One reader offers a rejoinder in which he makes a fairly common mistake, writing:
Okay, let's assume the prize is behind door #2. There are actually four scenarios, not three, that can occur in the game.1. Contestant picks door # 1, Monte elimnates door #3, switching wins2. Contestant pics door #2, Monte eliminates door # 1, switching loses.3. Contestant pics door #2, Monte eliminates door # 3, switching loses.4. Contestant pics door #3, Monte elimnates door #1, switching wins.
No other combination of contestant and Monte actions can occur.
So there are actually four, not three scenarios that can play out. Two win. Two lose.
I've seen many people submit to this fallacy. First of all, the host's name is Monty, not Monte. Show some respect—the guy's a mathematical genius.
Second, the probability of initially picking the prize is 1/3. Sure, on those occasions when you're fortunate enough to select the prize by your initial guess, gray-eyed Monty will nix one empty door half of the time, and another empty door the other half of the time; the probability of each being 1/6 and adding to 1/3. Parsing it this way, in one-sixth of the trials you pick the prize and Monty opens an empty door, and in one-sixth of the trials you pick the prize and he opens another empty door. The fact still remains that you'll only pick the prize (initially) one-third of the time. So you better switch my friend!
