Populating a departure date field which is after the arrival date - sql-server

Question

Step 1
Arrival Date (Already generated) – 1.35 Million Times
Step 2
Randomise a number between 0 and 1
Step 3
Use the Randomised number produced above to create the script below
UPDATE BOOKINGS
SET DepartureDate
CASE WHEN RAND() Result = Between 0 and 0.3 = Departure Date will be 2 Nights Later
CASE WHEN RAND() Result = Between 0.3 and 0.4 = Departure Date will be 3 Nights Later
CASE WHEN RAND ()Result >0.4 = Departure Date will be either 1,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28 Nights Later

Answer 1

Do not use RAND() with a changing seed. It makes for terribly randomized data.
To get to your solution you need to create "buckets" of possible values. 3 days is supposed to happen in 10% of the cases; that makes the smallest bucket, so we need ten buckets. 2 days goes into 3 buckets. The other values go into 2 buckets each. then just use modulo to select one of the 10 buckets like this:
CREATE TABLE dbo.booking(Id INT IDENTITY(1,1) PRIMARY KEY CLUSTERED,days INT);
GO
INSERT INTO dbo.booking(days)
SELECT TOP(100000) 0 FROM sys.columns A,sys.columns B,sys.columns C,sys.columns D;
GO
UPDATE b
SET days = rndm.days
FROM dbo.booking b
CROSS APPLY (
SELECT days
FROM (VALUES(0,2),(1,2),(2,2),(3,3),(4,1),(5,1),(6,4),(7,4),(8,28),(9,28))dn(n,days)
WHERE n = ABS(CHECKSUM(NEWID(),b.Id))%10
)rndm;
GO
SELECT days,COUNT(1) cnt
FROM dbo.booking
GROUP BY days;
GO
EDIT: Updated code to not use case statement.

Answer 2

Just to let you know the final solution I used was:
UPDATE BOOKINGS
SET DepartureDate =
DATEADD(day,
CASE WHEN Rand(CHECKSUM(NEWID())) BETWEEN 0 and 0.3 THEN 2 ELSE
CASE WHEN Rand(CHECKSUM(NEWID())) BETWEEN 0.3 and 0.5 THEN 3 ELSE
Round(Rand(CHECKSUM(NEWID())) * 28,0) END END,ArrivalDate)
Thanks
Wayne