Re: flip a coin

From: *** T. Winter (***.Winter_at_cwi.nl)
Date: 06/08/04 

Date: Tue, 8 Jun 2004 23:15:56 GMT

In article <FFkxc.2214$1c4.2035@fed1read06> "Richard Henry" <rphenry@home.com> writes:
> "G. A. Edgar" <edgar@math.ohio-state.edu.invalid> wrote in message
> news:080620041014501988%edgar@math.ohio-state.edu.invalid...
> > I would say it means this: We go until the first time we get it,
> > counting the number of trials. That number is a random variable.
> > What is the expectation of that random variable?
> >
> > So, I am asking for the "mean" while you are asking for the "median".
> > I think by convention when we say "expected number" for something,
> > it is assumed that the mean is required.
>
> I think that is a misapplication of the definition of "median".

Indeed. In an unbounded set the median does not exist (there should be
an equal number of outcomes smaller and larger than the median, disregarding
probability). But indeed, in mathematics the definition of expectation is:
   sum n_i.p_i
where n_i is the i-th outcome and p_i is the probability.

Consider throwing a coin until you have head, what is the number of throws
you need?
The 50% rule (cited above) gives 1.
The expectation is sum n/2^n = 2 (easily shown).
There is no median.

The distinction between the 50% rule and the expectation is important when
you play for money. Casinos flourish on that distinction (not only that,
of course). 

-- 
*** t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~***/
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