Percentages

The key is going to be getting the right percentage of each number. To begin with each number (1 to 10) has a 10% percent chance of begin picked. If you set the first five numbers to 50% then each would have to exist in the "pick list" twice as much as a number with a 25% chance and so on . . .

Here is what I came up with . . .

VB Code:

Option Explicit
 
Dim i As Integer
Dim x As Integer
Dim z As Integer
Dim intNumberEntries As Integer
Dim intSum As Integer
 
Private Sub Form_Load()
    List1.Clear
    List2.Clear
    For i = 1 To 10
        List1.AddItem i
        List2.AddItem 10
    Next i
    Text1.Text = ""
    Text2.Text = ""
    Debug.Print List1.ListCount & " " & List2.ListCount
End Sub
 
Private Sub List1_Click()
    List2.Selected(List1.ListIndex) = True
    Text1.Text = List2.List(List1.ListIndex)
    Text1.SetFocus
    Text1.SelStart = 0
    Text1.SelLength = Len(Text1.Text)
End Sub
 
Private Sub List2_Click()
    List1.Selected(List2.ListIndex) = True
End Sub
 
Private Sub Text1_Change()
    If Text1.Text = "" Then Exit Sub
    List2.List(List1.ListIndex) = Text1.Text
End Sub
 
Private Sub Command1_Click()
    'Here is where it gets interesting!
    List3.Clear
    intSum = 0
    For i = 0 To List2.ListCount - 1
        intSum = intSum + List2.List(i)
    Next i
    For i = 0 To List1.ListCount - 1
        intNumberEntries = ((List2.List(i)) / 100) * intSum
        For x = 1 To intNumberEntries
            List3.AddItem List1.List(i)
        Next x
    Next i
    Randomize
    'now lets get that number!
    z = (Rnd(1) * List3.ListCount) + 1
    Text2.Text = List3.List(z)
End Sub

and here is the form . . .