Quote:
Originally posted by soup sandwich
True, but I do know what card is in my hand (how else could I have flipped 50 non-ace of spade cards over) as well as what card is in your hand, just as Monty knows where the car is.
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If you do know what card is in your hand, and it isn't the ace of spades, then the chances that the one remaining card is are 52/52.
The canard in all of these accounts lies in the knowing serial discounting. You no longer have a valid chained probability equation. They are all separate, unconnected scenarios.
Oops. I mean, um, multiple vibrators are good.
