OK, no good deed goes unpunished! This approach works quite well, and
indeed, the second expression, at least 1, gives great results.
Now the second part of the question: Show that we get at least 1 with
"high probability" by choosing t to be (3/2)^n * poly(n) .. where n is
the original n in n-flips of the highly biased (2/3 heads) coin.
If n is 10, for example, using (3/2)^n I get 0.632175881706547, but if
I include a polynomial factor of n I get 0.999958413014979 ..
interesting! I'd call that "high probability"!
So what I lack here is a common technique for deriving a value of t
delivering high probability of success. In this case one that would
point to (3/2)^n * poly(n) with the 2/3 biased coin.
-- Owen
On May 4, 2010, at 4:30 PM, George Duncan wrote:
Owen,
Yes, you are right, it is (2/3)^n, call it p.
Then the probability of exactly 1 such happening in t repetitions of
the n-flip experiment is tp(1-p)^(t-1) (that's a binomial
probability),
or if you mean at least 1 happening, the probability is 1-(1-p)^t
(i.e., 1 - prob of no such happenings).
George
On Tue, May 4, 2010 at 4:18 PM, Owen Densmore <[email protected]>
wrote:
My probability is failing me. Could someone answer this?
I have a very biased coin that comes up 2/3 heads, 1/3 tails. I
want to do an experiment of n coin flips.
The probability that all are heads is (2/3)^n, right?
What I'm interested is the related question: Lets suppose I repeat
the experiment t times. How likely am I to get all heads once in a
series of t sets of n flips?
-- Owen
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============================================================
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Meets Fridays 9a-11:30 at cafe at St. John's College
lectures, archives, unsubscribe, maps at http://www.friam.org
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Meets Fridays 9a-11:30 at cafe at St. John's College
lectures, archives, unsubscribe, maps at http://www.friam.org
