Quote:
Originally posted by bilmore
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.
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I think we're talking past each other.
I know which of the 52 is the ace of spades.
You, who does not know which card is the ace of spades, choose a card. Your chances are 1/52 of it being the ace of spades.
I, who knows where the ace of spades is, remove 50 cards and declare that the ace of spades is either your card or the one last face down card.
The odds of your card being the ace of spades remain 1/52. The odds of the other card being the ace of spades is 51/52.
The odds of multiple vibrators being good are 52/52.
