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Path: utzoo!watmath!clyde!cbosgd!ihnp4!qantel!dual!lll-crg!seismo!cmcl2!csd2!rosenblg
From: rosenblg@csd2.UUCP (Gary J. Rosenblum)
Newsgroups: net.math
Subject: probability formula needed
Message-ID: <3210001@csd2.UUCP>
Date: Thu, 5-Sep-85 17:15:00 EDT
Article-I.D.: csd2.3210001
Posted: Thu Sep  5 17:15:00 1985
Date-Received: Wed, 11-Sep-85 04:47:48 EDT
Organization: New York University
Lines: 29


i've come across a "magic trick" that i need help finding out why it 
works.  actually, i need the probability formula.  here's the 
description:

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?

please reply either to me or the net.  thanks!!

					gary rosenblum

(ihnp4!cmcl2!rosenblg  or  rosenblg@csd2)