM
(
SELECT
book_a.book_id AS book_a_id,
book_b.book_id AS book_b_id,
COUNT(DISTINCT data_a.person_id) AS bought_a,
COUNT(DISTINCT data_b.person_id) AS bought_b
FROM
book AS book_a
CROSS JOIN
book AS book_b
INNER JOIN
data AS data_a
ON data_a.book_id = book_a.book_id
LEFT JOIN
data AS data_b
ON data_b.book_id = book_b.book_id
GROUP BY
book_a.book_id,
book_b.book_id
)
stats