Re: prob. problem


"garycarson" <garycarson@xxxxxxxxxxxxxxxxxxxxxxx> wrote in message
news:v2n3m6xj0p.ln2@xxxxxxxxxxxxxxxx
On Aug 20 2009 3:37 PM, eleaticus wrote:

"PSmith" <psmipsmi@xxxxxxxxxxx> wrote in message

news:5f6c2303-d481-4fa9-b5f4-908243f2e606@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On Aug 20, 5:01 pm, John the Savage <savage0...@xxxxxxxxx> wrote:

I'm now totally convinced that the probabilities are equal. I still
think they are 1/51, not 1/52, but I could be wrong about that, too.

Pete:
If the probability is the same for those two cards (ace of spades and
two of clubs), it is going to be the same for all possible cards.
Since there are 52 possible cards, and the probabilities have to add
up to 1, then the probability of each one had better be 1/52.

e:
If it were first ace, then either, say, the 2c or 3h, the situation
would
be very different, with a much simpler structure.

It only seems like that to you because you complicated the structure
unnecessarily. The structure is really the same for 2c, 3h because the
the only thing that really matters is that the cards are placed randomly.

Gary, you are off your rocker.

The "balance" of 2c early vs As first ace is not directly accessible by that
p(position) crap.

however, the 2c, 3h situation depends only on p(position).

It's often a good idea to use the Law of Total Probability and break down
a problem into a sum of conditional probabilities, but that's not the
situation here. Whenever you're faced with a problem and breaking it into
a sum of conditional probabilities starts looking like it's
computationally nitty then it's usually a good idea to back up and look at
the problem directly, without all the conditioning.

Above.

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