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From: lambert@boring.UUCP
Newsgroups: net.math
Subject: Re: probability formula needed
Message-ID: <6621@boring.UUCP>
Date: Mon, 9-Sep-85 23:16:42 EDT
Article-I.D.: boring.6621
Posted: Mon Sep  9 23:16:42 1985
Date-Received: Wed, 11-Sep-85 20:14:35 EDT
References: <3210001@csd2.UUCP>
Reply-To: lambert@mcvax.UUCP (Lambert Meertens)
Organization: CWI, Amsterdam
Lines: 35
Summary: 88.7%
Apparently-To: rnews@mcvax.LOCAL

> pick any three cards from a deck of (52) playing cards, ignoring the
> suit.  if you get a duplicate, pick another one.  find all the 
> combinations of pairs you can have.  for example, if you pick the 
> king, seven and three, you can have 
> 
> k 7, 7 k, 3 7, 7 3, k 3, 3 k  (clearly 6 choices).
> 
> now, return the cards to the deck, shuffle to your heart's content, and
> start dealing out the cards in any order.  what will happen in most of the 
> cases is that 2 consecutive cards will be one of your choices.  in other 
> words, if you have dealt out
> 
> q a 2 3 1 k 4 7 3, 
> 
> you would "win" because the last 2 cards were 7 and 3, one of your choices.  
> things like "k k" don't count, but "k k 3" is fine.
> 
> does anyone know how to figure out the probability of this problem?


                                             2404226057
My computations resulted in a probability of ---------- for "winning"
                                             2710423730
(assuming that the shuffling leaves a completely random permutation of the
deck), which is a tiny bit above 88.7%
Can anyone corroborate this?

     Lambert Meertens
     ...!{seismo,okstate,garfield,decvax,philabs}!lambert@mcvax.UUCP
     CWI (Centre for Mathematics and Computer Science), Amsterdam
-- 

     Lambert Meertens
     ...!{seismo,okstate,garfield,decvax,philabs}!lambert@mcvax.UUCP
     CWI (Centre for Mathematics and Computer Science), Amsterdam