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	<title>Comments on: The Well&#8217;s Running Dry</title>
	<link>http://blog.gaudyspuds.com/2008/03/01/the-wells-running-dry/</link>
	<description>The Blog Of A New Homosapian</description>
	<pubDate>Tue, 06 Jan 2009 12:11:59 +0000</pubDate>
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		<title>By: Auntie Ipswitch</title>
		<link>http://blog.gaudyspuds.com/2008/03/01/the-wells-running-dry/#comment-20</link>
		<author>Auntie Ipswitch</author>
		<pubDate>Sun, 02 Mar 2008 02:41:47 +0000</pubDate>
		<guid>http://blog.gaudyspuds.com/2008/03/01/the-wells-running-dry/#comment-20</guid>
		<description>My dear Gaudyspud,

Don't be disappointed if you don't see more than two people reading at the same time.  How do I know -- I did a back-of-the-envelope calculation.  Suppose that everyone is awake for 12 hours (I'm thinking of your dad here) and that the average reading time is 5 minutes.  So there are 144 5-minute periods per day for a reader.  The odds of two people choosing the same time are about 1/144!!!  In fact, if you have six readers, the odds of two people reading at the same time are 

&lt;i&gt;1 - (144/144)*(143/144)*(142/144)*(141/144)*(140/144)*(139/144) = 0.10&lt;/i&gt;

For a similar situation see &lt;a href="http://www.maa.org/mathland/mathtrek_11_23_98.html" rel="nofollow"&gt;this article&lt;/a&gt; about the odds of any two people having the same birthday.

Auntie Ipswitch

PS I'd like to think I'm smart but then I remember I married your Uncle Chappaquiddick</description>
		<content:encoded><![CDATA[<p>My dear Gaudyspud,</p>
<p>Don&#8217;t be disappointed if you don&#8217;t see more than two people reading at the same time.  How do I know &#8212; I did a back-of-the-envelope calculation.  Suppose that everyone is awake for 12 hours (I&#8217;m thinking of your dad here) and that the average reading time is 5 minutes.  So there are 144 5-minute periods per day for a reader.  The odds of two people choosing the same time are about 1/144!!!  In fact, if you have six readers, the odds of two people reading at the same time are </p>
<p><i>1 - (144/144)*(143/144)*(142/144)*(141/144)*(140/144)*(139/144) = 0.10</i></p>
<p>For a similar situation see <a href="http://www.maa.org/mathland/mathtrek_11_23_98.html" rel="nofollow">this article</a> about the odds of any two people having the same birthday.</p>
<p>Auntie Ipswitch</p>
<p>PS I&#8217;d like to think I&#8217;m smart but then I remember I married your Uncle Chappaquiddick</p>
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