{"id":577,"date":"2012-04-30T11:33:48","date_gmt":"2012-04-30T16:33:48","guid":{"rendered":"http:\/\/linesoftangency.wordpress.com\/?p=577"},"modified":"2013-08-03T01:18:10","modified_gmt":"2013-08-03T01:18:10","slug":"label-maker","status":"publish","type":"post","link":"http:\/\/blog.chrislusto.com\/?p=577","title":{"rendered":"Label Maker"},"content":{"rendered":"<p>If you've perused this blog, you know that I love probability.\u00a0 I was fortunate enough to see <a href=\"http:\/\/patternsinpractice.wordpress.com\/2011\/05\/05\/counting-with-polynomials\/\" target=\"_blank\">Al Cuoco<\/a> and Alicia Chiasson give a really cool presentation at this year's <a href=\"http:\/\/www.nctm.org\/\" target=\"_blank\">NCTM <\/a>conference about exploring the probabilities of dice sums geometrically and algebraically.\u00a0 Wheelhouse.\u00a0 After we got done looking at some student work and pictures of distributions, Al nonchalantly threw out the following question:<\/p>\n<blockquote><p>Is it possible to change the integer labels on two dice [from the standard 1,2,3,4,5,6] such that the distribution of sums remains unchanged?<\/p><\/blockquote>\n<p>Of course he was much cooler than that.\u00a0 I've significantly nerded up the language for the sake of brevity and clarity.\u00a0 Still, good question, right?\u00a0 And of course since our teacher has posed this tantalizing challenge, we know that the answer is yes, and now it's up to us to fill in the details.\u00a0 Thusly:<\/p>\n<p>First let's make use of the Cuoco\/Chiasson observation that we can represent the throw of a standard die with the polynomial<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='http:\/\/blog.chrislusto.com\/wp-content\/plugins\/latex\/cache\/tex_a41468af17177b200c0abf14d5183d0d.gif' style='vertical-align: middle; border: none;' class='tex' alt=\"\" \/><\/span><script type='math\/tex;  mode=display'><\/script><\/p><\/p>\n<p>When we do it this way, the exponents represent the label values for each face, and the coefficients represent frequencies of each label landing face up (relative to the total sample space).\u00a0 This is neither surprising, nor super helpful.\u00a0 Each \"sum\" occurs once out of the six possible.\u00a0 We knew this already.<\/p>\n<p>What <em>is<\/em> super helpful is that we can include <em>n<\/em> dice in our toss by expanding <em>n<\/em> factors of P(x).\u00a0 For two dice (the number in question), that looks like<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='http:\/\/blog.chrislusto.com\/wp-content\/plugins\/latex\/cache\/tex_35b60d35ab2e3d940cf724104a6468e0.gif' style='vertical-align: middle; border: none;' class='tex' alt=\"\" \/><\/span><script type='math\/tex;  mode=display'><\/script><\/p><\/p>\n<p>You can easily confirm that this jibes with the standard diagram.\u00a0 For instance the sum of 7 shows up most often (6 out of 36 times), which helps casinos make great heaps of money off of bettors on the come.\u00a0 Take a moment.\u00a0 Compare.<\/p>\n<p><a href=\"http:\/\/blog.chrislusto.com\/wp-content\/uploads\/2012\/04\/dice.png\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-589\" title=\"Dice Sums\" alt=\"\" src=\"http:\/\/blog.chrislusto.com\/wp-content\/uploads\/2012\/04\/dice.png\" width=\"500\" height=\"375\" srcset=\"http:\/\/blog.chrislusto.com\/wp-content\/uploads\/2012\/04\/dice.png 500w, http:\/\/blog.chrislusto.com\/wp-content\/uploads\/2012\/04\/dice-300x225.png 300w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><\/a><\/p>\n<p>Okay, so now we know that the standard labels yield the standard distribution of sums.\u00a0 The question, though, is whether there are any <strong>other<\/strong> labels that do so as well.\u00a0 Here's where some abstract algebra comes in handy.\u00a0 Let's assume that there are, in fact, dice out there who satisfy this property.\u00a0 We can represent those with polynomials as well.\u00a0 We know that the coefficient on each term must still be 1 (each face will still come up 1 out of 6 times), but we don't yet know about the exponents (labels).\u00a0 So let's say the labels on the two dice are, respectively<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='http:\/\/blog.chrislusto.com\/wp-content\/plugins\/latex\/cache\/tex_32eabcc43a60325e616d399fda579717.gif' style='vertical-align: middle; border: none;' class='tex' alt=\"\" \/><\/span><script type='math\/tex;  mode=display'><\/script><\/p> and <p style='text-align:center;'><span class='MathJax_Preview'><img src='http:\/\/blog.chrislusto.com\/wp-content\/plugins\/latex\/cache\/tex_4160a860d539abb2dbdd5e3a7ef2ee30.gif' style='vertical-align: middle; border: none;' class='tex' alt=\"\" \/><\/span><script type='math\/tex;  mode=display'><\/script><\/p>.<\/p>\n<p>If we want the same exact sum distribution, it had better be true that<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='http:\/\/blog.chrislusto.com\/wp-content\/plugins\/latex\/cache\/tex_2c7b1ed13eda06b27c1cdfaf0e43d86b.gif' style='vertical-align: middle; border: none;' class='tex' alt=\"\" \/><\/span><script type='math\/tex;  mode=display'><\/script><\/p>.<\/p>\n<p>For future convenience (trust me), let's call the first polynomial factor on the right hand side <strong>Q(x)<\/strong>.\u00a0 Great!\u00a0 Now we just have to figure out what all the a's and b's are.\u00a0 It helps that our polynomials belong to the ring <strong>Z[x]<\/strong>, which is a unique factorization domain.\u00a0 A little factoring practice will show us that<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='http:\/\/blog.chrislusto.com\/wp-content\/plugins\/latex\/cache\/tex_0fd69b245d927a6674bf5b36c13cd4f3.gif' style='vertical-align: middle; border: none;' class='tex' alt=\"\" \/><\/span><script type='math\/tex;  mode=display'><\/script><\/p>.<\/p>\n<p>We just have to rearrange these irreducible factors to get the answer we're looking for.\u00a0 Due to a theorem that is too long and frightening to reproduce here [<em>waves hands frantically<\/em>], we know that the unique factorization of Q(x)---our polynomial with unknown exponents---must be of the form<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='http:\/\/blog.chrislusto.com\/wp-content\/plugins\/latex\/cache\/tex_3e7a40a77e4c4b14a59976da6078ec22.gif' style='vertical-align: middle; border: none;' class='tex' alt=\"\" \/><\/span><script type='math\/tex;  mode=display'><\/script><\/p>,<\/p>\n<p>where <em>s<\/em>, <em>t<\/em>, <em>u<\/em>, and <em>v<\/em><em><\/em> are all either 0, 1, or 2.\u00a0 So that's good news, not too many possibilities to check.\u00a0 In fact, we can make our lives a little easier.\u00a0 First of all, notice that Q(1) must equal 6.\u00a0 Right?\u00a0 Each throw of that single die must yield each of the 6 faces with equal probability.\u00a0 But then substituting 1 into the factored form gives us<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='http:\/\/blog.chrislusto.com\/wp-content\/plugins\/latex\/cache\/tex_566c5e69242f39af80a07bf8e4d7bc26.gif' style='vertical-align: middle; border: none;' class='tex' alt=\"\" \/><\/span><script type='math\/tex;  mode=display'><\/script><\/p><\/p>\n<p>Clearly this means that <em>t<\/em> and <em>u<\/em> have to be 1, and we just have to nail down <em>s<\/em> and <em>v<\/em>.\u00a0 Well, if we take a look at Q(0), we also quickly realize that <em>s<\/em> can't be 0.\u00a0 It can't be 2 either, because, if <em>s<\/em> is 2, then the smallest sum we could obtain on our dice would be 3---which is absolutely no good at all.\u00a0 So\u00a0<em>s<\/em> is 1 as well.\u00a0 Let's see what happens in our three remaining cases, when <em>u<\/em> is 0, 1, and 2:<\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='http:\/\/blog.chrislusto.com\/wp-content\/plugins\/latex\/cache\/tex_9646f947d82a694a4c32f373a35edf0d.gif' style='vertical-align: middle; border: none;' class='tex' alt=\"\" \/><\/span><script type='math\/tex;  mode=display'><\/script><\/p><\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='http:\/\/blog.chrislusto.com\/wp-content\/plugins\/latex\/cache\/tex_c537a75606d0b6243a7dc5a0f87d7db6.gif' style='vertical-align: middle; border: none;' class='tex' alt=\"\" \/><\/span><script type='math\/tex;  mode=display'><\/script><\/p><\/p>\n<p><p style='text-align:center;'><span class='MathJax_Preview'><img src='http:\/\/blog.chrislusto.com\/wp-content\/plugins\/latex\/cache\/tex_e92778eceee9cfe2f9664eccd8fa5687.gif' style='vertical-align: middle; border: none;' class='tex' alt=\"\" \/><\/span><script type='math\/tex;  mode=display'><\/script><\/p><\/p>\n<p>Check out those strange and beautiful labels!\u00a0 We can mark up the first die with the exponents from the <em>u <\/em>= 0 case, and the second die with the <em>u<\/em> = 2 case.\u00a0 When we multiply those two polynomials together we get back P(x)<sup>2<\/sup>, which is precisely what we needed (check if you like)!\u00a0 Our other option, of course, is to label two dice with the <em>u <\/em>=1 case, which corresponds to a standard die.\u00a0 And, thanks to unique factorization, we can be sure that <strong>there are no other cases<\/strong>.\u00a0 Not only have we found some different labels, we've found\u00a0all of them!<\/p>\n<p>If the a's on the first die are (1,2,2,3,3,4), then the b's end up being (1,3,4,5,6,8), and vice versa.\u00a0 And, comfortingly, if the a's on the first die are (1,2,3,4,5,6), then so are the b's on the second one.<\/p>\n<p>Two dice with the <em>u<\/em> = 1 label are what you find at every craps table in the country.\u00a0 One die of each of the other labels forms a pair of <a href=\"http:\/\/en.wikipedia.org\/wiki\/Sicherman_dice\" target=\"_blank\">Sicherman dice<\/a>, and they are the only other dice that yield the same sum distribution.\u00a0 You could drop Sicherman dice in the middle of Vegas, and nobody would notice.\u00a0 At least in terms of money changing hands.\u00a0 The pit boss might take exception.\u00a0 Come to think of it, I cannot stress how important it is that you <strong>not<\/strong> attempt to switch out dice in Vegas.\u00a0 Your spine is also uniquely factorable...into irreducible vertebrae.<\/p>\n<p>*This whole proof has been cribbed from <em>Contemporary Abstract Algebra<\/em> (2nd ed.), by <a href=\"http:\/\/en.wikipedia.org\/wiki\/Joseph_Gallian\" target=\"_blank\">Joseph A. Gallian<\/a>.\u00a0 If you want the whole citation, click his name and scroll down.*<\/p>\n","protected":false},"excerpt":{"rendered":"<p>If you've perused this blog, you know that I love probability.\u00a0 I was fortunate enough to see Al Cuoco and Alicia Chiasson give a really cool presentation at this year's NCTM conference about exploring the probabilities of dice sums geometrically and algebraically.\u00a0 Wheelhouse.\u00a0 After we got done looking at some student work and pictures of [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[5,14,32,46],"class_list":["post-577","post","type-post","status-publish","format-standard","hentry","category-math-musing","tag-algebra","tag-dice","tag-math","tag-probability"],"_links":{"self":[{"href":"http:\/\/blog.chrislusto.com\/index.php?rest_route=\/wp\/v2\/posts\/577","targetHints":{"allow":["GET"]}}],"collection":[{"href":"http:\/\/blog.chrislusto.com\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"http:\/\/blog.chrislusto.com\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"http:\/\/blog.chrislusto.com\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/blog.chrislusto.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=577"}],"version-history":[{"count":2,"href":"http:\/\/blog.chrislusto.com\/index.php?rest_route=\/wp\/v2\/posts\/577\/revisions"}],"predecessor-version":[{"id":726,"href":"http:\/\/blog.chrislusto.com\/index.php?rest_route=\/wp\/v2\/posts\/577\/revisions\/726"}],"wp:attachment":[{"href":"http:\/\/blog.chrislusto.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=577"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"http:\/\/blog.chrislusto.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=577"},{"taxonomy":"post_tag","embeddable":true,"href":"http:\/\/blog.chrislusto.com\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=577"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}