More teasing

Here's an interesting variant on the problem:

Each door has 1/3rd chance. Assume that Monty lets you pick TWO doors at the beginning, giving you a clear 2/3rds chance of winning. Then he shows you one of the doors you picked was bogus and offers a switch. Do you switch? If not, why is this any different than the original question? The doors and the choice are the exact same thing now.

OK, let's count pathways to the prize.

If you pick two doors, there's a 1/3 chance the prize is behind the third. In that case, it doesn't matter which of the doors you picked Monty opens. If you switch, you win.

Mathematically, there is a 2/3 chance that the prize is behind one of the doors you picked. If we assume Monty always opens the door that doesn't have a prize behind it, that means the prize will inevitably be behind the other door.

Again, in two out of three cases, Monty has no choice, and in the third, his choice makes no difference in the outcome.

The only difference is, in this case, you start out with a 2/3 chance of having won, and if you switch, you are opting for the remaining chance. You are basically exchanging your chance of having won in the first place for your chance of having lost.