OK, there's a glimmer of understanding starting here... Because the magician is allowed to see all the cards, he has a 51/52 chance of finding the ace so he has skewed the probability in his favor. I only have a 1/52 chance based on my initial random selection. So at this point I'm offered a...
In the beginning the magician is allowed to select from and eliminate all but one card, thus giving him a 51/52 chance of winning and I have a 1/52 chance. You're saying that once he eliminates 50 of the cards that I still have a 1/52 chance of having the right card.
I submit that it is now a...
OK, here is the same situation presented to us in different words. The crux of the matter (and the most important) is that in each case we are presented with a new problem in the end, with a new set of variables.
In the case of the three doors, your first choice is 1 out of three, then you...
I've read all these comments and theories and I understand what's been said, but to me it boils down to this:
When you make your initial choice, you would of course have a 33% chance of winning. But, no matter which door you choose, the host is going to open one of the losing doors, then...
I look at this problem more simply. I initially have a 33% chance of finding the car. Regardless of which of the three doors I choose, the host always has a 100% chance of being able to reveal a door with one of the two goats. Because he knows where they are, and regardless of whether I...