|
Let me be the first person to fully solve this dilemma.
The 2/3 chance on switching holds, but only if the game show host does the same thing every time, regardless of the choice of the contestant.
If that is the case, when you choose A [1/3 chance] instead of (B or C) [2/3 chance], the host by removing an empty door from (B or C) and giving you the option to switch, effectively simply gives you the option of switching to (B or C) combined. There is always at least one empty door among (B or C), so the host is always able to remove one, and the act of removing one door is actually insignificant.
Think of it this way: (B or C) has a 2/3 chance, and at least one of (B or C) is empty. By pointing out one of both as empty, the combined chance for (B or C) does not change. All it changes is merge the chances of both doors into one.
Now, that is pretty simple. A problem arises when we give the host the option of not revealing one door.
If the host has the option of not revealing a door and indeed not always does reveal a door, and he has knowledge of which door is the right one, he might intentionally try to take your choice away from the right one by giving you a "more logical" alternative. Or, on the other hand, he might try and guide you towards the right choice.
Simply put, the 2/3 chance on switching is valid if and only if the elimination of a door and the subsequent option of switching are introduced completely independent of the validity of the original choice.
__________________
/(bb|[^b]{2})/
|