>Therefore, and this the counter intuitive part, your odds are still 1 in 3. Nothing else has changed.

>Hence switching increases your odds of winning to 1 in 2 at this point (or 2 in 3 overall).

The "hence" is where you lose me, Ed. Revealing one goat would indeed boost your odds to 1 in 2. But how does *switching* from the original unknown to the remaining unknown raise them to 2 in 3 at the moment you switch? You still have no idea what's behind either remaining door.

As you explain it, it almost sounds as though "1 in 3" at the start is somehow mystically equivalent to "2 in 3" later. But "one" never used to equal "two" in my book.

I'll do DMcG's experiment later. But if it works, I still won't understand how.

Suppose I *wanted* a goat. My chances at the start would be a pretty good 2 in 3. Then Monty opens a door, which I didn't select, with a goat behind it. My chances of winning a goat now have slipped to 1 in 2. Would they increase to 2 in 3 if I switched to the one remaining door?