Return number of rows based on column value

For Postgres you can use:

select guid, g.i
from the_table
  cross join lateral generate_series(1, entries) as g(i);

For interest's sake, here is a solution for SQL Server 2016 that uses the built-in [pseudo]random number generator to pick a winner for you.

First, the data:

CREATE TABLE #t
(ID int,
Entries int)

INSERT #t
VALUES
(1,1),
(2,5),
(3,2),
(4,7)

Then the code:

DECLARE @r float = RAND()

SELECT TOP 1 ID, @r
FROM (
    SELECT ID,
    Entries,
    SUM(Entries) OVER(ORDER BY ID) / CONVERT(float,SUM(Entries) OVER()) AS RunningFraction
    FROM #t
) RF
WHERE RunningFraction > @r
ORDER BY ID

You can skip the variable declaration and just use WHERE RunningFraction > RAND(), but this format makes it easier to test the functionality.

Using Sql Server (2016 in my case), this gave me the answer you wanted - (not fully tested and probably many other methods).

This technique uses a Sql Server Recursive Common Table Expression.

set nocount on
Declare @GuidEntries table (Guid int, Entries int)
insert into @GuidEntries values(1, 1)
insert into @GuidEntries values(2, 5)
insert into @GuidEntries values(3, 2)
insert into @GuidEntries values(4, 7)

;WITH cteGuidEntries  AS
(
    SELECT guid, entries from @GuidEntries
    UNION ALL
    SELECT guid, entries - 1
    FROM cteGuidEntries
    WHERE entries > 1
)
SELECT guid from cteGuidEntries 
order by guid