http://test.causeweb.org/wiki/chance/index.php?title=Chance_News_49&feed=atom&action=historyChance News 49 - Revision history2014-04-25T09:01:00ZRevision history for this page on the wikiMediaWiki 1.18.1http://test.causeweb.org/wiki/chance/index.php?title=Chance_News_49&diff=15212&oldid=prevBill Peterson: /* Meteorite hits boy */2012-02-20T17:25:01Z<p><span class="autocomment">Meteorite hits boy</span></p>
<table class='diff diff-contentalign-left'>
<col class='diff-marker' />
<col class='diff-content' />
<col class='diff-marker' />
<col class='diff-content' />
<tr valign='top'>
<td colspan='2' style="background-color: white; color:black;">← Older revision</td>
<td colspan='2' style="background-color: white; color:black;">Revision as of 17:25, 20 February 2012</td>
</tr><tr><td colspan="2" class="diff-lineno">Line 240:</td>
<td colspan="2" class="diff-lineno">Line 240:</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>==Meteorite hits boy==</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>==Meteorite hits boy==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="background: #ffa; color:black; font-size: smaller;"><div>[http://www.telegraph.co.uk/scienceandtechnology/science/space/5511619/14-year-old-hit-by-30000-mph-space-meteorite.html "14-year-old hit by 30,000 mph space meteorite"], The Telegraph, June 12, 2009<br></div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>[http://www.telegraph.co.uk/scienceandtechnology/science/space/5511619/14-year-old-hit-by-30000-mph-space-meteorite.html "14-year-old hit by 30,000 mph space meteorite"], <ins class="diffchange diffchange-inline">''</ins>The Telegraph<ins class="diffchange diffchange-inline">''</ins>, June 12, 2009<br></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>Gerrit Blank survived a direct hit to his hand by a meteorite as it hurtled to Earth at "more than 30,000 miles per hour".<br></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Gerrit Blank survived a direct hit to his hand by a meteorite as it hurtled to Earth at "more than 30,000 miles per hour".<br></div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_49&diff=15211&oldid=prevBill Peterson: /* Meteorite hits boy */2012-02-20T17:24:29Z<p><span class="autocomment">Meteorite hits boy</span></p>
<table class='diff diff-contentalign-left'>
<col class='diff-marker' />
<col class='diff-content' />
<col class='diff-marker' />
<col class='diff-content' />
<tr valign='top'>
<td colspan='2' style="background-color: white; color:black;">← Older revision</td>
<td colspan='2' style="background-color: white; color:black;">Revision as of 17:24, 20 February 2012</td>
</tr><tr><td colspan="2" class="diff-lineno">Line 250:</td>
<td colspan="2" class="diff-lineno">Line 250:</td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>From [http://www.wired.com/images_blogs/wiredscience/2009/06/meteorite-nearmisses.jpg Wired magazine], some meteorite "near misses" in history:</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>From [http://www.wired.com/images_blogs/wiredscience/2009/06/meteorite-nearmisses.jpg Wired magazine], some meteorite "near misses" in history:</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><del class="diffchange diffchange-inline"><center> </del>http://www.wired.com/images_blogs/wiredscience/2009/06/meteorite-nearmisses.jpg <del class="diffchange diffchange-inline"></center></del></div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>http://www.wired.com/images_blogs/wiredscience/2009/06/meteorite-nearmisses.jpg  </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>Discussion<br></div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div><ins class="diffchange diffchange-inline">'''</ins>Discussion<ins class="diffchange diffchange-inline">'''</ins><br></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>1.  How do you think the speed of 30,000 miles per hour was determined?<br>  </div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>1.  How do you think the speed of 30,000 miles per hour was determined?<br>  </div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_49&diff=13289&oldid=prevBill Peterson: /* A new record in craps. */2010-09-30T01:43:17Z<p><span class="autocomment">A new record in craps.</span></p>
<table class='diff diff-contentalign-left'>
<col class='diff-marker' />
<col class='diff-content' />
<col class='diff-marker' />
<col class='diff-content' />
<tr valign='top'>
<td colspan='2' style="background-color: white; color:black;">← Older revision</td>
<td colspan='2' style="background-color: white; color:black;">Revision as of 01:43, 30 September 2010</td>
</tr><tr><td colspan="2" class="diff-lineno">Line 260:</td>
<td colspan="2" class="diff-lineno">Line 260:</td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Submitted by Gregory Kohs<br></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Submitted by Gregory Kohs<br></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>==A new record in craps<del class="diffchange diffchange-inline">.</del>==</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>==A new record in craps==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>[http://www.time.com/time/nation/article/0,8599,1901663,00.html Holy Craps! How a Gambling Grandma Broke the Record] <br></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>[http://www.time.com/time/nation/article/0,8599,1901663,00.html Holy Craps! How a Gambling Grandma Broke the Record] <br></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Time.com, 29 May 2009<br></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Time.com, 29 May 2009<br></div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_49&diff=11211&oldid=prevBill Peterson: /* A new record in craps. */2009-06-25T15:24:26Z<p><span class="autocomment">A new record in craps.</span></p>
<table class='diff diff-contentalign-left'>
<col class='diff-marker' />
<col class='diff-content' />
<col class='diff-marker' />
<col class='diff-content' />
<tr valign='top'>
<td colspan='2' style="background-color: white; color:black;">← Older revision</td>
<td colspan='2' style="background-color: white; color:black;">Revision as of 15:24, 25 June 2009</td>
</tr><tr><td colspan="2" class="diff-lineno">Line 351:</td>
<td colspan="2" class="diff-lineno">Line 351:</td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>You can find a discussion of this problem in  [http://blogs.wsj.com/numbersguy/crunching-the-numbers-on-a-craps-record-703/ Crunching the numbers on a craps record] from Carl Bialik's Wall Street Journal column, &quot;The Numbers Guy&quot;.  He explains why the sevening-out event is complicated, and reports on various attempts to solve the problem.  In the final update, he reports that Keith Crank of the American Statistical Association used a Markov chain analysis (which presumably corresponds to what we did above) to find a probability of 1 in 5.6 billion.  Earlier in the article he cites simulation results from Professor Michael Shackleford, referencing Shackleford's [http://wizardofodds.com/askthewizard/askcolumns/askthewizard81.html Wizard of Odds] website.  Apart from the simulation results, Shackleford presents a recursive computation scheme that is equivalent to the Markov chain.  You can also read there interesting historical notes on craps records.</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>You can find a discussion of this problem in  [http://blogs.wsj.com/numbersguy/crunching-the-numbers-on-a-craps-record-703/ Crunching the numbers on a craps record] from Carl Bialik's Wall Street Journal column, &quot;The Numbers Guy&quot;.  He explains why the sevening-out event is complicated, and reports on various attempts to solve the problem.  In the final update, he reports that Keith Crank of the American Statistical Association used a Markov chain analysis (which presumably corresponds to what we did above) to find a probability of 1 in 5.6 billion.  Earlier in the article he cites simulation results from Professor Michael Shackleford, referencing Shackleford's [http://wizardofodds.com/askthewizard/askcolumns/askthewizard81.html Wizard of Odds] website.  Apart from the simulation results, Shackleford presents a recursive computation scheme that is equivalent to the Markov chain.  You can also read there interesting historical notes on craps records.</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>Returning to the Time.com piece, we read &quot;The average number of dice rolls before sevening out? Eight.&quot;  The true value is about 8.5, but it doesn't follow from their original analysis, which asked only how long it takes to roll a seven (<del class="diffchange diffchange-inline">for which </del>the expected number of rolls is 6).  <del class="diffchange diffchange-inline">Going back to </del>the Markov chain, the expected time to absorption is the sum of the row zero entries in the fundamental matrix <math>N = (I-Q)^{-1}</math>.  Computing this gives an expected 8.5 rolls to seven-out.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>Returning to the Time.com piece, we read &quot;The average number of dice rolls before sevening out? Eight.&quot;  The true value is about 8.5, but it doesn't follow from their original analysis, which asked only how long it takes to roll a seven (the expected number of rolls is 6).  <ins class="diffchange diffchange-inline">Using </ins>the Markov chain <ins class="diffchange diffchange-inline">model</ins>, the expected time to absorption is the sum of the row zero entries in the fundamental matrix <math>N = (I-Q)^{-1}</math>.  Computing this gives an expected 8.5 rolls to seven-out.</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>Time also interviewed Professor Thomas Cover of Stanford, who pointed out that while the probability of the event in question is small, one needs to remember that there are many people playing craps at any time, all of whom are in principle contending for the record!  Shackleford estimates their are about 50 million craps turns per year in the US, giving about a 1% chance that a feat like Demauro's would occur in a given year.</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Time also interviewed Professor Thomas Cover of Stanford, who pointed out that while the probability of the event in question is small, one needs to remember that there are many people playing craps at any time, all of whom are in principle contending for the record!  Shackleford estimates their are about 50 million craps turns per year in the US, giving about a 1% chance that a feat like Demauro's would occur in a given year.</div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_49&diff=7852&oldid=prevBill Peterson: /* A new record in craps. */2009-06-25T15:23:01Z<p><span class="autocomment">A new record in craps.</span></p>
<table class='diff diff-contentalign-left'>
<col class='diff-marker' />
<col class='diff-content' />
<col class='diff-marker' />
<col class='diff-content' />
<tr valign='top'>
<td colspan='2' style="background-color: white; color:black;">← Older revision</td>
<td colspan='2' style="background-color: white; color:black;">Revision as of 15:23, 25 June 2009</td>
</tr><tr><td colspan="2" class="diff-lineno">Line 347:</td>
<td colspan="2" class="diff-lineno">Line 347:</td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>The sevened-out state is absorbing;  hence the form of row 4.  The probabilities for the other rows are easily computed. From state 0, the chance of rolling  2, 3, 7, 11, or 12 is (1+2+6+2+1)/36 = 12/36 which keeps the chain in state 0.  The chance of rolling 4 or 10 is (3+3)/36 which leads to state 1;  similar calculations hold for transitions to states 2 and 3.  Next, from state 1, the chance of reproducing the point is 3/36, which leads to state 0; the chance of rolling a 7 is 6/36 which leads to state 4;  otherwise the chain remains in state 1.  Rows 2 and 3 are analogous to row 1.</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>The sevened-out state is absorbing;  hence the form of row 4.  The probabilities for the other rows are easily computed. From state 0, the chance of rolling  2, 3, 7, 11, or 12 is (1+2+6+2+1)/36 = 12/36 which keeps the chain in state 0.  The chance of rolling 4 or 10 is (3+3)/36 which leads to state 1;  similar calculations hold for transitions to states 2 and 3.  Next, from state 1, the chance of reproducing the point is 3/36, which leads to state 0; the chance of rolling a 7 is 6/36 which leads to state 4;  otherwise the chain remains in state 1.  Rows 2 and 3 are analogous to row 1.</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>To analyze the chain, we use standard absorbing chain theory, following notation from Grinstead and Snell's [http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/book.html Introduction to Probability] text.  (Concise notes from a recent Dartmouth course are [http://www.math.dartmouth.edu/archive/m20x06/public_html/Lecture14.pdf here]).  The leading 4x4 submatrix corresponding to the transient states is denoted <math>Q</math>.  The probabilities of still being in particular transient states given by entries in the vector <math>(1,0,0,0)Q^{153}</math>, and the sum of these probabilities is the chance of not having sevened out after 153 rolls.  It is easy to implement this computation iteratively, giving <math>1.788824 \times 10^{-10}</math>, which approximately 1 in 5.59 billion.  This certainly is a small probability, but much larger than the 1 in 1.56 trillion from Time.com.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>To analyze the chain, we use standard absorbing chain theory, following notation from Grinstead and Snell's [http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/book.html Introduction to Probability] text.  (Concise notes from a recent Dartmouth course are [http://www.math.dartmouth.edu/archive/m20x06/public_html/Lecture14.pdf here]).  The leading 4x4 submatrix corresponding to the transient states is denoted <math>Q</math>.  The probabilities of still being in particular transient states <ins class="diffchange diffchange-inline">after 153 rolls is </ins>given by entries in the vector <math>(1,0,0,0)Q^{153}</math>, and the sum of these probabilities is the chance of not having sevened out after 153 rolls.  It is easy to implement this computation iteratively, giving <math>1.788824 \times 10^{-10}</math>, which approximately 1 in 5.59 billion.  This certainly is a small probability, but much larger than the 1 in 1.56 trillion from Time.com.</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>You can find a discussion of this problem in  [http://blogs.wsj.com/numbersguy/crunching-the-numbers-on-a-craps-record-703/ Crunching the numbers on a craps record] from Carl Bialik's Wall Street Journal column, &quot;The Numbers Guy&quot;.  He explains why the sevening-out event is complicated, and reports on various attempts to solve the problem.  In the final update, he reports that Keith Crank of the American Statistical Association used a Markov chain analysis (which presumably corresponds to what we did above) to find a probability of 1 in 5.6 billion.  Earlier in the article he cites simulation results from Professor Michael Shackleford, referencing Shackleford's [http://wizardofodds.com/askthewizard/askcolumns/askthewizard81.html Wizard of Odds] website.  Apart from the simulation results, Shackleford presents a recursive computation scheme that is equivalent to the Markov chain.  You can also read there interesting historical notes on craps records.</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>You can find a discussion of this problem in  [http://blogs.wsj.com/numbersguy/crunching-the-numbers-on-a-craps-record-703/ Crunching the numbers on a craps record] from Carl Bialik's Wall Street Journal column, &quot;The Numbers Guy&quot;.  He explains why the sevening-out event is complicated, and reports on various attempts to solve the problem.  In the final update, he reports that Keith Crank of the American Statistical Association used a Markov chain analysis (which presumably corresponds to what we did above) to find a probability of 1 in 5.6 billion.  Earlier in the article he cites simulation results from Professor Michael Shackleford, referencing Shackleford's [http://wizardofodds.com/askthewizard/askcolumns/askthewizard81.html Wizard of Odds] website.  Apart from the simulation results, Shackleford presents a recursive computation scheme that is equivalent to the Markov chain.  You can also read there interesting historical notes on craps records.</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>Returning to the Time.com piece, <del class="diffchange diffchange-inline">which notes </del>&quot;The average number of dice rolls before sevening out? Eight.&quot;  The true value is about 8.5, but it doesn't follow from their original analysis, which asked only how long it takes to roll a seven (the <del class="diffchange diffchange-inline">average </del>number of rolls is 6).  Going back to the Markov chain, the expected time to absorption is the sum of the row zero entries in the fundamental matrix <math>N = (I-Q)^{-1}</math>.  Computing this gives an expected 8.5 rolls to seven-out.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>Returning to the Time.com piece, <ins class="diffchange diffchange-inline">we read </ins>&quot;The average number of dice rolls before sevening out? Eight.&quot;  The true value is about 8.5, but it doesn't follow from their original analysis, which asked only how long it takes to roll a seven (<ins class="diffchange diffchange-inline">for which </ins>the <ins class="diffchange diffchange-inline">expected </ins>number of rolls is 6).  Going back to the Markov chain, the expected time to absorption is the sum of the row zero entries in the fundamental matrix <math>N = (I-Q)^{-1}</math>.  Computing this gives an expected 8.5 rolls to seven-out.</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>Time also interviewed Professor Thomas Cover of Stanford, who pointed out that while the probability of the event in question is small, one needs to remember that there are many people playing craps at any time, all of whom are in principle contending for the record!  Shackleford estimates their are about 50 million craps turns per year in the US, giving about a 1% chance that a feat like Demauro's would occur in a given year.</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Time also interviewed Professor Thomas Cover of Stanford, who pointed out that while the probability of the event in question is small, one needs to remember that there are many people playing craps at any time, all of whom are in principle contending for the record!  Shackleford estimates their are about 50 million craps turns per year in the US, giving about a 1% chance that a feat like Demauro's would occur in a given year.</div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_49&diff=7851&oldid=prevBill Peterson: /* A new record in craps. */2009-06-24T21:59:39Z<p><span class="autocomment">A new record in craps.</span></p>
<table class='diff diff-contentalign-left'>
<col class='diff-marker' />
<col class='diff-content' />
<col class='diff-marker' />
<col class='diff-content' />
<tr valign='top'>
<td colspan='2' style="background-color: white; color:black;">← Older revision</td>
<td colspan='2' style="background-color: white; color:black;">Revision as of 21:59, 24 June 2009</td>
</tr><tr><td colspan="2" class="diff-lineno">Line 353:</td>
<td colspan="2" class="diff-lineno">Line 353:</td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Returning to the Time.com piece, which notes &quot;The average number of dice rolls before sevening out? Eight.&quot;  The true value is about 8.5, but it doesn't follow from their original analysis, which asked only how long it takes to roll a seven (the average number of rolls is 6).  Going back to the Markov chain, the expected time to absorption is the sum of the row zero entries in the fundamental matrix <math>N = (I-Q)^{-1}</math>.  Computing this gives an expected 8.5 rolls to seven-out.</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Returning to the Time.com piece, which notes &quot;The average number of dice rolls before sevening out? Eight.&quot;  The true value is about 8.5, but it doesn't follow from their original analysis, which asked only how long it takes to roll a seven (the average number of rolls is 6).  Going back to the Markov chain, the expected time to absorption is the sum of the row zero entries in the fundamental matrix <math>N = (I-Q)^{-1}</math>.  Computing this gives an expected 8.5 rolls to seven-out.</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>Time also interviewed Professor Thomas Cover of Stanford, who pointed out that while the probability of the event in question is small, one needs to remember that there are many people playing craps at any time, all of whom are in principle contending for the record!  Shackleford estimates their are about 50 million craps turns per year in the US, giving about a 1% chance that a feat like Demauro's would occur in a year.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>Time also interviewed Professor Thomas Cover of Stanford, who pointed out that while the probability of the event in question is small, one needs to remember that there are many people playing craps at any time, all of whom are in principle contending for the record!  Shackleford estimates their are about 50 million craps turns per year in the US, giving about a 1% chance that a feat like Demauro's would occur in a <ins class="diffchange diffchange-inline">given </ins>year.</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 QUESTION:<br></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>DISCUSSION QUESTION:<br></div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_49&diff=7827&oldid=prevBill Peterson: /* A new record in craps. */2009-06-24T21:58:49Z<p><span class="autocomment">A new record in craps.</span></p>
<table class='diff diff-contentalign-left'>
<col class='diff-marker' />
<col class='diff-content' />
<col class='diff-marker' />
<col class='diff-content' />
<tr valign='top'>
<td colspan='2' style="background-color: white; color:black;">← Older revision</td>
<td colspan='2' style="background-color: white; color:black;">Revision as of 21:58, 24 June 2009</td>
</tr><tr><td colspan="2" class="diff-lineno">Line 353:</td>
<td colspan="2" class="diff-lineno">Line 353:</td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Returning to the Time.com piece, which notes &quot;The average number of dice rolls before sevening out? Eight.&quot;  The true value is about 8.5, but it doesn't follow from their original analysis, which asked only how long it takes to roll a seven (the average number of rolls is 6).  Going back to the Markov chain, the expected time to absorption is the sum of the row zero entries in the fundamental matrix <math>N = (I-Q)^{-1}</math>.  Computing this gives an expected 8.5 rolls to seven-out.</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Returning to the Time.com piece, which notes &quot;The average number of dice rolls before sevening out? Eight.&quot;  The true value is about 8.5, but it doesn't follow from their original analysis, which asked only how long it takes to roll a seven (the average number of rolls is 6).  Going back to the Markov chain, the expected time to absorption is the sum of the row zero entries in the fundamental matrix <math>N = (I-Q)^{-1}</math>.  Computing this gives an expected 8.5 rolls to seven-out.</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>Time also interviewed Professor Thomas Cover of Stanford, who pointed out that while the probability of the event in question is small, one needs to remember that there are many people playing craps at any time, all of whom are in principle contending for the record!  Shackleford estimates their are about 50 million craps turns per year in the US, giving about a 1% chance that <del class="diffchange diffchange-inline">such an event </del>would occur in a year.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>Time also interviewed Professor Thomas Cover of Stanford, who pointed out that while the probability of the event in question is small, one needs to remember that there are many people playing craps at any time, all of whom are in principle contending for the record!  Shackleford estimates their are about 50 million craps turns per year in the US, giving about a 1% chance that <ins class="diffchange diffchange-inline">a feat like Demauro's </ins>would occur in a year.</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 QUESTION:<br></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>DISCUSSION QUESTION:<br></div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_49&diff=7826&oldid=prevBill Peterson: /* A new record in craps. */2009-06-24T21:55:44Z<p><span class="autocomment">A new record in craps.</span></p>
<table class='diff diff-contentalign-left'>
<col class='diff-marker' />
<col class='diff-content' />
<col class='diff-marker' />
<col class='diff-content' />
<tr valign='top'>
<td colspan='2' style="background-color: white; color:black;">← Older revision</td>
<td colspan='2' style="background-color: white; color:black;">Revision as of 21:55, 24 June 2009</td>
</tr><tr><td colspan="2" class="diff-lineno">Line 265:</td>
<td colspan="2" class="diff-lineno">Line 265:</td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Claire Suddath</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Claire Suddath</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><del class="diffchange diffchange-inline">The article reports that on </del>May 23, a New Jersey woman named Patricia Demauro set a new world record for the longest turn at craps without &quot;sevening out&quot; by rolling the dice 154 consecutive times.  Unfortunately, the <del class="diffchange diffchange-inline">lead paragraph </del>misstates the probability calculation needed to compute the odds of her feat<del class="diffchange diffchange-inline">.</del></div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div><ins class="diffchange diffchange-inline">On </ins>May 23, a New Jersey woman named Patricia Demauro set a new world record for the longest turn at craps without &quot;sevening out&quot; by rolling the dice 154 consecutive times.  Unfortunately, the <ins class="diffchange diffchange-inline">article </ins>misstates the probability calculation needed to compute the odds of her feat<ins class="diffchange diffchange-inline">:</ins></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><blockquote></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div><blockquote></div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_49&diff=7825&oldid=prevBill Peterson: /* A new record in craps. */2009-06-23T12:59:19Z<p><span class="autocomment">A new record in craps.</span></p>
<table class='diff diff-contentalign-left'>
<col class='diff-marker' />
<col class='diff-content' />
<col class='diff-marker' />
<col class='diff-content' />
<tr valign='top'>
<td colspan='2' style="background-color: white; color:black;">← Older revision</td>
<td colspan='2' style="background-color: white; color:black;">Revision as of 12:59, 23 June 2009</td>
</tr><tr><td colspan="2" class="diff-lineno">Line 319:</td>
<td colspan="2" class="diff-lineno">Line 319:</td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div></center></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div></center></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 this illustration, the shooter first establishes a point, namely 5, on the fourth roll.  This moves the chain to state 2, where it remains until the shooter reproduces a the 5 on the <del class="diffchange diffchange-inline">eight </del>roll.  This starts another sequence of come-out attempts.  A point of 4 is established on the tenth roll, and the shooter sevens-out on the twelfth roll, ending her turn.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>In this illustration, the shooter first establishes a point, namely 5, on the fourth roll.  This moves the chain to state 2, where it remains until the shooter reproduces a the 5 on the <ins class="diffchange diffchange-inline">eighth </ins>roll.  This starts another sequence of come-out attempts.  A point of 4 is established on the tenth roll, and the shooter sevens-out on the twelfth roll, ending her turn.</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>The probability transition matrix ''P'' for the Markov chain is given below:</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>The probability transition matrix ''P'' for the Markov chain is given below:</div></td></tr>
</table>Bill Petersonhttp://test.causeweb.org/wiki/chance/index.php?title=Chance_News_49&diff=7792&oldid=prevBill Peterson: /* A new record in craps. */2009-06-23T12:58:54Z<p><span class="autocomment">A new record in craps.</span></p>
<table class='diff diff-contentalign-left'>
<col class='diff-marker' />
<col class='diff-content' />
<col class='diff-marker' />
<col class='diff-content' />
<tr valign='top'>
<td colspan='2' style="background-color: white; color:black;">← Older revision</td>
<td colspan='2' style="background-color: white; color:black;">Revision as of 12:58, 23 June 2009</td>
</tr><tr><td colspan="2" class="diff-lineno">Line 345:</td>
<td colspan="2" class="diff-lineno">Line 345:</td></tr>
<tr><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div></center></div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div></center></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>The sevened-out state is absorbing;  hence the form of row 4.  The probabilities for the other rows are easily computed. From state 0, the chance of rolling  2, 3, 7, 11, or 12 is (1+2+6+2+1)/36 = 12/36 which keeps the chain in state 0.  The chance of rolling 4 or 10 is (3+3)/36 which leads to state 1.  <del class="diffchange diffchange-inline">From </del>state 1, the chance of reproducing the point is 3/36, which leads to state 0; the chance of rolling a 7 is 6/36 which leads to state 4;  otherwise the chain remains in state 1.  Rows 2 and 3 are analogous to row 1.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>The sevened-out state is absorbing;  hence the form of row 4.  The probabilities for the other rows are easily computed. From state 0, the chance of rolling  2, 3, 7, 11, or 12 is (1+2+6+2+1)/36 = 12/36 which keeps the chain in state 0.  The chance of rolling 4 or 10 is (3+3)/36 which leads to state 1<ins class="diffchange diffchange-inline">;  similar calculations hold for transitions to states 2 and 3</ins>.  <ins class="diffchange diffchange-inline">Next, from </ins>state 1, the chance of reproducing the point is 3/36, which leads to state 0; the chance of rolling a 7 is 6/36 which leads to state 4;  otherwise the chain remains in state 1.  Rows 2 and 3 are analogous to row 1.</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>To analyze the chain, we use standard absorbing chain theory, following notation from Grinstead and Snell's [http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/book.html Introduction to Probability] text.  (Concise notes from a recent Dartmouth course are [http://www.math.dartmouth.edu/archive/m20x06/public_html/Lecture14.pdf here]).  The leading 4x4 submatrix corresponding to the transient states is denoted <math>Q</math>.  The probabilities of still being in particular transient states given by entries in the vector <math>(1,0,0,0)Q^{153}</math>, and the sum of these probabilities is the chance of not having sevened out after 153 rolls.  It is easy to implement this computation iteratively, giving <math>1.788824 \times 10^{-10}</math>, which approximately 1 in 5.59 billion.  This certainly is a small probability, but much larger than the 1 in 1.56 trillion from Time.com.</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>To analyze the chain, we use standard absorbing chain theory, following notation from Grinstead and Snell's [http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/book.html Introduction to Probability] text.  (Concise notes from a recent Dartmouth course are [http://www.math.dartmouth.edu/archive/m20x06/public_html/Lecture14.pdf here]).  The leading 4x4 submatrix corresponding to the transient states is denoted <math>Q</math>.  The probabilities of still being in particular transient states given by entries in the vector <math>(1,0,0,0)Q^{153}</math>, and the sum of these probabilities is the chance of not having sevened out after 153 rolls.  It is easy to implement this computation iteratively, giving <math>1.788824 \times 10^{-10}</math>, which approximately 1 in 5.59 billion.  This certainly is a small probability, but much larger than the 1 in 1.56 trillion from Time.com.</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>You can find a discussion of this problem in  [http://blogs.wsj.com/numbersguy/crunching-the-numbers-on-a-craps-record-703/ Crunching the numbers on a craps record] from Carl Bialik's Wall Street Journal column, &quot;The Numbers Guy&quot;.  He explains why the sevening-out event is complicated, and reports on various attempts to solve the problem.  In the final update, he reports that Keith Crank of the American Statistical Association used a Markov chain analysis (which presumably corresponds to what we did above) to find a probability of 1 in 5.6 billion.  Earlier in the article he cites simulation results from Professor Michael Shackleford, referencing <del class="diffchange diffchange-inline">Shakleford</del>'s [http://wizardofodds.com/askthewizard/askcolumns/askthewizard81.html Wizard of Odds] website.  Apart from the simulation results, Shackleford presents a recursive computation scheme that is equivalent to the Markov chain.  You can also read there interesting historical notes on craps records.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>You can find a discussion of this problem in  [http://blogs.wsj.com/numbersguy/crunching-the-numbers-on-a-craps-record-703/ Crunching the numbers on a craps record] from Carl Bialik's Wall Street Journal column, &quot;The Numbers Guy&quot;.  He explains why the sevening-out event is complicated, and reports on various attempts to solve the problem.  In the final update, he reports that Keith Crank of the American Statistical Association used a Markov chain analysis (which presumably corresponds to what we did above) to find a probability of 1 in 5.6 billion.  Earlier in the article he cites simulation results from Professor Michael Shackleford, referencing <ins class="diffchange diffchange-inline">Shackleford</ins>'s [http://wizardofodds.com/askthewizard/askcolumns/askthewizard81.html Wizard of Odds] website.  Apart from the simulation results, Shackleford presents a recursive computation scheme that is equivalent to the Markov chain.  You can also read there interesting historical notes on craps records.</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>Returning to the Time.com piece, which notes &quot;The average number of dice rolls before sevening out? Eight.&quot;<del class="diffchange diffchange-inline">. </del> The true value is about 8.5, but it doesn't follow from their original analysis, which asked only how long it takes to roll a seven (the average number of rolls is 6).  Going back to the Markov chain, the expected time to absorption is the sum of the row zero entries in the fundamental matrix <math>N = (I-Q)^{-1}</math>.  Computing this gives an expected 8.5 rolls to seven-out.</div></td><td class='diff-marker'>+</td><td style="background: #cfc; color:black; font-size: smaller;"><div>Returning to the Time.com piece, which notes &quot;The average number of dice rolls before sevening out? Eight.&quot;  The true value is about 8.5, but it doesn't follow from their original analysis, which asked only how long it takes to roll a seven (the average number of rolls is 6).  Going back to the Markov chain, the expected time to absorption is the sum of the row zero entries in the fundamental matrix <math>N = (I-Q)^{-1}</math>.  Computing this gives an expected 8.5 rolls to seven-out.</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>Time also interviewed Professor Thomas Cover of Stanford, who pointed out that while the probability of the event in question is small, one needs to remember that there are many people playing craps at any time, all of whom are in principle contending for the record!  Shackleford estimates their are about 50 million craps turns per year in the US, giving about a 1% chance that such an event would occur in a year.</div></td><td class='diff-marker'> </td><td style="background: #eee; color:black; font-size: smaller;"><div>Time also interviewed Professor Thomas Cover of Stanford, who pointed out that while the probability of the event in question is small, one needs to remember that there are many people playing craps at any time, all of whom are in principle contending for the record!  Shackleford estimates their are about 50 million craps turns per year in the US, giving about a 1% chance that such an event would occur in a year.</div></td></tr>
</table>Bill Peterson