Conference rusure::math

1201.0. "Impressive coin flips (from usenet)" by AITG::DERAMO (Dan D'Eramo, nice person) Sun Mar 04 1990 13:29

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From: ph600fji@sdcc3.ucsd.edu (Sir Six)
Newsgroups: sci.math
Subject: Impressive Coin Flips
Message-ID: <8136@sdcc6.ucsd.edu>
Date: 3 Mar 90 05:03:39 GMT
Sender: news@sdcc6.ucsd.edu
Organization: University of California, San Diego
Lines: 45
 
Suppose we have a coin, weighted to give heads with a probability p.
We flip it f times, and get h heads.  We define P as the probability
of getting h heads in f flips:
 
            h     f-h     f!
P(p,h,f) = p (1-p)     --------
                       f!(f-h)!
 
 
Furthermore, we define the "impressiveness" of <H,F> to be the
probability of flipping fewer than H heads with F flips, for a coin
weighted to give heads with probability p:
 
          H-1
I (H,F) = SUM P(p,h,F)
 p        h=0
 
 
I (0,F) = 0 for all F, p.
 p
 
 
Now, with two sets of flips, <H1,F1> and <H2,F2>, we want to
determine which is "more impressive."  Unfortunately, we don't know
the weight of the coin, so p is an unknown.  Suppose we define
 
     1    /H1   H2\
p =  - * | -- + -- |
     2    \F1   F2/
 
Then we define the "More impressive" operator:
 
MI(<H1,F1>,<H2,F2>) is true iff I (H1,F1) > I (H2,F2) with p defined
as above.                        p           p
 
We can define "Less impressive" and "as impressive" operators
similarly.  Now, to the question:
 
 
I want to know whether MI is transitive:  I.e., if
MI(<H1,F1>,<H2,F2>) and MI(<H2,F2>,<H3,F3>), does it follow that
MI(<H1,F1>,<H3,F3>)?
 
 
Don't ask me where I come up with these.  I wish I knew.

    I'm pretty sure it's not transitive.  Now to find a counterexample....

> I want to know whether MI is transitive:  I.e., if
> MI(<H1,F1>,<H2,F2>) and MI(<H2,F2>,<H3,F3>), does it follow that
> MI(<H1,F1>,<H3,F3>)?

No.  Example:

	MI(<11,19>,< 7,12>)
	MI(< 7,12>,< 2, 3>)
	MI(< 2, 3>,<11,19>)

(Dan, could you send this counterexample back to the poser?  Thanks.)