Re: Running it twice question

In the case of one out, I guess I was mistaken, the trials are mutually
exclusive, but not independent.



___| reply |__________________________________________________________

garycarson <garycarson@xxxxxxxxxxxxxxxxx> wrote:
They aren't identical trials, just independent trials.

http://en.wikipedia.org/wiki/Statistical_independence

"In probability theory, to say that two events are independent
intuitively
means that knowing whether one of them occurs makes it neither more
probable
nor less probable that the other occurs."

If an event is the dealing of two cards from the stub, or more
specifically
the identity of two cards so dealt, then per the above the events
(trials)
are not independent. From the same article,

"Two events A and B are independent iff P(A ? B) = P(A)P(B).

Here A ? B is the intersection of A and B, that is, it is the event
that
both events A and B occur." [In order to avoid special font
requirements I
have substituted "?" for the inverted-U set intersection symbol.]

Considering the one-out case as an easy example, then if A is the one
out
with some other card, and B is the one out with yet another card, then
each
of A and B has a non-zero probability but their intersection does not,
so
the criterion of independence is not met.

Alas, I am unfamiliar with the concept of identical trials.


--
For mail, please use my surname where indicated:
steve@xxxxxxxxxxxxxxxxxx (Steve Brecher)

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