Here is just one more way of looking at the 3 doors problem:

Say every contestant chooses not to switch.  Then they will win, on average,
1/3 of the time, which is their probability of winning when they first
picked their door, even if Monty reveals one of the empty doors.

The converse to not switching, provided everything else stays the same
(Monty always reveals an empty door), is switching.  The converse
probability to 1/3 (not switching) is 1 - 1/3, or 2/3 (switching).  It is
better to switch.  Marilyn was correct.

I verified this with the following VB program:

   Dim w As Integer, p As Integer, h As Integer, i As Integer, x As Integer

   Randomize

   x = 0
   For i = 1 To 20000           '20000 iterations
      w = Int(Rnd * 3 + 1)              'winning door (1-2-3)
      p = Int(Rnd * 3 + 1)              'player's pick (1-2-3)
      h = 1                             'host's pick
      If (h = w) Or (h = p) Then        'make sure host's pick
         h = 2                          '  is not the winning
         If (h = w) Or (h = p) Then     '  door, or the player's
            h = 3                       '  door.
         End If
      End If
      '* p = 6 - h - p          'switch
      p = p                             'no switch
      If p = w Then             'if player won,
         x = x + 1              'then tally
      End If
   Next i

   Debug.Print x

switching produced numbers in the low 13000 range
not switching produced numbers in the high 6000 range

Curt