{"id":229,"date":"2012-07-28T17:27:02","date_gmt":"2012-07-28T17:27:02","guid":{"rendered":"http:\/\/alexboisvert.com\/musings\/?p=229"},"modified":"2012-07-28T17:27:02","modified_gmt":"2012-07-28T17:27:02","slug":"a-probability-problem","status":"publish","type":"post","link":"https:\/\/alexboisvert.com\/musings\/2012\/07\/28\/a-probability-problem\/","title":{"rendered":"A Probability Problem"},"content":{"rendered":"<p>[latexpage]<br \/>\nWhile I&#8217;m at it, I might as well write up the problem that inspired <a href=\"http:\/\/t.co\/DKOrw5oo\">the last post<\/a>.  Here it is:<\/p>\n<blockquote><p>Flip a coin until you get consecutive heads, then stop.  Let X represent the number of flips it took.  Then flip another coin until you get <strong>three<\/strong> consecutive heads.  Let Y represent the total number of flips there.  Find Pr(X>Y).<\/p><\/blockquote>\n<p><!--more--><br \/>\nWe started this problem in the previous post.  There we found that \\(Pr(X=k)=F_{k-1}\/2^k\\) where \\(F_k\\) is the k-th Fibonacci number.  From the homework problem in the last post, we can get that \\(Pr(X>k)=F_{k+2}\/2^k\\).  And the distribution of Y is fairly easy to derive along the same lines as X: it turns out to be<br \/>\n\\[Pr(Y=k)=\\frac{G_{k-2}}{2^k}\\]<br \/>\nwhere \\(G_k\\) is the k-th <a href=\"http:\/\/mathworld.wolfram.com\/TribonacciNumber.html\">Tribonacci number<\/a>.<\/p>\n<p>So!<br \/>\n\\[Pr(X>Y)=\\sum_{k=3}^{\\infty}Pr(X>k)Pr(Y=k)=\\sum_{k=3}^{\\infty}\\frac{F_{k+2}G_{k-2}}{2^{2k}}\\]<br \/>\nwhich is approximately 0.21.  Can anyone find a nice closed form solution of this?  There are closed form solutions for the Fibonacci and Tribonacci numbers but they&#8217;re complicated.  There are also matrix representations, which might help.  Any ideas?<\/p>\n","protected":false},"excerpt":{"rendered":"<p>[latexpage] While I&#8217;m at it, I might as well write up the problem that inspired the last post. Here it is: Flip a coin until you get consecutive heads, then stop. Let X represent the number of flips it took. &hellip; <a href=\"https:\/\/alexboisvert.com\/musings\/2012\/07\/28\/a-probability-problem\/\">Continue reading <span class=\"meta-nav\">&rarr;<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[9],"tags":[],"class_list":["post-229","post","type-post","status-publish","format-standard","hentry","category-math"],"_links":{"self":[{"href":"https:\/\/alexboisvert.com\/musings\/wp-json\/wp\/v2\/posts\/229","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/alexboisvert.com\/musings\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/alexboisvert.com\/musings\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/alexboisvert.com\/musings\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/alexboisvert.com\/musings\/wp-json\/wp\/v2\/comments?post=229"}],"version-history":[{"count":5,"href":"https:\/\/alexboisvert.com\/musings\/wp-json\/wp\/v2\/posts\/229\/revisions"}],"predecessor-version":[{"id":234,"href":"https:\/\/alexboisvert.com\/musings\/wp-json\/wp\/v2\/posts\/229\/revisions\/234"}],"wp:attachment":[{"href":"https:\/\/alexboisvert.com\/musings\/wp-json\/wp\/v2\/media?parent=229"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/alexboisvert.com\/musings\/wp-json\/wp\/v2\/categories?post=229"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/alexboisvert.com\/musings\/wp-json\/wp\/v2\/tags?post=229"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}