{"id":9224,"date":"2023-09-24T14:32:51","date_gmt":"2023-09-24T12:32:51","guid":{"rendered":"https:\/\/www.mathweb.fr\/euclide\/?p=9224"},"modified":"2024-05-30T14:51:02","modified_gmt":"2024-05-30T12:51:02","slug":"le-paradoxe-des-anniversaires","status":"publish","type":"post","link":"https:\/\/www.mathweb.fr\/euclide\/2023\/09\/24\/le-paradoxe-des-anniversaires\/","title":{"rendered":"Le paradoxe des anniversaires"},"content":{"rendered":"\n<p class=\"is-style-Paragraph-paragraph\">Le paradoxe des anniversaires est un c\u00e9l\u00e8bre probl\u00e8me math\u00e9matiques qui peut rev\u00eatir plusieurs formes. Par exemple, il peut se libeller ainsi:<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"is-style-Paragraph-paragraph\">Combien de personnes faut-il r\u00e9unir dans une pi\u00e8ce pour avoir une probabilit\u00e9 d&#8217;au moins 50 % que deux d&#8217;entre elles aient la m\u00eame date d&#8217;anniversaire ?<\/p>\n<\/blockquote>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Nous allons r\u00e9pondre \u00e0 cette question en utilisant au passage le langage Python.<\/p>\n\n\n\n<!--more-->\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Le paradoxe des anniversaires a \u00e9t\u00e9 popularis\u00e9 gr\u00e2ce aux travaux du math\u00e9maticien danois <strong>Richard von Mises<\/strong> au <em>d\u00e9but du XXe si\u00e8cle<\/em> (&#8220;Probability, Statistics, and Truth&#8221;, 1928) et d&#8217;autres math\u00e9maticiens semble-t-il.<\/p>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Le paradoxe a \u00e9t\u00e9 popularis\u00e9 dans le milieu acad\u00e9mique. Il est souvent utilis\u00e9 pour enseigner les concepts de probabilit\u00e9 et d&#8217;intuition probabiliste. Il est \u00e9galement un sujet int\u00e9ressant de discussion et d&#8217;exp\u00e9rimentation dans de nombreux contextes. En effet, il montre comment les gens ont souvent du mal \u00e0 estimer correctement les probabilit\u00e9s. En particulier, lorsqu&#8217;il s&#8217;agit de situations impliquant de grandes combinaisons possibles.<\/p>\n\n\n<div class=\"wp-block-image is-style-rounded\">\n<figure class=\"aligncenter size-medium is-resized\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2023\/09\/paradoxe-anniversaires.jpg\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"300\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2023\/09\/paradoxe-anniversaires-300x300.jpg\" alt=\"paradoxe anniversaires\" class=\"wp-image-9225\" style=\"width:304px;height:304px\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2023\/09\/paradoxe-anniversaires-300x300.jpg 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2023\/09\/paradoxe-anniversaires-160x160.jpg 160w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2023\/09\/paradoxe-anniversaires-150x150.jpg 150w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2023\/09\/paradoxe-anniversaires.jpg 512w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/figure>\n<\/div>\n\n\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_82_2 counter-hierarchy ez-toc-counter ez-toc-white ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Au menu sur cette page...<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.mathweb.fr\/euclide\/2023\/09\/24\/le-paradoxe-des-anniversaires\/#Paradoxe_des_anniversaires_simplification_du_probleme_a_3_personnes\" >Paradoxe des anniversaires: simplification du probl\u00e8me \u00e0 3 personnes<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.mathweb.fr\/euclide\/2023\/09\/24\/le-paradoxe-des-anniversaires\/#Paradoxe_des_anniversaires_generalisation_mathematique\" >Paradoxe des anniversaires: g\u00e9n\u00e9ralisation math\u00e9matique<\/a><ul class='ez-toc-list-level-3' ><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.mathweb.fr\/euclide\/2023\/09\/24\/le-paradoxe-des-anniversaires\/#Python_a_la_rescousse\" >Python \u00e0 la rescousse<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Paradoxe_des_anniversaires_simplification_du_probleme_a_3_personnes\"><\/span>Paradoxe des anniversaires: simplification du probl\u00e8me \u00e0 3 personnes<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Avant tout, clarifions une chose: par &#8220;date d&#8217;anniversaire&#8221;, nous devons entendre &#8220;jour d&#8217;anniversaire&#8221;. De plus, consid\u00e9rons qu&#8217;il y a 365 jours dans une ann\u00e9e pour simplifier les calculs. Au diable les ann\u00e9es bissextiles et peu nous chaut des individus n\u00e9s les 29 f\u00e9vrier!&#8230; <\/p>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Consid\u00e9rons un groupe de trois personnes; notons-les \\(p_1\\), \\(p_2\\) et \\(p_3\\). D\u00e9terminons la probabilit\u00e9 qu&#8217;au moins deux des trois personnes aient la m\u00eame date d&#8217;anniversaire.<\/p>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">L&#8217;\u00e9nonc\u00e9 comporte &#8220;la probabilit\u00e9 qu&#8217;au moins&#8230;&#8221;; cela nous pousse \u00e0 consid\u00e9rer l&#8217;\u00e9v\u00e9nement contraire. En effet, il est plus simple de d\u00e9terminer la probabilit\u00e9 qu&#8217;aucune personne n&#8217;aient la m\u00eame date de naissance.<\/p>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Consid\u00e9rons \\(p_1\\). La probabilit\u00e9 que \\(p_2\\) n&#8217;ait pas la m\u00eame date d&#8217;anniversaire est \\(\\frac{364}{365}\\). La probabilit\u00e9 que \\(p_3\\) n&#8217;ait pas la m\u00eame date d&#8217;anniversaire que \\(p_1\\) et \\(p_2\\) est \\(\\frac{363}{365}\\). On peut donc dire que la probabilit\u00e9 que ces trois personnes n&#8217;aient pas la m\u00eame date d&#8217;anniversaire est:$$\\frac{364}{365}\\times\\frac{363}{365}.$$<\/p>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">La probabilit\u00e9 qui nous int\u00e9resse est donc:$$1-\\frac{364}{365}\\times\\frac{363}{365}\\approx0,008.$$ Autant dire que l&#8217;on est loin des 50% avec trois personnes&#8230; Mais on s&#8217;en doutais hein ?<\/p>\n\n\n<div class=\"wp-block-image is-style-rounded\">\n<figure class=\"aligncenter size-medium\"><a href=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2023\/09\/paradoxe-anniversaires-2.jpg\" data-fancybox=\"gallery\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"300\" src=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2023\/09\/paradoxe-anniversaires-2-300x300.jpg\" alt=\"paradoxe anniversaires\" class=\"wp-image-9226\" srcset=\"https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2023\/09\/paradoxe-anniversaires-2-300x300.jpg 300w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2023\/09\/paradoxe-anniversaires-2-160x160.jpg 160w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2023\/09\/paradoxe-anniversaires-2-150x150.jpg 150w, https:\/\/www.mathweb.fr\/euclide\/wp-content\/uploads\/2023\/09\/paradoxe-anniversaires-2.jpg 512w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><\/figure>\n<\/div>\n\n\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Paradoxe_des_anniversaires_generalisation_mathematique\"><\/span>Paradoxe des anniversaires: g\u00e9n\u00e9ralisation math\u00e9matique<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Consid\u00e9rons d\u00e9sormais que plus de trois personnes sont dans une m\u00eame pi\u00e8ce. Allez, on va dire qu&#8217;il y en a <em>n<\/em> &gt; 3. Soyons fous!<\/p>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Par un raisonnement analogue \u00e0 celui utilis\u00e9 pr\u00e9c\u00e9demment, on peut dire que la probabilit\u00e9 cherch\u00e9e est:$$p_n=1-\\frac{364}{365}\\times\\frac{363}{365}\\times\\cdots\\times\\frac{365-(n-1)}{365}=1-\\frac{364\\times363\\times\\cdots\\times(366-n)}{365^{n-1}}.$$<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"Python_a_la_rescousse\"><\/span>Python \u00e0 la rescousse<span class=\"ez-toc-section-end\"><\/span><\/h3>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Il nous faut maintenant trouver le premier entier <em>n<\/em> pour lequel \\(p_n\\) est sup\u00e9rieure \u00e0 0,5. Rien de plus simple avec notre ami Pyth.<\/p>\n\n\n\n<pre class=\"EnlighterJSRAW\" data-enlighter-language=\"python\" data-enlighter-theme=\"dracula\" data-enlighter-highlight=\"\" data-enlighter-linenumbers=\"\" data-enlighter-lineoffset=\"\" data-enlighter-title=\"\" data-enlighter-group=\"\">def proba(n):\n    p = 1\n    for i in range(1,n):\n        p *= (365-i)\/365\n        \n    return 1-p\n\nn = 3\nwhile proba(n) &lt;= 0.5:\n    print( f'Pour n={n}, la probabilit\u00e9 est \u00e9gale \u00e0 {proba(n)}.' )\n    n += 1\n    \nprint( f'Pour n={n}, la probabilit\u00e9 est \u00e9gale \u00e0 {proba(n)}.' )<\/pre>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Bon, l\u00e0, je vous ai pondu un programme d\u00e9taill\u00e9 donc il para\u00eet un peu long. Mais je voulais qu&#8217;il soit clair. On voit ainsi les \u00e9tapes.<\/p>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Et voici ce que ce programme nous recrache:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>Pour n=3, la probabilit\u00e9 est \u00e9gale \u00e0 0.008204165884781345.\nPour n=4, la probabilit\u00e9 est \u00e9gale \u00e0 0.016355912466550215.\nPour n=5, la probabilit\u00e9 est \u00e9gale \u00e0 0.02713557369979347.\nPour n=6, la probabilit\u00e9 est \u00e9gale \u00e0 0.040462483649111425.\nPour n=7, la probabilit\u00e9 est \u00e9gale \u00e0 0.056235703095975365.\nPour n=8, la probabilit\u00e9 est \u00e9gale \u00e0 0.07433529235166902.\nPour n=9, la probabilit\u00e9 est \u00e9gale \u00e0 0.09462383388916673.\nPour n=10, la probabilit\u00e9 est \u00e9gale \u00e0 0.11694817771107768.\nPour n=11, la probabilit\u00e9 est \u00e9gale \u00e0 0.14114137832173312.\nPour n=12, la probabilit\u00e9 est \u00e9gale \u00e0 0.1670247888380645.\nPour n=13, la probabilit\u00e9 est \u00e9gale \u00e0 0.19441027523242949.\nPour n=14, la probabilit\u00e9 est \u00e9gale \u00e0 0.2231025120049731.\nPour n=15, la probabilit\u00e9 est \u00e9gale \u00e0 0.25290131976368646.\nPour n=16, la probabilit\u00e9 est \u00e9gale \u00e0 0.2836040052528501.\nPour n=17, la probabilit\u00e9 est \u00e9gale \u00e0 0.3150076652965609.\nPour n=18, la probabilit\u00e9 est \u00e9gale \u00e0 0.3469114178717896.\nPour n=19, la probabilit\u00e9 est \u00e9gale \u00e0 0.37911852603153695.\nPour n=20, la probabilit\u00e9 est \u00e9gale \u00e0 0.41143838358058027.\nPour n=21, la probabilit\u00e9 est \u00e9gale \u00e0 0.443688335165206.\nPour n=22, la probabilit\u00e9 est \u00e9gale \u00e0 0.4756953076625503.\nPour n=23, la probabilit\u00e9 est \u00e9gale \u00e0 0.5072972343239857.<\/code><\/pre>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Ringo! Nous avons notre <em>n<\/em> minimal: c&#8217;est 23!<\/p>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Il faut donc au minimum 23 personnes pour que la probabilit\u00e9 que deux d&#8217;entre elles aient la m\u00eame date d&#8217;anniversaire! Dingue non ?<\/p>\n\n\n\n<p class=\"is-style-Paragraph-paragraph\">Ce r\u00e9sultat, pouvant para\u00eetre contre-intuitif, est \u00e0 l&#8217;origine de la qualification de ce probl\u00e8me comme un <em>paradoxe<\/em>. Cependant, \u00e0 mon avis, ce n&#8217;est pas \u00e0 proprement parl\u00e9 un paradoxe math\u00e9matique.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Le paradoxe des anniversaires est un c\u00e9l\u00e8bre probl\u00e8me math\u00e9matiques qui peut rev\u00eatir plusieurs formes. Par exemple, il peut se libeller ainsi: Combien de personnes faut-il r\u00e9unir dans une pi\u00e8ce pour avoir une probabilit\u00e9 d&#8217;au moins 50 % que deux d&#8217;entre elles aient la m\u00eame date d&#8217;anniversaire ? Nous allons r\u00e9pondre [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":9228,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6,5],"tags":[399,398],"class_list":["post-9224","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematiques","category-python","tag-anniversaire","tag-paradoxe"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.3 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Le paradoxe des anniversaires - Mathweb.fr<\/title>\n<meta name=\"description\" content=\"Le paradoxe des anniversaires est un c\u00e9l\u00e8bre probl\u00e8me math\u00e9matiques qui peut rev\u00eatir plusieurs formes. 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