http://test.causeweb.org/wiki/chance/index.php?title=Chance_News_80&feed=atom&action=historyChance News 80 - Revision history2014-04-24T01:22:04ZRevision history for this page on the wikiMediaWiki 1.18.1http://test.causeweb.org/wiki/chance/index.php?title=Chance_News_80&diff=15907&oldid=prevBill Peterson: /* Marilyn slips up on a drug testing question */2012-05-09T19:09:17Z<p><span class="autocomment">Marilyn slips up on a drug testing question</span></p>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>:by Marilyn vos Savant, ''Parade'', 2 January 2012</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>:by Marilyn vos Savant, ''Parade'', 2 January 2012</div></td></tr>
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<tr><td class='diff-marker'>−</td><td style="background: #ffa; color:black; font-size: smaller;"><div>Here Marilyn suggests that the original problem may have been ambiguous.  Nevertheless, she does print corrections from Jerry and another reader. She also swears off eggnog, and promises a followup on January 22.  Check back here for the update<del class="diffchange diffchange-inline">!</del></div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>Here Marilyn suggests that the original problem may have been ambiguous.  Nevertheless, she does print corrections from Jerry and another reader. She also swears off eggnog, and promises a followup on January 22.  Check back here for the <ins class="diffchange diffchange-inline">[http://test.causeweb.org/wiki/chance/index.php/Chance_News_81#Marilyn.27s_correction_on_the_drug-testing_problem </ins>update<ins class="diffchange diffchange-inline">].</ins></div></td></tr>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>==Impact of large philanthropists on research==</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>==Impact of large philanthropists on research==</div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_80&diff=15068&oldid=prevBill Peterson: /* Surprising dreidel outcome */2012-02-08T20:30:49Z<p><span class="autocomment">Surprising dreidel outcome</span></p>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>To the amazement everyone present, Alfred Lorini compiled a streak of 68 spins that included 56 gimels and zero shins.  According to the article, his great-nephew "used a binomial distribution and came up with 1-in-2.25 times 10 to the 22nd power for the order of magnitude."</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>To the amazement everyone present, Alfred Lorini compiled a streak of 68 spins that included 56 gimels and zero shins.  According to the article, his great-nephew "used a binomial distribution and came up with 1-in-2.25 times 10 to the 22nd power for the order of magnitude."</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="background: #ffa; color:black; font-size: smaller;"><div>In fact, <del class="diffchange diffchange-inline">the figure reported </del>is the binomial probability for obtaining exactly 56 successes (gimels) in 68 trials (spins) with success probability 1/4;  this assumes independent spins of a perfectly balanced dreidel.  But there are two problems here.  First, we should ask for the chance of 56 ''or more'' successes.  This adjustment does change the order of magnitude of the probability, which becomes 1 in <math>2.09 \times 10^{22}</math>.  However,  the binomial description is not correct because it allows the non-gimel rolls to be nun, hei or shin.  Mr. Lorini's feat was more unusual in that the non-gimels did not include any shins.  Thus we really need to consider a multinomial situation, with categories (gimel, shin, neither), for which the probabilities are (1/4, 1/4, 1/2).  The chance of 56 or more gimels in 68 rolls, with zero shins, is then calculated as 1 in <math>2.62 \times 10^{24}</math>.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>In fact, <ins class="diffchange diffchange-inline">this </ins>is the binomial probability for obtaining exactly 56 successes (gimels) in 68 trials (spins) with success probability 1/4;  this assumes independent spins of a perfectly balanced dreidel.  But there are two problems here.  First, we should ask for the chance of 56 ''or more'' successes.  This adjustment does change the order of magnitude of the probability, which becomes 1 in <math>2.09 \times 10^{22}</math>.  However,  the binomial description is not correct because it allows the non-gimel rolls to be nun, hei or shin.  Mr. Lorini's feat was more unusual in that the non-gimels did not include any shins.  Thus we really need to consider a multinomial situation, with categories (gimel, shin, neither), for which the probabilities are (1/4, 1/4, 1/2).  The chance of 56 or more gimels in 68 rolls, with zero shins, is then calculated as 1 in <math>2.62 \times 10^{24}</math>.</div></td></tr>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>'''Discussion'''<br></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>'''Discussion'''<br></div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_80&diff=15067&oldid=prevBill Peterson: /* Surprising dreidel outcome */2012-02-08T20:29:21Z<p><span class="autocomment">Surprising dreidel outcome</span></p>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>To the amazement everyone present, Alfred Lorini compiled a streak of 68 spins that included 56 gimels and zero shins.  According to the article, his great-nephew "used a binomial distribution and came up with 1-in-2.25 times 10 to the 22nd power for the order of magnitude."</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>To the amazement everyone present, Alfred Lorini compiled a streak of 68 spins that included 56 gimels and zero shins.  According to the article, his great-nephew "used a binomial distribution and came up with 1-in-2.25 times 10 to the 22nd power for the order of magnitude."</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="background: #ffa; color:black; font-size: smaller;"><div>In fact, the figure <del class="diffchange diffchange-inline">given </del>is the binomial probability for obtaining exactly 56 successes (gimels) in 68 trials (spins) with success probability 1/4;  this assumes independent spins of a perfectly balanced dreidel.  But there are two problems here.  First, we should ask for the chance of 56 ''or more'' successes.  This adjustment does change the order of magnitude of the probability, which becomes 1 in <math>2.09 \times 10^{22}</math>.  However,  the binomial description is not correct because it allows the non-gimel rolls to be nun, hei or shin.  Mr. Lorini's feat was more unusual in that the non-gimels did not include any shins.  Thus we really need to consider a multinomial situation, with categories (gimel, shin, neither), for which the probabilities are (1/4, 1/4, 1/2).  The chance of 56 or more gimels in 68 rolls, with zero shins, is then calculated as 1 in <math>2.62 \times 10^{24}</math>.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>In fact, the figure <ins class="diffchange diffchange-inline">reported </ins>is the binomial probability for obtaining exactly 56 successes (gimels) in 68 trials (spins) with success probability 1/4;  this assumes independent spins of a perfectly balanced dreidel.  But there are two problems here.  First, we should ask for the chance of 56 ''or more'' successes.  This adjustment does change the order of magnitude of the probability, which becomes 1 in <math>2.09 \times 10^{22}</math>.  However,  the binomial description is not correct because it allows the non-gimel rolls to be nun, hei or shin.  Mr. Lorini's feat was more unusual in that the non-gimels did not include any shins.  Thus we really need to consider a multinomial situation, with categories (gimel, shin, neither), for which the probabilities are (1/4, 1/4, 1/2).  The chance of 56 or more gimels in 68 rolls, with zero shins, is then calculated as 1 in <math>2.62 \times 10^{24}</math>.</div></td></tr>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>'''Discussion'''<br></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>'''Discussion'''<br></div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_80&diff=15066&oldid=prevBill Peterson: /* Winning the fight against crime by putting your head in the sand */2012-02-08T20:28:09Z<p><span class="autocomment">Winning the fight against crime by putting your head in the sand</span></p>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>==Winning the fight against crime by putting your head in the sand==</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>==Winning the fight against crime by putting your head in the sand==</div></td></tr>
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<tr><td class='diff-marker'>−</td><td style="background: #ffa; color:black; font-size: smaller;"><div>[http://www.nytimes.com/2011/12/31/nyregion/nypd-leaves-offenses-unrecorded-to-keep-crime-rates-down.html Police Tactic: Keeping Crime Reports Off the Books] Al Baker and Joseph Goldstein, <del class="diffchange diffchange-inline">The </del>New York Times, December 30, 2011.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>[http://www.nytimes.com/2011/12/31/nyregion/nypd-leaves-offenses-unrecorded-to-keep-crime-rates-down.html Police Tactic: Keeping Crime Reports Off the Books] Al Baker and Joseph Goldstein, <ins class="diffchange diffchange-inline">''</ins>New York Times<ins class="diffchange diffchange-inline">''</ins>, December 30, 2011.</div></td></tr>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Police officers are joining just about every other profession in trying to skew the statistics to make themselves look good.</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Police officers are joining just about every other profession in trying to skew the statistics to make themselves look good.</div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_80&diff=15065&oldid=prevBill Peterson: /* Surprising dreidel outcome */2012-02-08T20:25:54Z<p><span class="autocomment">Surprising dreidel outcome</span></p>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div></blockquote></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div></blockquote></div></td></tr>
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<tr><td class='diff-marker'>−</td><td style="background: #ffa; color:black; font-size: smaller;"><div>To the amazement everyone present, Alfred Lorini compiled a streak of 68 spins that included 56 gimels and zero shins.  <del class="diffchange diffchange-inline">His </del>great-nephew "used a binomial distribution and came up with 1-in-2.25 times 10 to the 22nd power for the order of magnitude."</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>To the amazement everyone present, Alfred Lorini compiled a streak of 68 spins that included 56 gimels and zero shins.  <ins class="diffchange diffchange-inline">According to the article, his </ins>great-nephew "used a binomial distribution and came up with 1-in-2.25 times 10 to the 22nd power for the order of magnitude."</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="background: #ffa; color:black; font-size: smaller;"><div>In fact, <del class="diffchange diffchange-inline">this </del>is the binomial probability for obtaining exactly 56 successes (gimels) in 68 trials (spins) with success probability 1/4;  this assumes independent spins of a perfectly balanced dreidel.  But there are two problems here.  First, we should ask for the chance of 56 or more successes.  This adjustment does change the order of magnitude of the probability, which becomes 1 in <math>2.09 \times 10^{22}</math>.  However,  the binomial description is not correct because it allows the non-gimel rolls to be nun, hei or shin.  Mr. Lorini's feat was more unusual in that the non-gimels did not include any shins.  Thus we really need to consider a multinomial situation, with categories (gimel, shin, neither), for which the probabilities are (1/4, 1/4, 1/2).  The chance of 56 or more gimels in 68 rolls, with zero shins, is then calculated as 1 in <math>2.62 \times 10^{24}</math>.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>In fact, <ins class="diffchange diffchange-inline">the figure given </ins>is the binomial probability for obtaining exactly 56 successes (gimels) in 68 trials (spins) with success probability 1/4;  this assumes independent spins of a perfectly balanced dreidel.  But there are two problems here.  First, we should ask for the chance of 56 <ins class="diffchange diffchange-inline">''</ins>or more<ins class="diffchange diffchange-inline">'' </ins>successes.  This adjustment does change the order of magnitude of the probability, which becomes 1 in <math>2.09 \times 10^{22}</math>.  However,  the binomial description is not correct because it allows the non-gimel rolls to be nun, hei or shin.  Mr. Lorini's feat was more unusual in that the non-gimels did not include any shins.  Thus we really need to consider a multinomial situation, with categories (gimel, shin, neither), for which the probabilities are (1/4, 1/4, 1/2).  The chance of 56 or more gimels in 68 rolls, with zero shins, is then calculated as 1 in <math>2.62 \times 10^{24}</math>.</div></td></tr>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>'''Discussion'''<br></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>'''Discussion'''<br></div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_80&diff=14904&oldid=prevBill Peterson: /* Surprising dreidel outcome */2012-01-18T16:01:58Z<p><span class="autocomment">Surprising dreidel outcome</span></p>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>To the amazement everyone present, Alfred Lorini compiled a streak of 68 spins that included 56 gimels and zero shins.  His great-nephew "used a binomial distribution and came up with 1-in-2.25 times 10 to the 22nd power for the order of magnitude."</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>To the amazement everyone present, Alfred Lorini compiled a streak of 68 spins that included 56 gimels and zero shins.  His great-nephew "used a binomial distribution and came up with 1-in-2.25 times 10 to the 22nd power for the order of magnitude."</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="background: #ffa; color:black; font-size: smaller;"><div>In fact, this is the binomial probability for obtaining exactly 56 successes in 68 trials with success probability 1/4 <del class="diffchange diffchange-inline">(assuming </del>independent spins of a perfectly balanced dreidel<del class="diffchange diffchange-inline">)</del>.  <del class="diffchange diffchange-inline">There </del>are two problems here.  First, we should ask for the chance of 56 or more successes.  This adjustment does change the order of magnitude of the probability, which becomes 1 in <math>2.09 \times 10^{22}</math>.  However,  the binomial description is not correct because it allows the non-gimel rolls to be nun, hei or shin.  Mr. Lorini's feat was more unusual in that the non-gimels did not include any shins.  Thus we really need to consider a multinomial situation, with categories (gimel, shin, neither), for which the probabilities are (1/4, 1/4, 1/2).  The chance of 56 or more gimels in 68 rolls, with zero shins, is then calculated as 1 in <math>2.62 \times 10^{24}</math>.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>In fact, this is the binomial probability for obtaining exactly 56 successes <ins class="diffchange diffchange-inline">(gimels) </ins>in 68 trials <ins class="diffchange diffchange-inline">(spins) </ins>with success probability 1/4<ins class="diffchange diffchange-inline">;  this assumes </ins>independent spins of a perfectly balanced dreidel.  <ins class="diffchange diffchange-inline">But there </ins>are two problems here.  First, we should ask for the chance of 56 or more successes.  This adjustment does change the order of magnitude of the probability, which becomes 1 in <math>2.09 \times 10^{22}</math>.  However,  the binomial description is not correct because it allows the non-gimel rolls to be nun, hei or shin.  Mr. Lorini's feat was more unusual in that the non-gimels did not include any shins.  Thus we really need to consider a multinomial situation, with categories (gimel, shin, neither), for which the probabilities are (1/4, 1/4, 1/2).  The chance of 56 or more gimels in 68 rolls, with zero shins, is then calculated as 1 in <math>2.62 \times 10^{24}</math>.</div></td></tr>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>'''Discussion'''<br></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>'''Discussion'''<br></div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_80&diff=14903&oldid=prevBill Peterson: /* Surprising dreidel outcome */2012-01-18T15:59:34Z<p><span class="autocomment">Surprising dreidel outcome</span></p>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>To the amazement everyone present, Alfred Lorini compiled a streak of 68 spins that included 56 gimels and zero shins.  His great-nephew "used a binomial distribution and came up with 1-in-2.25 times 10 to the 22nd power for the order of magnitude."</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>To the amazement everyone present, Alfred Lorini compiled a streak of 68 spins that included 56 gimels and zero shins.  His great-nephew "used a binomial distribution and came up with 1-in-2.25 times 10 to the 22nd power for the order of magnitude."</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="background: #ffa; color:black; font-size: smaller;"><div>In fact, this is the binomial probability for obtaining exactly 56 successes in 68 trials with success probability 1/4 (assuming independent spins of a perfectly balanced dreidel).  There are two problems here.  First, we should ask for the chance of 56 or more successes.  This adjustment does change the order of magnitude of the probability, which becomes 1 in <math>2.09 \times 10^{22}</math>.  However,  the binomial description is not correct because it allows the non-gimel rolls to be nun, hei or shin.  Mr. Lorini's feat was more unusual in that the non-gimels did not include shins.  Thus we really need to consider a multinomial situation, with categories (gimel, shin, neither), for which the probabilities are (1/4, 1/4, 1/2).  The chance of 56 or more gimels in 68 rolls, with zero shins, is then calculated as 1 in <math>2.62 \times 10^{24}</math>.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>In fact, this is the binomial probability for obtaining exactly 56 successes in 68 trials with success probability 1/4 (assuming independent spins of a perfectly balanced dreidel).  There are two problems here.  First, we should ask for the chance of 56 or more successes.  This adjustment does change the order of magnitude of the probability, which becomes 1 in <math>2.09 \times 10^{22}</math>.  However,  the binomial description is not correct because it allows the non-gimel rolls to be nun, hei or shin.  Mr. Lorini's feat was more unusual in that the non-gimels did not include <ins class="diffchange diffchange-inline">any </ins>shins.  Thus we really need to consider a multinomial situation, with categories (gimel, shin, neither), for which the probabilities are (1/4, 1/4, 1/2).  The chance of 56 or more gimels in 68 rolls, with zero shins, is then calculated as 1 in <math>2.62 \times 10^{24}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>'''Discussion'''<br></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>'''Discussion'''<br></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="background: #ffa; color:black; font-size: smaller;"><div>In an effort to describe the order of magnitude of the answer, the article reports that the figure was 22.5 billion times 1 trillion.  Do you think this helps the lay person to understand it?  Can you suggest an alternative?</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>In an effort to describe the order of magnitude of the <ins class="diffchange diffchange-inline">(originally reported) </ins>answer, the article reports that the figure was 22.5 billion times 1 trillion.  Do you think this helps the lay person to understand it?  Can you suggest an alternative?</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Submitted by Bill Peterson, based on a suggestion from Adam Peterson</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Submitted by Bill Peterson, based on a suggestion from Adam Peterson</div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_80&diff=14902&oldid=prevBill Peterson at 01:35, 18 January 20122012-01-18T01:35:18Z<p></p>
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<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div><ins style="color: red; font-weight: bold; text-decoration: none;">December 23, 2011 to January 17, 2012</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div><ins style="color: red; font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>==Quotations==</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>==Quotations==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>“The role of context. ....  The focus on variability naturally gives statistics a particular content that sets it apart from from mathematics itself and from other mathematical sciences, but there is more than just content that distinguishes statistical thinking from mathematics.  Statistics requires a different kind of thinking, because <i>data are not just numbers, they are numbers with a context</i>.  ....  <i>In mathematics, context obscures structure.  ....  In data analysis, context provides meaning.</i>  ....  [A]lthough statistics cannot prosper without mathematics, the converse fails.”<br></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>“The role of context. ....  The focus on variability naturally gives statistics a particular content that sets it apart from from mathematics itself and from other mathematical sciences, but there is more than just content that distinguishes statistical thinking from mathematics.  Statistics requires a different kind of thinking, because <i>data are not just numbers, they are numbers with a context</i>.  ....  <i>In mathematics, context obscures structure.  ....  In data analysis, context provides meaning.</i>  ....  [A]lthough statistics cannot prosper without mathematics, the converse fails.”<br></div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_80&diff=14896&oldid=prevBill Peterson: /* Surprising dreidel outcome */2012-01-18T01:27:40Z<p><span class="autocomment">Surprising dreidel outcome</span></p>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>To the amazement everyone present, Alfred Lorini compiled a streak of 68 spins that included 56 gimels and zero shins.  His great-nephew "used a binomial distribution and came up with 1-in-2.25 times 10 to the 22nd power for the order of magnitude."</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>To the amazement everyone present, Alfred Lorini compiled a streak of 68 spins that included 56 gimels and zero shins.  His great-nephew "used a binomial distribution and came up with 1-in-2.25 times 10 to the 22nd power for the order of magnitude."</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="background: #ffa; color:black; font-size: smaller;"><div>In fact, this is the binomial probability for obtaining exactly 56 successes in 68 trials with success probability 1/4 (assuming independent spins of a perfectly balanced dreidel).  There are two problems here.  First, we should ask for the chance of 56 or more successes.  This <del class="diffchange diffchange-inline">correction </del>does change the order of magnitude of the probability, which becomes 1 in <math>2.09 \times 10^{22}</math>.  However,<del class="diffchange diffchange-inline">observe that </del>the binomial description allows the non-gimel rolls to be nun, hei or shin.  Mr. Lorini's feat was more unusual in that the non-gimels did not include shins.  Thus we really need to consider a multinomial situation, with categories (gimel, shin, neither), for which the probabilities are (1/4, 1/4, 1/2).  The chance of 56 or more gimels in 68 rolls, with zero shins, is then calculated as 1 in <math>2.62 \times 10^{24}</math>.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>In fact, this is the binomial probability for obtaining exactly 56 successes in 68 trials with success probability 1/4 (assuming independent spins of a perfectly balanced dreidel).  There are two problems here.  First, we should ask for the chance of 56 or more successes.  This <ins class="diffchange diffchange-inline">adjustment </ins>does change the order of magnitude of the probability, which becomes 1 in <math>2.09 \times 10^{22}</math>.  However, <ins class="diffchange diffchange-inline"> </ins>the binomial description <ins class="diffchange diffchange-inline">is not correct because it </ins>allows the non-gimel rolls to be nun, hei or shin.  Mr. Lorini's feat was more unusual in that the non-gimels did not include shins.  Thus we really need to consider a multinomial situation, with categories (gimel, shin, neither), for which the probabilities are (1/4, 1/4, 1/2).  The chance of 56 or more gimels in 68 rolls, with zero shins, is then calculated as 1 in <math>2.62 \times 10^{24}</math>.</div></td></tr>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>'''Discussion'''<br></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>'''Discussion'''<br></div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_80&diff=14895&oldid=prevBill Peterson: /* Surprising dreidel outcome */2012-01-18T01:26:52Z<p><span class="autocomment">Surprising dreidel outcome</span></p>
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<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>To the amazement everyone present, Alfred Lorini compiled a streak of 68 spins that included 56 gimels and zero shins.  His great-nephew "used a binomial distribution and came up with 1-in-2.25 times 10 to the 22nd power for the order of magnitude."</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>To the amazement everyone present, Alfred Lorini compiled a streak of 68 spins that included 56 gimels and zero shins.  His great-nephew "used a binomial distribution and came up with 1-in-2.25 times 10 to the 22nd power for the order of magnitude."</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="background: #ffa; color:black; font-size: smaller;"><div>In fact, this is the binomial probability for obtaining exactly 56 successes in 68 trials with success probability 1/4 (assuming independent spins of a perfectly balanced dreidel).  There are two problems here.  First, we should ask for the chance of 56 or more successes.  This correction does change the order of magnitude of the <del class="diffchange diffchange-inline">answer</del>.  However,observe that the binomial description allows the non-gimel rolls to be nun, hei or shin.  Mr. Lorini's feat was more unusual in that the non-gimels did not include shins.  Thus we really need to consider a multinomial situation, with categories (gimel, shin, neither), for which the probabilities are (1/4, 1/4, 1/2).  The chance of 56 or more gimels in 68 rolls, with zero shins, is then calculated as 1 in <math>2.62 \times 10^{24}</math>.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>In fact, this is the binomial probability for obtaining exactly 56 successes in 68 trials with success probability 1/4 (assuming independent spins of a perfectly balanced dreidel).  There are two problems here.  First, we should ask for the chance of 56 or more successes.  This correction does change the order of magnitude of the <ins class="diffchange diffchange-inline">probability, which becomes 1 in <math>2.09 \times 10^{22}</math></ins>.  However,observe that the binomial description allows the non-gimel rolls to be nun, hei or shin.  Mr. Lorini's feat was more unusual in that the non-gimels did not include shins.  Thus we really need to consider a multinomial situation, with categories (gimel, shin, neither), for which the probabilities are (1/4, 1/4, 1/2).  The chance of 56 or more gimels in 68 rolls, with zero shins, is then calculated as 1 in <math>2.62 \times 10^{24}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>'''Discussion'''<br></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>'''Discussion'''<br></div></td></tr>
</table>Bill Peterson