The Monty Hall Problem

Feeling uninspired these days.  So much to do, so little time in which to do it.  Days like these call for random and pointless posts, so here I go.

The Monty Hall Problem

You are given a choice: select any one of three doors.  You are told in advance that one of the doors conceals a prize, while the other two contain nothing.

You select a door, and then the chipper game show host opens one of the doors you did not select, always to reveal nothing.  Now you have the choice to switch, or stay with your original choice.  What do you do?

Answer

Like I said, uninspired.  Overdone and boring for that matter.  But in any case… for some reason this problem was not immediately obvious to me when first I considered it.  Although I quickly wanted to switch doors, it was for the wrong reason and with the wrong excepted gain in value.  I had figured my odds of guessing were going from 1/3 to 1/2, which is certainly an improvement, but its not actually what is going on in this problem.  It wasn’t too hard to convince myself of the correct answer, so here it is.

Yes, you always want to switch doors.  In fact, doing so doubles your chances of winning the prize.  The reason for this gain is because the game show host always shows you an empty door, and in doing so, removes duplication from the system.  After he pulls this trick, and the only options available are one car and one door with nothing.  By eliminating duplication, the host has guaranteed that in switching, you always end up with the opposite prize to what you selected originally.

Simple enough, yes?  If you happened to choose a car the first time, switching always lands you with nothing.  If you happened to select a losing door the first time, switching always wins you the prize.  And that is the key.  You can’t switch from a losing door to a losing door, you can only switch to the winning one.

So going in what were the odds?  Two out of three doors landed you on nothing, and one out of three doors land you with a car.  Since switching also swaps your prize as well, two out of three times you will switch from nothing to the car, and only one out of three times will you switch from the car to nothing.  Your odds went from one in three to two in three.

Another way to think about this is to expand the problem to many more doors, say 50.  You select any one of the 50 door, and then the host opens 48 other doors to reveal nothing.  All you have left is the one you picked originally, and one other door in the lineup.  Now it is obvious you should switch, and your odds are much better than pure random one in two.  In this example, 49 out of 50 times you would have picked a door with nothing, and in every one of those cases the switch gets you the opposite prize (the car), so with the switch your odds go from 1/50 to 49/50.

People tend to get caught up by the instantaneous odds, and have trouble convincing themselves the choice to switch or not has any impact on the outcome.  But to the contrary, this is a perfect example of how context can provide meaning and insight to an otherwise colorless math problem.  Here, the relationships created between the original probabilities and the new outcomes are critical.

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