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From: msp@warwick.UUCP (Mike Paterson)
Newsgroups: net.math
Subject: Re: probability formula needed
Message-ID: <140@ubu.warwick.UUCP>
Date: Fri, 20-Sep-85 11:56:43 EDT
Article-I.D.: ubu.140
Posted: Fri Sep 20 11:56:43 1985
Date-Received: Wed, 25-Sep-85 07:43:56 EDT
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Organization: Computer Science, Warwick University, UK
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Xpath: warwick ubu

The probability needed can be viewed as the probability of getting 
two adjacent picture cards with different values (e.g. JK) in a 
random deal. The probability of failure seems to be the coefficient 
of x^38 in
	41!40!/52!.e^x.(24+24x+12x^2+x^3)^3

My pocket calculator is smoking but the answer looks like
	49220186227/447219915450  (about .110058129)

This would give a winning percentage of a tiny bit below 89.0% 
compared with Lambert Meertens' 88.7%. Any other offers?
					Mike Paterson