For one of my sites (Eeliz.com) I needed to randomly position an image either left or right. For this I used the following random number function :

Public Function RandomNum(ByVal MaxNum As Integer, _
Optional ByVal MinNum As Integer = 0) As Integer

‘initialise the number generator
Dim r As New Random(System.DateTime.Now.Millisecond)

’some validation on the values passed in
‘on exception will return 0

If MinNum > MaxNum Then
Dim x As Integer = MinNum
MinNum = MaxNum
MaxNum = x
End If

Return r.Next(MinNum, MaxNum)

End Function

For my purpose I call the function like this – Random(2) 0 this will return either 0 or 1 randomly
You could do this – Random (21,9) – this will return a number between 10 and 20 at random