Standard Monty Hall Problem: Contrary to what you may be thinking this is the easy one for me to understand. Once it was explained, the 2/3 odds make perfect sense. The trouble comes with the related "Monty Fall" problem: The probabilities quoted for this are 50% win chance whether you stick or switch, rather than the 2/3 win chance for switching with the original Monty Hall problem. I ran 10,000 simulations of this game with an Excel spreadsheet. The prize was put behind a random door, the contestant picked a random door with even probability, the host picked a door from the other two at random with even probability, then the contestant randomly decided to switch or not with even probability. Games were declared either Win, Lose, or Void (if the host picked the prize door). On games that were not declared Void, switching had a 50% win record. As a check, I ran the exact same simulation with the host pick changed to reflect the original Monty Hall problem, and it resulted in the expected 2/3 chance of winning with switching. How do I make sense of this? Why should the host's intentions matter when the mechanics of the situation (Contestant chooses door, host ends up revealing goat) are identical? One thing I noticed is that all situation that end up Void in Monty Fall would end up as wins for switching in Monty Hall, as the host's pick would be constrained by the presence of the prize in the choose-able doors, which may lead to those situations being "double-counted" in the Monty Fall scenario as opposed to Monty Fall. However, I still can't get past the fact that that explanation is too Deepak Chopra-ish. The power of human intention to affect the laws of probability, and nonsense like that. I still don't get why two situations with the same objective facts (Contestant chooses door, host reveals goat) can retroactively change probabilities if you know the host's intention. Can anybody help explain that?