Monty Hall problem

[wy125]

I didn't look too much at your code but I did follow what you were saying. You correctly noted that the host intentionally chooses to open a door where it is known that that car is not located. This makes it favorable to switch doors when the host offers. The fact that the probabilities are not equal in this scenario is what makes this problem so difficult. Most people see the probability as equal. If the host had simply chosen a door at random to open (ignoring the location of the car), revealing a goat, then the probability that the car is behind the door you chose is 1/2 and switching would do you no good (if this were really happening the host would reveal the car 1/3 of the time and the game would be over right there). This is the scenario most people imagine and why most people get it wrong and that's precisely why, as you say, 'the host is ignored as a variable' -- it makes the problem less obvious, which is the point of telling these problems in the first place.

BTW, your prob. of 1/3 and 1/3 isn't so crazy. You probably counted all trials, even those that resulted in the host revealing the car. You should get 1/3 win by keeping same door, 1/3 win by switching, and 1/3 host selects car and game over. But if you limit the trials to those where the host doesn't reveal the car, then it's 50/50 keeping the door or switching.