<?xml version="1.0" encoding="UTF-8"?><rss version="2.0"
	xmlns:content="http://purl.org/rss/1.0/modules/content/"
	xmlns:dc="http://purl.org/dc/elements/1.1/"
	xmlns:atom="http://www.w3.org/2005/Atom"
	xmlns:sy="http://purl.org/rss/1.0/modules/syndication/"
	xmlns:itunes="http://www.itunes.com/dtds/podcast-1.0.dtd"
xmlns:rawvoice="http://www.rawvoice.com/rawvoiceRssModule/"
	>
<channel>
	<title>Comments on: Probability Paradoxes</title>
	<atom:link href="http://www.stubbornmule.net/2010/06/probability-paradoxes/feed/" rel="self" type="application/rss+xml" />
	<link>http://www.stubbornmule.net/2010/06/probability-paradoxes/</link>
	<description>Obstinately objective</description>
	<lastBuildDate>Wed, 08 Feb 2012 01:46:02 +0000</lastBuildDate>
	<sy:updatePeriod>hourly</sy:updatePeriod>
	<sy:updateFrequency>1</sy:updateFrequency>
	<generator>http://wordpress.org/?v=3.3.1</generator>
	<item>
		<title>By: The Monty Hall Problem</title>
		<link>http://www.stubbornmule.net/2010/06/probability-paradoxes/comment-page-1/#comment-8261</link>
		<dc:creator>The Monty Hall Problem</dc:creator>
		<pubDate>Mon, 21 Jun 2010 05:35:53 +0000</pubDate>
		<guid isPermaLink="false">http://www.stubbornmule.net/?p=2997#comment-8261</guid>
		<description>[...] order. But now that your brains are well rested, it is time to move on to the next puzzle from the Probability Paradoxes post: the Monty Hall problem. If you thought that the first puzzle was controversial, you [...]</description>
		<content:encoded><![CDATA[<p>[...] order. But now that your brains are well rested, it is time to move on to the next puzzle from the Probability Paradoxes post: the Monty Hall problem. If you thought that the first puzzle was controversial, you [...]</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Bruce</title>
		<link>http://www.stubbornmule.net/2010/06/probability-paradoxes/comment-page-1/#comment-8255</link>
		<dc:creator>Bruce</dc:creator>
		<pubDate>Sun, 20 Jun 2010 01:00:53 +0000</pubDate>
		<guid isPermaLink="false">http://www.stubbornmule.net/?p=2997#comment-8255</guid>
		<description>SM: did you also see these posts: Tuesday’s Child and Eliminating the irrelevant?

The answer is obviously dependent on the probabilities associated with the interpretation of statements, and as you rightfully show using symmetry and exclusion (in eliminating the irrelevant), the probabilities vary.

I understand the derivation of your formula, but am not confident of a proof (35 years since I did senior math).</description>
		<content:encoded><![CDATA[<p>SM: did you also see these posts: Tuesday’s Child and Eliminating the irrelevant?</p>
<p>The answer is obviously dependent on the probabilities associated with the interpretation of statements, and as you rightfully show using symmetry and exclusion (in eliminating the irrelevant), the probabilities vary.</p>
<p>I understand the derivation of your formula, but am not confident of a proof (35 years since I did senior math).</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Stubborn Mule</title>
		<link>http://www.stubbornmule.net/2010/06/probability-paradoxes/comment-page-1/#comment-8250</link>
		<dc:creator>Stubborn Mule</dc:creator>
		<pubDate>Sat, 19 Jun 2010 09:23:41 +0000</pubDate>
		<guid isPermaLink="false">http://www.stubbornmule.net/?p=2997#comment-8250</guid>
		<description>In which case, the best solution I can come up with so far has 13 trips, which seems like a lot. Not sure whether I can do better.</description>
		<content:encoded><![CDATA[<p>In which case, the best solution I can come up with so far has 13 trips, which seems like a lot. Not sure whether I can do better.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Plunko</title>
		<link>http://www.stubbornmule.net/2010/06/probability-paradoxes/comment-page-1/#comment-8249</link>
		<dc:creator>Plunko</dc:creator>
		<pubDate>Sat, 19 Jun 2010 07:28:02 +0000</pubDate>
		<guid isPermaLink="false">http://www.stubbornmule.net/?p=2997#comment-8249</guid>
		<description>Dear Stubby,
                            Yes, of course...since when do self-respecting cannibals let a chance go by? 

And while we&#039;re at it, don&#039;t forget that the GFC proved that most banks (especially the Yank ones) are heavily populated with cannibals! Err, apologies and greetings to the memsahib!!</description>
		<content:encoded><![CDATA[<p>Dear Stubby,<br />
                            Yes, of course&#8230;since when do self-respecting cannibals let a chance go by? </p>
<p>And while we&#8217;re at it, don&#8217;t forget that the GFC proved that most banks (especially the Yank ones) are heavily populated with cannibals! Err, apologies and greetings to the memsahib!!</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Stubborn Mule</title>
		<link>http://www.stubbornmule.net/2010/06/probability-paradoxes/comment-page-1/#comment-8246</link>
		<dc:creator>Stubborn Mule</dc:creator>
		<pubDate>Sat, 19 Jun 2010 06:16:23 +0000</pubDate>
		<guid isPermaLink="false">http://www.stubbornmule.net/?p=2997#comment-8246</guid>
		<description>&lt;strong&gt;Plunko:&lt;/strong&gt; does having one cannibal and one missionary on a bank while the canoe is stopped at the same bank with a cannibal in it count as having more cannibals than missionaries in the same place?</description>
		<content:encoded><![CDATA[<p><strong>Plunko:</strong> does having one cannibal and one missionary on a bank while the canoe is stopped at the same bank with a cannibal in it count as having more cannibals than missionaries in the same place?</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Stubborn Mule</title>
		<link>http://www.stubbornmule.net/2010/06/probability-paradoxes/comment-page-1/#comment-8245</link>
		<dc:creator>Stubborn Mule</dc:creator>
		<pubDate>Sat, 19 Jun 2010 05:21:56 +0000</pubDate>
		<guid isPermaLink="false">http://www.stubbornmule.net/?p=2997#comment-8245</guid>
		<description>&lt;strong&gt;Bruce:&lt;/strong&gt; did you also see these posts: &lt;a href=&quot;http://www.stubbornmule.net/2010/06/tuesdays-child-2/&quot; rel=&quot;nofollow&quot;&gt;Tuesday’s Child&lt;/a&gt; and &lt;a href=&quot;http://www.stubbornmule.net/2010/06/eliminating-the-irrelevant/&quot; rel=&quot;nofollow&quot;&gt;Eliminating the irrelevant&lt;/a&gt;?</description>
		<content:encoded><![CDATA[<p><strong>Bruce:</strong> did you also see these posts: <a href="http://www.stubbornmule.net/2010/06/tuesdays-child-2/" rel="nofollow">Tuesday’s Child</a> and <a href="http://www.stubbornmule.net/2010/06/eliminating-the-irrelevant/" rel="nofollow">Eliminating the irrelevant</a>?</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Bruce</title>
		<link>http://www.stubbornmule.net/2010/06/probability-paradoxes/comment-page-1/#comment-8244</link>
		<dc:creator>Bruce</dc:creator>
		<pubDate>Sat, 19 Jun 2010 05:04:46 +0000</pubDate>
		<guid isPermaLink="false">http://www.stubbornmule.net/?p=2997#comment-8244</guid>
		<description>“A man says to you “I have two children,  at least one is a boy born on a Tuesday”. What is the probability the man has two boys?”

Why is the solution not thus?
1. P(1st is b) = .5 * P(2nd is b) = .5 -&gt; .25
2. P(1st is b) = .5 * P(2nd is g) = .5  -&gt; .25
3. P(1st is g) = .5 * P(2nd is b) = .5  -&gt; .25
4. P(1st is g) = .5 * P(2nd is g) = .5  -&gt; .25

Therefore, P(2 boys) = 0.25/0.75= 0.333? 

Because the information presented changes all possible outcomes to:
1. P(1st is b) = 1 * P(2nd is b) = .5 -&gt; .5
2. P(1st is b) = .5 * P(2nd is b) = 1 -&gt; .5
3. P(1st is b) = 1 * P(2nd is g) = .5  -&gt; .5
4. P(1st is g) = .5 * P(2nd is b) = 1  -&gt; .5
And these reduce to 
P(2 boys) = 0.5</description>
		<content:encoded><![CDATA[<p>“A man says to you “I have two children,  at least one is a boy born on a Tuesday”. What is the probability the man has two boys?”</p>
<p>Why is the solution not thus?<br />
1. P(1st is b) = .5 * P(2nd is b) = .5 -&gt; .25<br />
2. P(1st is b) = .5 * P(2nd is g) = .5  -&gt; .25<br />
3. P(1st is g) = .5 * P(2nd is b) = .5  -&gt; .25<br />
4. P(1st is g) = .5 * P(2nd is g) = .5  -&gt; .25</p>
<p>Therefore, P(2 boys) = 0.25/0.75= 0.333? </p>
<p>Because the information presented changes all possible outcomes to:<br />
1. P(1st is b) = 1 * P(2nd is b) = .5 -&gt; .5<br />
2. P(1st is b) = .5 * P(2nd is b) = 1 -&gt; .5<br />
3. P(1st is b) = 1 * P(2nd is g) = .5  -&gt; .5<br />
4. P(1st is g) = .5 * P(2nd is b) = 1  -&gt; .5<br />
And these reduce to<br />
P(2 boys) = 0.5</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Plunko</title>
		<link>http://www.stubbornmule.net/2010/06/probability-paradoxes/comment-page-1/#comment-8239</link>
		<dc:creator>Plunko</dc:creator>
		<pubDate>Fri, 18 Jun 2010 06:41:17 +0000</pubDate>
		<guid isPermaLink="false">http://www.stubbornmule.net/?p=2997#comment-8239</guid>
		<description>G&#039;day SM,
                      Seems to me the punters are easily confused by paradoxes!
1. The man with two boys paradox is about timing. Before he started procreating, the man had a 1/2 chance of a son first up, and a 1/2 chance of a son second time around - total chance of 2 boys is 1/2 x 1/2 = 1/4. But now we know he has produced 2 kids, one of whom was a boy, the only question left is what is the probability of the second being a boy. Since none of the other facts affects this outcome it&#039;s 1/2 (and the Tuesday bit is irrelevant). 
2. The quiz show host job is also only about timing. Before the contestant picks a door he/she has 1/3 chance of the car (note that some might prefer a goat). 
After the host reveals the goat behind one door, the chances of picking a car from the other 2 doors is 1/2, but there is no information for the contestant which makes one door more likely to hold the car than the other, unless it be that the host is subtly trying to steer the contestant away from the door already chosen, which probably means that&#039;s where the car is ... Never trust a game show host!
3. Doesn&#039;t matter what was in the first envelope - all it does is tell you how much you stand to gain or lose. The chances are still 1/2.

Now as your devoted followers seem to love a paradox, you might want to hit them with a conundrum:
Three cannibals and three missionaries are travelling together. They come to a river which they all must cross. They only have a 2-man canoe, which is their only means of crossing the river. Only one of the cannibals can row, but all of the missionaries can row. At no time or place can the missionaries be outnumbered by cannibals, or they will be eaten. How do they cross the river, and how many trips does it take?

Maybe a prize of a chaff bag of mule droppings to the solution with the least number of trips?? Second prize, 2 bags, etc...</description>
		<content:encoded><![CDATA[<p>G&#8217;day SM,<br />
                      Seems to me the punters are easily confused by paradoxes!<br />
1. The man with two boys paradox is about timing. Before he started procreating, the man had a 1/2 chance of a son first up, and a 1/2 chance of a son second time around &#8211; total chance of 2 boys is 1/2 x 1/2 = 1/4. But now we know he has produced 2 kids, one of whom was a boy, the only question left is what is the probability of the second being a boy. Since none of the other facts affects this outcome it&#8217;s 1/2 (and the Tuesday bit is irrelevant).<br />
2. The quiz show host job is also only about timing. Before the contestant picks a door he/she has 1/3 chance of the car (note that some might prefer a goat).<br />
After the host reveals the goat behind one door, the chances of picking a car from the other 2 doors is 1/2, but there is no information for the contestant which makes one door more likely to hold the car than the other, unless it be that the host is subtly trying to steer the contestant away from the door already chosen, which probably means that&#8217;s where the car is &#8230; Never trust a game show host!<br />
3. Doesn&#8217;t matter what was in the first envelope &#8211; all it does is tell you how much you stand to gain or lose. The chances are still 1/2.</p>
<p>Now as your devoted followers seem to love a paradox, you might want to hit them with a conundrum:<br />
Three cannibals and three missionaries are travelling together. They come to a river which they all must cross. They only have a 2-man canoe, which is their only means of crossing the river. Only one of the cannibals can row, but all of the missionaries can row. At no time or place can the missionaries be outnumbered by cannibals, or they will be eaten. How do they cross the river, and how many trips does it take?</p>
<p>Maybe a prize of a chaff bag of mule droppings to the solution with the least number of trips?? Second prize, 2 bags, etc&#8230;</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Stubborn Mule</title>
		<link>http://www.stubbornmule.net/2010/06/probability-paradoxes/comment-page-1/#comment-8232</link>
		<dc:creator>Stubborn Mule</dc:creator>
		<pubDate>Fri, 18 Jun 2010 01:34:58 +0000</pubDate>
		<guid isPermaLink="false">http://www.stubbornmule.net/?p=2997#comment-8232</guid>
		<description>&lt;strong&gt;Bruce:&lt;/strong&gt; so, what would you conclude if the puzzle was phrased as:

&quot;A man says to you “I have two children, &lt;em&gt; at least&lt;/em&gt; one is a boy born on a Tuesday”. What is the probability the man has two boys?&quot;</description>
		<content:encoded><![CDATA[<p><strong>Bruce:</strong> so, what would you conclude if the puzzle was phrased as:</p>
<p>&#8220;A man says to you “I have two children, <em> at least</em> one is a boy born on a Tuesday”. What is the probability the man has two boys?&#8221;</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Bruce</title>
		<link>http://www.stubbornmule.net/2010/06/probability-paradoxes/comment-page-1/#comment-8231</link>
		<dc:creator>Bruce</dc:creator>
		<pubDate>Fri, 18 Jun 2010 01:15:24 +0000</pubDate>
		<guid isPermaLink="false">http://www.stubbornmule.net/?p=2997#comment-8231</guid>
		<description>Absolutely. 
&quot;At least one landed heads&quot; is unambiguous. 
&quot;One landed heads&quot; is ambiguous.</description>
		<content:encoded><![CDATA[<p>Absolutely.<br />
&#8220;At least one landed heads&#8221; is unambiguous.<br />
&#8220;One landed heads&#8221; is ambiguous.</p>
]]></content:encoded>
	</item>
</channel>
</rss>

