Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10 5/3/83; site umcp-cs.UUCP
Path: utzoo!linus!decvax!harpo!seismo!rlgvax!cvl!umcp-cs!israel
From: israel@umcp-cs.UUCP
Newsgroups: net.math,net.misc,net.rec.bridge
Subject: Re: simple (?) statistics problem solved
Message-ID: <1996@umcp-cs.UUCP>
Date: Fri, 19-Aug-83 17:57:26 EDT
Article-I.D.: umcp-cs.1996
Posted: Fri Aug 19 17:57:26 1983
Date-Received: Sat, 20-Aug-83 01:54:14 EDT
References: <1814@rabbit.UUCP> <903@utcsstat.UUCP>
Organization: Univ. of Maryland, Computer Science Dept.
Lines: 41


	To begin with, you have an equal chance of picking a 2-gold, a
	2-silver or a 1-of-each cabinet.  You eliminate one of these
	when you make your choice.  I toss this one out of the window
	and say that you have a 50% chance of either cabinet.  The
	answer says that I should not have tossed the other cabinet out
	of the window.  Why is that silly cabinet relavant to the
	question?

	Laura Creighton		utzoo!utcsstat!laura

You CAN throw the all-silver cabinet out the window.  The problem that
arises is that the probability is not 50% just because you have two
cabinets.  To use an analogy:

I have three BIG cardboard boxes.  I fill the first with one dollar
bills to the brim.  In the second I put a single dollar bill and fill
the rest of the box with shredded "National Enquirer".  The third
box I fill with dimes.  You reach into a box and pull out the first
object you touch.  It is a one dollar bill.  Now it obviously was
not the third box (unless someone who needed some change came by).
It could be the first box (which was filled with one dollar bills).
It could even be the second box since that one had a single one
dollar bill in it.  Since there are two boxes left, would you
say that it was equal chance of it being either box?  Of course
not since it should be pretty obvious that the one-dollar bill
was much more likely to come from the first box and not the
second.

In the same fashion, when you find a gold coin, it is more likely
that it come from the cabinet with two gold coins, since it has
twice as many gold coins to find as the second cabinet does.  Since
every gold coin in that cabinet has a gold coin in the opposite
drawer and no other gold coin outside that cabinet has a gold
coin in an opposite drawer, the solution is equivalent to the
probability of the chosen coin being in the first cabinet.
-- 

~~~ Bruce
Computer Science Dept., University of Maryland
{rlgvax,seismo}!umcp-cs!israel (Usenet)    israel.umcp-cs@Udel-Relay (Arpanet)