The Monty Hall Problem

There can't be many novels that include much mathematics, but Mark Haddon's "The curious incident of the dog in the night" breaks the mold.

My first reaction to the Monty Hall Problem is that there are some errors in the way the problem is solved in the novel. First is shown a purely mathematical solution, which I felt was flawed because it seems to treat the action of the game show host as a probability, whereas in some cases it is a foregone conclusion which of the remaining doors he will open. Second is a graphical solution which looks convincing, but my feeling was that there are two options here which are identical, and that oversimplifies the problem.

I look at it this way. There are really two games that you could play, and that the concept of changing your selection is a red herring. The first game is when you have 3 doors to choose from, the second is when you have two doors to choose from. If you don't play the first game (or at least forget about it once the host has opened his door) and consider the second game a case of choosing between two doors, rather than deciding whether to change your original choice, I can't see how the odds are anything but even.

I think that the conflicts in this problem boil down to this. There is a difference at the last stage of the game between making a random choice and making a switch from a previous choice. So what are the probabilities that these options will result in the same outcome?

Keith M Ellis has written a really good discourse about this problem, along with references and links.