What is interesting is that out of the six possible scenarios, three result in winning and three result in losing, which supports the 50/50 solution.
Nope.
Say you picked door #1 and another door is opened with nada behind it.
For the first three games, you choose #1 and "switch" each time, for the second three games, you choose #1 and "stay" each time, and the
host always opens a loser.
Here are the results.
Switch each time:
GAME 1: PRIZE NADA NADA ~ Switch and you lose.
GAME 2: NADA PRIZE NADA ~ Switch and you win.
GAME 3: NADA NADA PRIZE ~ Switch and you win.
............................................................
Stay each time:
GAME 4: PRIZE NADA NADA ~ Stay and you win.
GAME 5: NADA PRIZE NADA ~ Stay and you lose.
GAME 6: NADA NADA PRIZE ~ Stay and you lose.
When you switch, you win 2/3 of the time and lose 1/3, but when you don't switch, you only win 1/3 of the time and lose 2/3.
