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Path: utzoo!utcs!utcsstat!larry
From: larry@utcsstat.UUCP (larry)
Newsgroups: net.math
Subject: Re: prob. formula: Close but no cigar
Message-ID: <2252@utcsstat.UUCP>
Date: Mon, 7-Oct-85 09:08:04 EDT
Article-I.D.: utcsstat.2252
Posted: Mon Oct  7 09:08:04 1985
Date-Received: Mon, 7-Oct-85 10:22:43 EDT
Organization: U. of Toronto, Canada
Lines: 25

>In trying to find the probability of a certain occurence, it is often
>easier to try to find the probability that it does NOT occur. So, in
>trying to find the probability that two card values X and Y occur
>consecutively in a deck, assume X is at position i in the deck.  The
>probability that a Y does NOT occur at position i-1 is p1=47/51 (the
>number of non Y's over the number of cards left).  The probability
>that a Y also does not occur at i+1 is p2=p1*46/50 (since we used up
>a non-Y at i-1).  The probability that this happens for all 8 cards
>is p2**8 which equals about .267.  Thus the probability that NO X
>and Y will be consecutive is about 26.7%. (There is probably a slight
>statistical variation at the beginning and end of the deck.)
>
>Ken Doyle
>
>
No!  Using p2**8 assumes that all these events are INDEPENDENT
which they are not.  The enumeration for all the possible arrangements
of the eight cards still persists as a messy problem.

        Larry Wasserman

-- 

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