| >Does this mean there's an even chance of the "counter" (whatever that is)
being white or black ? It seems like this would matter.
Yes, it does. Dodgson's answers depend on unstated a priori assumptions like
this which aren't necessarily sound. They are (barely) acceptable as a
convention, but modern conventions would usually require more explicitness.
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| > 5. A bag contains one counter, known to be either white or black.
> A white counter is put in, the bag shaken, and a counter drawn out,
> which proves to be white. What is now the chance of drawing a
> white counter?
If I initially believed the probability of the bag holding a white counter to
be p, then we have the following cases:
p/2: I draw the same counter I put in, and the remaining one is white.
(1-p)/2: I draw the same counter I put in, and the remaining one is black.
p/2: I draw the original counter (white), and the one I put in remains.
(1-p)/2: I draw the original counter (black), and the one I put in remains.
The fourth case is eliminated, since the counter I drew proved to be white.
Of the remaining cases, the probability that a white counter remains in the
bag is
2p / (1+p)
If p=1/2, then we have a 2/3 chance that a white counter remains, which makes
sense because the four cases above are all equally likely.
-- Bill
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