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Q: Probability for mathtalk ( Answered 5 out of 5 stars,   0 Comments )
Question  
Subject: Probability for mathtalk
Category: Science > Math
Asked by: racecar-ga
List Price: $5.00
Posted: 11 May 2004 12:15 PDT
Expires: 10 Jun 2004 12:15 PDT
Question ID: 344734
Hi mathtalk,

You recently answered a probability question from milburn-ga (339435).
 It is very similar to a problem I have tried and failed to solve:
what is the probability that in going through the two decks, no two
cards match in rank?  That is, what is the answer to milburn's
question if you don't care about suit?  The question arose in a
slightly different scenario:  there's a very simple solitaire game
where you turn over each card in a shuffled deck one at a time, saying
"ace, two, three, .... king, ace, two... ".  You lose as soon as you
say the name of the card you just turned up, and you win if you make
it through the whole 52 card deck.  I wanted to calculate the
probability of winning.  Can you help?

Clarification of Question by racecar-ga on 11 May 2004 14:07 PDT
The numerical probability is ~0.0163

Request for Question Clarification by mathtalk-ga on 11 May 2004 18:33 PDT
Hi, racecar-ga:

A fellow Researcher, richard-ga, pointed out to me that he'd
previously posted this Question and gotten a five-star Answer from the
peerless pinkfreud-ga:

[Q: Odds of Winning A Simple Solitaire Game]
http://answers.google.com/answers/threadview?id=52175

Since then it seems the location at Dartmouth of the paper that gives
an analytical solution has moved:

[Frustration solitaire and rank-derangements] 
http://www.math.dartmouth.edu/~doyle/docs/rank/rank/rank.html 

The authors Doyle, Grinstead, and Snell give an exact rational value,
first computed by Doyle in 1994, which is approximately:

0.01623272746719463674...

If you are still game for me to try to add something to the
discussion, I'll go ahead and post some experiments at trying to
reproduce their computations.

regards, mathtalk-ga

Clarification of Question by racecar-ga on 12 May 2004 11:15 PDT
Thanks, the question has been answered, nothing further needed.  I
suppose the question fee could most fairly be divided three ways,
between mathtalk, richard, and pinkfreud.  But as that is not easily
possible, I guess it's up for grabs.
Answer  
Subject: Re: Probability for mathtalk
Answered By: pinkfreud-ga on 12 May 2004 15:48 PDT
Rated:5 out of 5 stars
 
I'm glad that richard, mathtalk, and I were able to provide useful info.

Mathtalk is egging me into posting the official answer, and I can't
resist being egged by an erudite colleague. ;-)

As mentioned above, I answered a similar question in 2002. My ability
to locate relevant information hung upon my awareness that the
solitaire game described is called "Frustration."

[Q: Odds of Winning A Simple Solitaire Game]
http://answers.google.com/answers/threadview?id=52175

Google search strategy:

Google Web Search: "frustration solitaire"
://www.google.com/search?hl=en&q=%22frustration+solitaire

Best regards,
pinkfreud
racecar-ga rated this answer:5 out of 5 stars

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