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Q: Continuation of 3-digit question for mathtalk ( No Answer,   1 Comment )
Question  
Subject: Continuation of 3-digit question for mathtalk
Category: Science > Math
Asked by: davidj-ga
List Price: $2.00
Posted: 09 Feb 2004 17:05 PST
Expires: 10 Feb 2004 22:21 PST
Question ID: 305184
Hi Mathtalk,

I closed the previous question without a detailed review of your
answer. Then I realized that your answer was not correct!

For the PPX condition, you give a probability of 0.006. However, PPX
cannot exist, because naturally with two placed digits, the third
cannot be a misplaced "X". It would have to be another P or a -.

Therefore, this skews all the rest of the data and therefore, none of
the previous answers are right with the exception of - - - and PPP
probabilities.

The data I have shows that a PP- has a probability of .028, a P-- or
better as .271, and XX- or better as .150. However, I don't know how
they arrived at those answers.

Please clarify your answers and let me know the proper probabilities. Thanks!

Request for Question Clarification by mathtalk-ga on 09 Feb 2004 18:56 PST
Hi, davidj-ga:

I realized that I may have incorrectly interpreted the treatment of
duplicate digits and posted a note on that original question earlier
tonight.

My thought was that a "guessed" digit could match more than one
position, hence the PPX possibility.

I'll be happy to post the Answers worked both ways on the other
Question if you want to close this one.

regards, mathtalk-ga

Request for Question Clarification by mathtalk-ga on 09 Feb 2004 22:52 PST
Hi, davidj-ga:

Please see the computations provided on your earlier Question.

These are the results with the revised scoring interpretation:

Pr( _ _ _ ) = 0.343

Pr( X _ _ ) = 0.315

Pr( X X _ ) = 0.069

Pr( X X X ) = 0.002

Pr( P _ _ ) = 0.192

Pr( P X _ ) = 0.048

Pr( P X X ) = 0.003

Pr( P P _ ) = 0.027

Pr( P P X ) = 0.000

Pr( P P P ) = 0.001
             ------
              1.000

Assuming the "scores" are ranked in the order shown, then "P P _" or
better would occur with chance 0.027 + 0.001 = 0.028, and "P _ _" or
better with chance 0.271.  With regard to the chance of "X X _" or
better, if this were to exclude these cases:

Pr( _ _ _ ) = 0.343
Pr( X _ _ ) = 0.315
Pr( P _ _ ) = 0.192
             ------
              0.850

then in that sense, the chance of "X X _" or better would be 0.150. 
However those assessments require that both "X X _" is not better than
"P _ _" and vice versa.

Hope this clears up the confusion!

regards, mathtalk-ga
Answer  
There is no answer at this time.

Comments  
Subject: Re: Continuation of 3-digit question for mathtalk
From: aht-ga on 09 Feb 2004 17:30 PST
 
Hi davidj-ga:

Thought I'd mention to you that, even after you accept and rate an
Answer, you are able to ask for further clarification within that
Answer, without needing to post a new question. The Researcher who
provided the Answer, will be informed by the Google Answers system
that you have requested additional clarification, and will respond
accordingly.

Good luck!

aht-ga
Google Answers Researcher

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