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==Against all odds?==
==Against all odds?==


113,527,276,681,000,000 to 1<br>
Those were the odds local golfer Jacqueline Gagne could make seven holes in one in 14 weeks.<br>
To estimate skill and luck percentages Sherman uses a statistical procedure called analysis of variance (ANOVA).  There  many discussion of ANOVA on the web.  One of these  [http://www.unc.edu/courses/2006spring/psyc/130/001/variance.htm Variance and the Design of Experimentsl] begins with the following simple example.
The Desert Sun<br>
Palm Springs<br>
Larry Gohannan<br>
April 28, 2007


Lee Sechrest suggested this would be a good example to discuss how we should react to a claim that an event with an unbelievable small probabillity has occurred.


Jacqueling Gagne is a 47 year old who retired from her job as a computer and moved to Rancho Mirage, California.  Here she continued her love of sports by playing golf four or five times a week.  Between January and May she made seven holes in one.  In the Desert Sun article we read


<blockquote>Before she recorded her eighth this week, Professor Mike McJilton of the College of the Desert Math and Science Department calculated the odds of making seven aces in approximately 65 rounds (five rounds a week for 13 weeks) as Gagne did is just under 114 million billion to 1. As a decimal number, the chance of making the seven aces is 0.0000000000000000088
The  data might look like this
</blockquote>
<p>&nbsp;</p>
 
<table width="70%" border="1">
McJilton used the poisson distribution presumeably as an opproximation to the binomial distribution. For this he needed an estimate for the probability that a reasonable golfer makes a hole in one in a round. There have been many attempts to estimate this.  You can read about this [http://golf.about.com/od/faqs/f/holeinoneodds.htm here]. It is said here that the Golf Digest hired Francis Scheid, the retired chairman of the math department at Boston University, to calculate the odds using the latest and best information available. The odds Scheid came up with  5,000 to 1 for a "low-handicapper," 12,000 to 1 for an "average player" were lower than any others estimate. Using the odds 1/5000 and  Mathematca it is easy to compute the probability of 7 success using the binomial distribution with p = 1/5000 and n  =  65.  Doing this gives McJilton’s result.
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    <td><div align="center">Treatment 1</div></td>
The next mention of calculating the odds was in the [http://sports.espn.go.com/golf/news/story?id=2881134ESPN sports news May 24, 2007]. By this time Gagne has made 10 holes in one in 16 weeks. Here we read
    <td><div align="center">Treatment 2</div></td>
 
  </tr>
<blockquote> We asked one of the world's foremost mathematicians, Joseph B. Keller, professor emeritus of mathematics and mechanical engineering at Stanford, to compute the odds of making 10 aces in 80 rounds (using a previous finding by Golf Digest that the odds of an average player making an ace in any given round are 1 in 5,000). Keller's answer: "Roughly two chances out of 10 followed by 24 zeros. This is the same chance as picking one particular molecule out of all the molecules in 50 gallons of air."
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    <td><div align="center">4</div></td>
By June 3 Jacqueline  had made 14 holes in one in the past 5 months.  Again using the Binomial model and Mathematica we find the probability of this event is .0000000000000000000000000000000000003
    <td><div align="center">7</div></td>
 
  </tr>
Of course people were suspicious of a fraud but they were able to find witnesses to all of her hole in ones.  Her authenticity gained further support when an ABC TV station sent a crew to interview her and she hit a hole in one while they were there. You can see this hole in one [http://abcnews.go.com/Video/playerIndex?id=3203645 here].
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    <td><div align="center">6</div></td>
There has been a lot of discussion about this event.  See in particular the discussion by Carl Bialick and the comments of his readers at the  [http://blogs.wsj.com/numbersguy/golfing-ace-jacqueline-gagne-strikes-again-133/ Numbers Guy Blog]. Carl  Bialik writes a column called The Numbers Guy for  the Wall Street Journal and maintains this Blog.
    <td><div align="center">5</div></td>
 
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We like the following remarks made by Professor Sechrest when he suggested this topic.
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    <td><div align="center">8</div></td>
<blockquote>The accounts of
    <td><div align="center">8</div></td>
the statistical estimation procedures were quite interesting and represented
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something of a tour-de-force in calculating probabilities.
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What seemed to me to be odd, however, is that no one recognized, at least not
    <td><div align="center">4</div></td>
explicitly, that the probability of such a record, once beyond a certain point,
    <td><div align="center">9</div></td>
ceases to be interesting. One simply reaches a point at which it is evident that
  </tr>
the woman has some special skill. Packing on further improbabilities does not
  <tr>
provide any new information. It is my understanding that she came to the
    <td><div align="center">5</div></td>
attention of people for her golfing prowess some time ago, i.e., before she
    <td><div align="center">7</div></td>
reached ten, further supporting the conclusion that she has more than luck
  </tr>
going for her.
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    <td><div align="center">3</div></td>
I have a pair of loaded dice that I use for demonstrations from time to time.
    <td><div align="center">9</div></td>
After I have rolled about six sevens out of, say, nine tries, everyone agrees
  </tr>
that the dice are loaded. Going ahead to roll another six sevens does not
</table>
impress anyone as remarkable.</blockquote>
 
 
 
===Discussion==
 
(1)  Of course we only hear about unusual events so we should really have a bigger sample space.  How could this be done for this event?
 
(2) Keith Devlin is reported to have said "The only justifible probability is 100% since the event happened.  What do you think about this?
 
(3)  What odds would you give for Jacqeuline getting a hole in one in her next round of play?

Revision as of 18:33, 4 September 2007

Against all odds?

To estimate skill and luck percentages Sherman uses a statistical procedure called analysis of variance (ANOVA). There many discussion of ANOVA on the web. One of these Variance and the Design of Experimentsl begins with the following simple example.


The data might look like this

 

Treatment 1
Treatment 2
4
7
6
5
8
8
4
9
5
7
3
9
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