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Dan
Posted: Thu Apr 10, 2008 1:23 pm
Guest
The March issue of the New York Times magazine "Play" contained an
article on the Steinbrenners. A sidebar by Jordan Ellenberg discussed
the probability calculation for a team winning a division.
Specifically, the article calculated the probability of the Yankees
winning the division over the Red Sox assuming that (1) the Red Sox
are actually two games better, (2) the effect of chance is normally
distributed, and (3) the standard deviation of wins is seven as: 1/
(7*sqrt(2*pi)) * Integrate[e^(-(x+2)^2/9Cool, {x: 0, Infinity}].
(Ellenberg then solves this integral to 0.3875...)

I assume this integral is derived from the normal density function,
but I'm too far removed from my college calculus to figure it out.
Any help in showing how the integral is derived from the normal
distribution density function would be appreciated.

--Dan
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Stan Devia
Posted: Thu Apr 10, 2008 8:21 pm
Guest
On Apr 11, 8:23 am, Dan <sport...@hotmail.com> wrote:
Quote:
The March issue of the New York Times magazine "Play" contained an
article on the Steinbrenners. A sidebar by Jordan Ellenberg discussed
the probability calculation for a team winning a division.
Specifically, the article calculated the probability of the Yankees
winning the division over the Red Sox assuming that (1) the Red Sox
are actually two games better, (2) the effect of chance is normally
distributed, and (3) the standard deviation of wins is seven as: 1/
(7*sqrt(2*pi)) * Integrate[e^(-(x+2)^2/9Cool, {x: 0, Infinity}].
(Ellenberg then solves this integral to 0.3875...)

I assume this integral is derived from the normal density function,
but I'm too far removed from my college calculus to figure it out.
Any help in showing how the integral is derived from the normal
distribution density function would be appreciated.

--Dan

It looks like Ellenberg has applied a z-score to the problem.

P(Z>z) = P(Z>[X-mu]/sigma) = P(Z>[2-0]/7) = P(Z>0.2857) = 1-0.6125 =
0.3875
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Paul Rubin
Posted: Fri Apr 11, 2008 7:41 am
Guest
Dan wrote:
Quote:
The March issue of the New York Times magazine "Play" contained an
article on the Steinbrenners. A sidebar by Jordan Ellenberg discussed
the probability calculation for a team winning a division.
Specifically, the article calculated the probability of the Yankees
winning the division over the Red Sox assuming that (1) the Red Sox
are actually two games better, (2) the effect of chance is normally
distributed, and (3) the standard deviation of wins is seven as: 1/
(7*sqrt(2*pi)) * Integrate[e^(-(x+2)^2/9Cool, {x: 0, Infinity}].
(Ellenberg then solves this integral to 0.3875...)

I assume this integral is derived from the normal density function,
but I'm too far removed from my college calculus to figure it out.
Any help in showing how the integral is derived from the normal
distribution density function would be appreciated.


The integrand is the density function of a normal distribution with mean
-2 and std. dev. 7. Given the limits of integration (0 to infinity),
the integral represents the probability that the random variable (which
is apparently the number of games the Yanks finish ahead of the Sox) is
greater than or equal to zero. The value of 0.3875 is correct for that
probability.

Separate from the no doubt dubious assumptions the author made to get
this far, the upper limit of infinity is curious but harmless (the
variable has an upper bound of 162, but the right tail above 162 has a
cumulative probability of something like 10 to the -121 power). More
curious is the lower limit of zero; this includes at least a piece of
the probability of a tie, which would likely result in the Sox winning a
playoff game, given that Bucky F. Dent is long retired. I would have
used 0.5 as the lower limit of integration, which drops the result to
around 0.36.

/Paul
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