Quote:
Originally Posted by CaptHayfever  Spoiler Below There are 3 possible situations.
(Case 1) Monty picks door 1.
This has a 2/3 chance of happening, & once it happens, your odds for switching are 2/3 by the original Monty Hall problem, since opening door 1 reveals nothing conclusive about doors 2 & 3.
2/3*2/3 = 4/9
(Case 2) Monty picks door 2 (1/3 chance of happening), & you had already picked door 2 (1/3 chance of happening).
You should switch to door 1 right now, 100%.
1/3*1/3*1 = 1/9
(Case 3) Monty picks door 2 (1/3 chance), & you had already picked door 3 (1/3 chance).
Again, switch to door 1, 100%.
1/3*1/3*1 = 1/9
(Case 4) Monty picks door 2, & you had already picked door 1.
The car is behind door 1; stay put.
Odds for switching = 0.
Total: 4/9+1/9+1/9+0 = 6/9 = 2/3. And remember, "I'm-a Luigi, number one!" |
Spoiler Below What if the car (in Case 4) is behind Door 3? Your saying that
D1: Car (Player)
D2: Zonk (Monty)
D3: Zonk
But it's still possible for this to happen:
D1: Zonk (Player)
D2: Zonk (Monty)
D3: Car
Remember that Monty doesn't pick the same door the Player picks. Therefore, it's still possible that the Car is behind Door Three.
And Your Case Two is impossible for the above reasoning.