One more time. Suppose there are five doors, one with a prize. Initially the odds of winning the prize are 1/5, of losing 4/5. A choice is made, the host reveals a losing door, and asks his question - switch doors or stick with your choice. It doesn't matter whether you stay with your choice or not. There are now four unopened doors and the odds of winning are 1/4. Repeat this again and the odds become 1/3, similar to the Monty Hall problem. Repeat it once more and now, with two doors left, one chosen and the other not, the odds of winning become 2/3 by switching doors? Seems not to make sense! That's what walking the dog gets you! It seems to me that regardless of how the question is posed and regardless of any previous choices made or revealed, in the end there are only two choices left with a 1/2 probability of selecting the winner.