But you are ignoring the fact that changing is a winner every time you initially choose a goat and your chances of initially choosing the goat is 2 out of 3. It's true that in any particular instance you don't know which door has the car, but the odds are only 50/50 if you ignore the knowledge that there are 2 goats and only 1 car when you initially choose.Penn wrote:Fine, you want pictures.Schultz wrote:I don't know how you can look at the purple boxes in the bottom row of Fred Barclay's diagram and still come to that conclusion.
[ I initially agreed with you guys, until I looked at that diagram. ]
It starts with one alteration, I labeled the doors 1, 2 and 3.
So if you choose door 1, you will be shown door 2 and that option is removed. You can't change your mind AWAY from door 3 since you didn't choose that. So your CURRENT possibilities are
Not a thing erased affects your CURRENT choice and the resulting odds. That stuff is GONE.
This same logic is applicable if your first choice is 2 or 3.
Here's an analogy: say you roll two dice and the goal is to predict the total count on the two. Any number between 2 and 12 is possible, so there are 11 possibilities. At any particular roll of the dice, you have no way to know what the count will be, but to say that all possibilities are equal is to ignore the fact that there are more ways to get a 7 than to get a 2 or a 12. Similarly, there are more ways to win by changing than by sticking, because there are more goats than cars.