The Monty Hall conundrum is a brain-teaser based on the TV game show “Let’s Make a Deal” in which a contestant is given a choice of three doors, with a prize behind one of them. The host then opens one of the remaining two doors* to reveal no prize, then gives the contestant the opportunity to switch to the remaining door. It has been shown that human nature is to stick with the original choice, but probability calculations prove that it is better to switch.

One of the most interesting aspects of this problem is that it is quite difficult to convince people that switching is better. The traditional explanation is this: At the beginning, the contestant has a 1/3 probability of winning and, therefore, a 2/3 probability of losing. What many people call “common sense” would say that, after the host opens one of the other doors, the probability is 50/50, but this is incorrect. Because the host will never open a door to reveal the prize (that is, the probability of the host “winning” is zero), the odds of the contestant winning remain 1/3 with the original door and, therefore, 2/3 by switching. This seems to be a difficult concept to grasp. I therefore propose an intermediate step in the explanation.

Assume that, instead of opening a door to reveal nothing, the host offers the contestant the opportunity to take both of the remaining doors instead of the original door. Clearly, having two doors instead of one increases the odds of winning. A little thought should convince you that there is no difference between switching to both remaining doors and switching to the one remaining door after the host opens the other. After the contestant switches, both doors will be opened. It does not matter in which order they are opened, nor whether the host or the contestant does the opening. And, believe it or not, it does not matter whether the decision to switch is made before or after the “losing” door of the two remaining is revealed. The odds remain 1/3 by staying with the original choice and 2/3 by switching.

*In 2/3 of the cases, the contestant has not chosen the door with the prize, and the host has no choice but to open the other door that has no prize. In 1/3 of the cases, the contestant has chosen the door with the prize, and the host must make a  choice of which door he reveals. Per the discussion in the comments, the host’s choice of which door to open in this case must be random (that is, unpredictable) to avoid giving the contestant more information and thereby changing the odds. If this choice is predictable, there is a possibility that the contestant may know with certainty that he must switch to win. With this scenario eliminated, the odds for the remaining scenarios become 50/50 between staying or switching. A random choice means that no additional information is given (we already knew that one of the doors unchosen by the contestant had no prize), and the odds do not change.

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