[tyleraland]

I only started to understand how to look at this problem once I realized the host was going to eliminate a bad door _no matter what was chosen_. Of course if you have already selected a bad door he won't eliminate it.

Another useful exercise is to stretch the number of doors to some large N and suppose the choice was posed repeatedly with a decreasing number of doors. If you can have anywhere from 2 to N guesses and as N decreases the number of bad doors decreases, it seems more obvious that you should always switch.

[ustolemyname]

Not switching until the end should work out better than switching multiple times.

4 doors, in 8ths to simplify fractions.

  2/8,2/8,2/8,2/8 => initial conditions, choose leftmost
  2/8,3/8,3/8 => random bad door removed, choose a random 3/8
  3/8,5/8 => remove another door.

Next choice is only 5/8, whereas if we had waited it would have been 6/8.