Le paradoxe des anniversaires \u2014 MathsVivantes<\/title>\n<link href=\"https:\/\/fonts.googleapis.com\/css2?family=Playfair+Display:ital,wght@0,400;0,700;1,400&family=Source+Serif+4:ital,wght@0,300;0,400;0,600;1,300;1,400&display=swap\" rel=\"stylesheet\">\n<style>\n :root {\n --noir: #0f0e0b;\n --creme: #f5f0e8;\n --rose: #a0254a;\n --rose-clair: #e8547a;\n --rose-pale: #fceef3;\n --ocre: #c8850a;\n --ocre-clair: #f0c060;\n --gris: #6b6560;\n --gris-clair: #e8e2d8;\n }\n\n * { margin: 0; padding: 0; box-sizing: border-box; }\n\n body {\n background-color: var(--creme);\n color: var(--noir);\n font-family: 'Source Serif 4', Georgia, serif;\n font-weight: 300;\n line-height: 1.8;\n }\n\n .hero {\n background: #1f0a16;\n color: var(--creme);\n padding: 80px 40px 70px;\n position: relative;\n overflow: hidden;\n }\n .hero::before {\n content: \"\ud83c\udf82 ? ! \ud83c\udf89\";\n position: absolute;\n top: 20px; right: 40px;\n font-size: 4rem;\n opacity: 0.06;\n letter-spacing: 0.3em;\n }\n .hero-inner { max-width: 760px; 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top: -10px;\n font-size: 3.5rem;\n color: var(--rose-clair);\n font-family: 'Playfair Display', serif;\n line-height: 1;\n }\n blockquote cite { display: block; font-size: 0.82rem; font-style: normal; letter-spacing: 0.1em; text-transform: uppercase; color: var(--gris); margin-top: 8px; }\n\n .conclusion { border-top: 2px solid var(--noir); margin-top: 3em; padding-top: 2em; }\n\n @media (max-width: 600px) {\n .hero { padding: 50px 20px 40px; }\n .article-body { padding: 40px 20px 60px; }\n .exercice { padding: 24px 20px; }\n .code-block { padding: 16px; font-size: 0.78rem; }\n }\n<\/style>\n<\/head>\n<body>\n\n<div class=\"hero\">\n <div class=\"hero-inner\">\n <div class=\"article-rubrique\">Probabilit\u00e9s \u00b7 Article #6<\/div>\n <h1>Le paradoxe des anniversaires :<br><em>la surprise math\u00e9matique<\/em><\/h1>\n <p class=\"hero-intro\">Dans une classe de 30 \u00e9l\u00e8ves, quelle est la probabilit\u00e9 que deux d’entre eux aient le m\u00eame anniversaire ? La plupart des gens r\u00e9pondent : \u00ab\u00a0tr\u00e8s faible\u00a0\u00bb. La r\u00e9alit\u00e9 : plus de 70%. Et dans une classe de 23 \u00e9l\u00e8ves seulement, c’est d\u00e9j\u00e0 50%. Comment est-ce possible ?<\/p>\n <div class=\"meta\">Par Odilon AKOWANOU \u00b7 MathsVivantes \u00b7 Lecture : 10 min<\/div>\n <\/div>\n<\/div>\n\n<article class=\"article-body\">\n\n <p>\n Imaginons la sc\u00e8ne. Tu entres dans ta classe de 3\u00e8me au CS Flambeau d’Afrique. 30 \u00e9l\u00e8ves sont assis devant toi. Tu lances un d\u00e9fi : <em>\u00ab\u00a0Je parie que deux \u00e9l\u00e8ves dans cette salle ont le m\u00eame jour d’anniversaire.\u00a0\u00bb<\/em> Les \u00e9l\u00e8ves sourient, incr\u00e9dules. Il y a 365 jours dans l’ann\u00e9e. Seulement 30 \u00e9l\u00e8ves. Les chances semblent minimes.\n <\/p>\n\n <p>\n Tu gagnes ton pari. Presque \u00e0 coup s\u00fbr.\n <\/p>\n\n <p>\n Ce r\u00e9sultat \u2014 qui d\u00e9fie toute intuition \u2014 s’appelle le <strong>paradoxe des anniversaires<\/strong>. Et comprendre pourquoi il est vrai, c’est comprendre l’une des erreurs les plus profondes que fait notre cerveau quand il raisonne sur les probabilit\u00e9s.\n <\/p>\n\n <h2>L’erreur de notre intuition<\/h2>\n\n <p>\n Quand on entend \u00ab\u00a0deux personnes sur 30 ont le m\u00eame anniversaire parmi 365 jours possibles\u00a0\u00bb, notre cerveau fait inconsciemment ce calcul : 30 \/ 365 \u2248 8%. Tr\u00e8s peu. Donc la probabilit\u00e9 semble faible.\n <\/p>\n\n <p>\n Mais c’est la mauvaise question ! On ne demande pas si <em>quelqu’un en particulier<\/em> a le m\u00eame anniversaire que toi. On demande si <em>n’importe quelle paire<\/em> dans le groupe partage un anniversaire. Et les paires possibles dans un groupe, \u00e7a se multipie tr\u00e8s vite.\n <\/p>\n\n <div class=\"encadre\">\n <span class=\"encadre-titre\">\ud83d\udd22 Combien de paires dans un groupe ?<\/span>\n <p>Dans un groupe de <strong>n<\/strong> personnes, le nombre de paires possibles est :<\/p>\n <p style=\"text-align:center; font-family:'Playfair Display',serif; font-size:1.2rem;\">C(n,2) = n \u00d7 (n-1) \/ 2<\/p>\n <p><strong>Pour 30 \u00e9l\u00e8ves :<\/strong> 30 \u00d7 29 \/ 2 = <strong>435 paires !<\/strong><\/p>\n <p>Ce ne sont pas 30 chances de co\u00efncidence \u2014 c’est <strong>435 chances<\/strong>. C’est \u00e7a l’intuition cl\u00e9.<\/p>\n <\/div>\n\n <h2>Le calcul : par le compl\u00e9ment<\/h2>\n\n <p>\n Le calcul direct serait complexe. On utilise une astuce classique en probabilit\u00e9s : calculer la probabilit\u00e9 de l’\u00e9v\u00e9nement <strong>contraire<\/strong> \u2014 que personne ne partage d’anniversaire \u2014 puis soustraire de 1.\n <\/p>\n\n <div class=\"definition\">\n <strong>R\u00e8gle du compl\u00e9ment :<\/strong> P(au moins une co\u00efncidence) = 1 – P(aucune co\u00efncidence). Quand l’\u00e9v\u00e9nement direct est compliqu\u00e9, on calcule son contraire, souvent plus simple.\n <\/div>\n\n <p>Calculons P(aucune co\u00efncidence) pour un groupe de n personnes :<\/p>\n\n <div class=\"etape\"><span class=\"etape-numero\">\u2460<\/span> Personne 1 : son anniversaire peut \u00eatre n’importe quel jour \u2192 365\/365 = 1<\/div>\n <div class=\"etape\"><span class=\"etape-numero\">\u2461<\/span> Personne 2 : doit \u00e9viter le jour de la personne 1 \u2192 364\/365<\/div>\n <div class=\"etape\"><span class=\"etape-numero\">\u2462<\/span> Personne 3 : doit \u00e9viter les 2 jours d\u00e9j\u00e0 pris \u2192 363\/365<\/div>\n <div class=\"etape\"><span class=\"etape-numero\">\u2463<\/span> Personne n : doit \u00e9viter les (n-1) jours d\u00e9j\u00e0 pris \u2192 (365-n+1)\/365<\/div>\n\n <p>Donc :<\/p>\n\n <div class=\"formule\">\n P(aucune co\u00efncidence) = (365\/365) \u00d7 (364\/365) \u00d7 (363\/365) \u00d7 … \u00d7 (365-n+1)\/365\n <\/div>\n\n <p>Et la probabilit\u00e9 d’au moins une co\u00efncidence vaut :<\/p>\n\n <div class=\"formule\">\n P(co\u00efncidence) = 1 – P(aucune co\u00efncidence)\n <\/div>\n\n <h2>Les r\u00e9sultats qui stup\u00e9fient<\/h2>\n\n <div class=\"table-container\">\n <table>\n <tr>\n <th>Nombre de personnes<\/th>\n <th>P(au moins une co\u00efncidence)<\/th>\n <th>Niveau de surprise<\/th>\n <\/tr>\n <tr><td>5<\/td><td>2,7%<\/td><td>\ud83d\ude10 Faible<\/td><\/tr>\n <tr><td>10<\/td><td>11,7%<\/td><td>\ud83d\ude10 Faible<\/td><\/tr>\n <tr><td>15<\/td><td>25,3%<\/td><td>\ud83e\udd14 Surprenant<\/td><\/tr>\n <tr><td>20<\/td><td>41,1%<\/td><td>\ud83d\ude32 Tr\u00e8s surprenant<\/td><\/tr>\n <tr class=\"highlight\"><td><strong>23<\/strong><\/td><td><strong>50,7%<\/strong><\/td><td>\ud83c\udfaf <strong>Seuil 50% !<\/strong><\/td><\/tr>\n <tr><td>30<\/td><td>70,6%<\/td><td>\ud83d\ude31 Choquant<\/td><\/tr>\n <tr><td>40<\/td><td>89,1%<\/td><td>\ud83d\ude31 Presque certain<\/td><\/tr>\n <tr><td>50<\/td><td>97,0%<\/td><td>\ud83d\udd25 Quasi certain<\/td><\/tr>\n <tr><td>60<\/td><td>99,4%<\/td><td>\u2705 Pratiquement s\u00fbr<\/td><\/tr>\n <\/table>\n <\/div>\n\n <p>Voici une visualisation de cette progression :<\/p>\n\n <div class=\"schema-container\">\n <div class=\"schema-titre\">\ud83d\udcca Probabilit\u00e9 de co\u00efncidence selon la taille du groupe<\/div>\n <svg width=\"100%\" viewBox=\"0 0 520 300\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\">\n <rect width=\"520\" height=\"300\" fill=\"white\"\/>\n \n <line x1=\"60\" y1=\"260\" x2=\"490\" y2=\"260\" stroke=\"#0f0e0b\" stroke-width=\"2\"\/>\n <line x1=\"60\" y1=\"260\" x2=\"60\" y2=\"20\" stroke=\"#0f0e0b\" stroke-width=\"2\"\/>\n \n <text x=\"260\" y=\"290\" font-size=\"11\" fill=\"#6b6560\" text-anchor=\"middle\">Nombre de personnes<\/text>\n <text x=\"15\" y=\"145\" font-size=\"11\" fill=\"#6b6560\" text-anchor=\"middle\" transform=\"rotate(-90,15,145)\">Probabilit\u00e9 (%)<\/text>\n \n <text x=\"40\" y=\"264\" font-size=\"9\" fill=\"#6b6560\">0%<\/text>\n <text x=\"35\" y=\"214\" font-size=\"9\" fill=\"#6b6560\">25%<\/text>\n <text x=\"35\" y=\"164\" font-size=\"9\" fill=\"#6b6560\">50%<\/text>\n <text x=\"35\" y=\"114\" font-size=\"9\" fill=\"#6b6560\">75%<\/text>\n <text x=\"35\" y=\"64\" font-size=\"9\" fill=\"#6b6560\">100%<\/text>\n \n <line x1=\"60\" y1=\"214\" x2=\"490\" y2=\"214\" stroke=\"#f0ebe3\" stroke-width=\"1\"\/>\n <line x1=\"60\" y1=\"164\" x2=\"490\" y2=\"164\" stroke=\"#e8547a\" stroke-width=\"1\" stroke-dasharray=\"4,3\" opacity=\"0.5\"\/>\n <line x1=\"60\" y1=\"114\" x2=\"490\" y2=\"114\" stroke=\"#f0ebe3\" stroke-width=\"1\"\/>\n <line x1=\"60\" y1=\"64\" x2=\"490\" y2=\"64\" stroke=\"#f0ebe3\" stroke-width=\"1\"\/>\n \n <text x=\"395\" y=\"160\" font-size=\"10\" fill=\"#e8547a\">\u2190 Seuil 50%<\/text>\n \n <polyline\n points=\"60,259 100,249 140,234 180,214 207,196 220,186 250,172 280,154 310,130 340,108 370,88 400,72 430,60 460,54 490,50\"\n fill=\"none\" stroke=\"#a0254a\" stroke-width=\"3\" stroke-linejoin=\"round\"\/>\n \n <polygon\n points=\"60,260 60,259 100,249 140,234 180,214 207,196 220,186 250,172 280,154 310,130 340,108 370,88 400,72 430,60 460,54 490,50 490,260\"\n fill=\"#a0254a\" opacity=\"0.08\"\/>\n \n <circle cx=\"207\" cy=\"196\" r=\"6\" fill=\"#a0254a\"\/>\n <line x1=\"207\" y1=\"196\" x2=\"207\" y2=\"260\" stroke=\"#a0254a\" stroke-width=\"1.5\" stroke-dasharray=\"4,3\"\/>\n <text x=\"195\" y=\"275\" font-size=\"9\" fill=\"#a0254a\" font-weight=\"bold\">23<\/text>\n \n <circle cx=\"250\" cy=\"172\" r=\"6\" fill=\"#e8547a\"\/>\n <text x=\"238\" y=\"168\" font-size=\"9\" fill=\"#e8547a\">30\u219270%<\/text>\n \n <text x=\"100\" y=\"274\" font-size=\"9\" fill=\"#6b6560\">10<\/text>\n <text x=\"180\" y=\"274\" font-size=\"9\" fill=\"#6b6560\">20<\/text>\n <text x=\"260\" y=\"274\" font-size=\"9\" fill=\"#6b6560\">30<\/text>\n <text x=\"340\" y=\"274\" font-size=\"9\" fill=\"#6b6560\">40<\/text>\n <text x=\"420\" y=\"274\" font-size=\"9\" fill=\"#6b6560\">50<\/text>\n <\/svg>\n <\/div>\n\n <h2>Calculons ensemble pour 23 personnes<\/h2>\n\n <p>\n V\u00e9rifions que pour 23 personnes, la probabilit\u00e9 d\u00e9passe effectivement 50% :\n <\/p>\n\n <div class=\"etape\"><span class=\"etape-numero\">\u2460<\/span> P(aucune co\u00efncidence, 23 personnes) = (365 \u00d7 364 \u00d7 363 \u00d7 … \u00d7 343) \/ 365\u00b2\u00b3<\/div>\n <div class=\"etape\"><span class=\"etape-numero\">\u2461<\/span> = (365!\/342!) \/ 365\u00b2\u00b3<\/div>\n <div class=\"etape\"><span class=\"etape-numero\">\u2462<\/span> \u2248 0,493<\/div>\n <div class=\"etape\"><span class=\"etape-numero\">\u2463<\/span> P(co\u00efncidence) = 1 – 0,493 = <strong>0,507 = 50,7%<\/strong> \u2713<\/div>\n\n <div class=\"formule\">Avec seulement 23 personnes \u2192 P(co\u00efncidence) > 50% !<\/div>\n\n <h2>Le code Python pour visualiser<\/h2>\n\n <div class=\"code-block\">\n <span class=\"code-label\">Python<\/span>\n<span class=\"code-comment\"># Paradoxe des anniversaires – Simulation et calcul exact<\/span>\n<span class=\"code-keyword\">import<\/span> numpy <span class=\"code-keyword\">as<\/span> np\n<span class=\"code-keyword\">import<\/span> matplotlib.pyplot <span class=\"code-keyword\">as<\/span> plt\n\n<span class=\"code-comment\"># — M\u00e9thode 1 : Calcul exact —<\/span>\n<span class=\"code-keyword\">def<\/span> <span class=\"code-function\">proba_anniversaire<\/span>(n):\n <span class=\"code-string\">\u00ab\u00a0\u00a0\u00bb\u00a0\u00bbCalcule la probabilit\u00e9 exacte de co\u00efncidence pour n personnes\u00a0\u00bb\u00a0\u00bb\u00a0\u00bb<\/span>\n p_aucune = <span class=\"code-number\">1.0<\/span>\n <span class=\"code-keyword\">for<\/span> i <span class=\"code-keyword\">in<\/span> <span class=\"code-function\">range<\/span>(n):\n p_aucune *= (<span class=\"code-number\">365<\/span> – i) \/ <span class=\"code-number\">365<\/span>\n <span class=\"code-keyword\">return<\/span> <span class=\"code-number\">1<\/span> – p_aucune\n\n<span class=\"code-comment\"># Afficher les r\u00e9sultats cl\u00e9s<\/span>\n<span class=\"code-keyword\">for<\/span> n <span class=\"code-keyword\">in<\/span> [<span class=\"code-number\">10<\/span>, <span class=\"code-number\">20<\/span>, <span class=\"code-number\">23<\/span>, <span class=\"code-number\">30<\/span>, <span class=\"code-number\">50<\/span>]:\n p = <span class=\"code-function\">proba_anniversaire<\/span>(n)\n <span class=\"code-function\">print<\/span>(<span class=\"code-string\">f\u00a0\u00bbn={n:2d} \u2192 P(co\u00efncidence) = {p:.1%}\u00a0\u00bb<\/span>)\n\n<span class=\"code-comment\"># — M\u00e9thode 2 : Simulation Monte Carlo —<\/span>\n<span class=\"code-keyword\">def<\/span> <span class=\"code-function\">simulation_anniversaire<\/span>(n, nb_simulations=<span class=\"code-number\">100000<\/span>):\n <span class=\"code-string\">\u00ab\u00a0\u00a0\u00bb\u00a0\u00bbSimule le paradoxe par tirage al\u00e9atoire\u00a0\u00bb\u00a0\u00bb\u00a0\u00bb<\/span>\n coincidences = <span class=\"code-number\">0<\/span>\n <span class=\"code-keyword\">for<\/span> _ <span class=\"code-keyword\">in<\/span> <span class=\"code-function\">range<\/span>(nb_simulations):\n anniversaires = np.random.<span class=\"code-function\">randint<\/span>(<span class=\"code-number\">1<\/span>, <span class=\"code-number\">366<\/span>, n)\n <span class=\"code-keyword\">if<\/span> <span class=\"code-function\">len<\/span>(<span class=\"code-function\">set<\/span>(anniversaires)) < n:\n coincidences += <span class=\"code-number\">1<\/span>\n <span class=\"code-keyword\">return<\/span> coincidences \/ nb_simulations\n\n<span class=\"code-comment\"># V\u00e9rification par simulation pour 23 personnes<\/span>\np_sim = <span class=\"code-function\">simulation_anniversaire<\/span>(<span class=\"code-number\">23<\/span>)\n<span class=\"code-function\">print<\/span>(<span class=\"code-string\">f\u00a0\u00bb\\nSimulation pour 23 personnes : {p_sim:.1%}\u00a0\u00bb<\/span>)\n<span class=\"code-function\">print<\/span>(<span class=\"code-string\">f\u00a0\u00bbCalcul exact pour 23 personnes : {proba_anniversaire(23):.1%}\u00a0\u00bb<\/span>)\n\n<span class=\"code-comment\"># — Graphique —<\/span>\nn_valeurs = <span class=\"code-function\">range<\/span>(<span class=\"code-number\">1<\/span>, <span class=\"code-number\">61<\/span>)\nprobas = [<span class=\"code-function\">proba_anniversaire<\/span>(n) <span class=\"code-keyword\">for<\/span> n <span class=\"code-keyword\">in<\/span> n_valeurs]\n\nplt.<span class=\"code-function\">figure<\/span>(figsize=(<span class=\"code-number\">10<\/span>, <span class=\"code-number\">6<\/span>))\nplt.<span class=\"code-function\">plot<\/span>(n_valeurs, probas, color=<span class=\"code-string\">‘#a0254a’<\/span>, linewidth=<span class=\"code-number\">2.5<\/span>)\nplt.<span class=\"code-function\">axhline<\/span>(y=<span class=\"code-number\">0.5<\/span>, color=<span class=\"code-string\">‘#e8547a’<\/span>, linestyle=<span class=\"code-string\">‘–‘<\/span>, label=<span class=\"code-string\">‘Seuil 50%’<\/span>)\nplt.<span class=\"code-function\">axvline<\/span>(x=<span class=\"code-number\">23<\/span>, color=<span class=\"code-string\">‘#c8850a’<\/span>, linestyle=<span class=\"code-string\">‘–‘<\/span>, label=<span class=\"code-string\">’23 personnes’<\/span>)\nplt.<span class=\"code-function\">fill_between<\/span>(n_valeurs, probas, alpha=<span class=\"code-number\">0.1<\/span>, color=<span class=\"code-string\">‘#a0254a’<\/span>)\nplt.<span class=\"code-function\">xlabel<\/span>(<span class=\"code-string\">‘Nombre de personnes dans le groupe’<\/span>)\nplt.<span class=\"code-function\">ylabel<\/span>(<span class=\"code-string\">‘Probabilit\u00e9 d\\’au moins une co\u00efncidence’<\/span>)\nplt.<span class=\"code-function\">title<\/span>(<span class=\"code-string\">‘Paradoxe des anniversaires’<\/span>)\nplt.<span class=\"code-function\">legend<\/span>()\nplt.<span class=\"code-function\">grid<\/span>(alpha=<span class=\"code-number\">0.3<\/span>)\nplt.<span class=\"code-function\">show<\/span>()\n <\/div>\n\n <div class=\"encadre\">\n <span class=\"encadre-titre\">\ud83d\udcbb R\u00e9sultat attendu<\/span>\n <p>n=10 \u2192 P(co\u00efncidence) = 11.7%<\/p>\n <p>n=20 \u2192 P(co\u00efncidence) = 41.1%<\/p>\n <p><strong>n=23 \u2192 P(co\u00efncidence) = 50.7%<\/strong><\/p>\n <p>n=30 \u2192 P(co\u00efncidence) = 70.6%<\/p>\n <p>n=50 \u2192 P(co\u00efncidence) = 97.0%<\/p>\n <p style=\"margin-top:8px; color:#a0254a\"><strong>Simulation pour 23 personnes : ~50.7% \u2713<\/strong><\/p>\n <\/div>\n\n <h2>Tester en classe \u2014 une exp\u00e9rience vivante<\/h2>\n\n <div class=\"encadre-histoire encadre\">\n <span class=\"encadre-titre\">\ud83c\udfeb Activit\u00e9 p\u00e9dagogique pour ta classe<\/span>\n <p>\n Voici comment utiliser ce paradoxe directement en classe, que ce soit en 3\u00e8me, en Terminale ou m\u00eame en premi\u00e8re ann\u00e9e de licence :\n <\/p>\n <p>\n <strong>\u00c9tape 1 :<\/strong> Pose la question avant d’expliquer \u2014 <em>\u00ab\u00a0Qui pense que deux \u00e9l\u00e8ves dans cette salle ont le m\u00eame anniversaire ?\u00a0\u00bb<\/em> Note les r\u00e9ponses. La majorit\u00e9 dira non.\n <\/p>\n <p>\n <strong>\u00c9tape 2 :<\/strong> Fais le tour de la classe, chaque \u00e9l\u00e8ve dit son mois et jour d’anniversaire. Arr\u00eate-toi d\u00e8s qu’une co\u00efncidence appara\u00eet. Dans 70% des classes de 30, \u00e7a arrivera.\n <\/p>\n <p>\n <strong>\u00c9tape 3 :<\/strong> Explique le calcul. Regarde les visages changer. C’est ce moment de surprise \u2014 cette rupture entre intuition et r\u00e9alit\u00e9 \u2014 qui cr\u00e9e l’envie d’apprendre.\n <\/p>\n <p>\n <strong>\u00c9tape 4 :<\/strong> Introduis la r\u00e8gle du compl\u00e9ment et le concept de paires. Tu viens d’enseigner les probabilit\u00e9s de fa\u00e7on inoubliable.\n <\/p>\n <\/div>\n\n <h2>Pourquoi notre cerveau se trompe ?<\/h2>\n\n <p>\n Ce paradoxe r\u00e9v\u00e8le un biais cognitif fondamental : notre cerveau est mauvais pour \u00e9valuer les probabilit\u00e9s qui impliquent des <strong>combinaisons<\/strong>. Nous pensons de fa\u00e7on lin\u00e9aire (30 personnes = 30 chances) alors que la r\u00e9alit\u00e9 est combinatoire (30 personnes = 435 paires).\n <\/p>\n\n <p>\n Ce m\u00eame biais nous fait sous-estimer la probabilit\u00e9 d’accidents rares qui se r\u00e9p\u00e8tent, surestimer nos chances \u00e0 la loterie, et mal \u00e9valuer les risques en m\u00e9decine. Comprendre les probabilit\u00e9s, c’est apprendre \u00e0 corriger son propre cerveau.\n <\/p>\n\n <blockquote>\n L’intuition est un outil formidable \u2014 sauf en probabilit\u00e9s, o\u00f9 elle nous trahit presque syst\u00e9matiquement.\n <cite>\u2014 Le\u00e7on des probabilit\u00e9s<\/cite>\n <\/blockquote>\n\n \n <div class=\"exercice\">\n <h3>\u270f\ufe0f Exercice corrig\u00e9<\/h3>\n <p>\n Dans un lyc\u00e9e de Cotonou, une salle de r\u00e9union accueille les d\u00e9l\u00e9gu\u00e9s de classe : <strong>15 \u00e9l\u00e8ves<\/strong> en tout.\n <\/p>\n <p><strong>a)<\/strong> Combien de paires d’\u00e9l\u00e8ves peut-on former dans ce groupe ?<\/p>\n <p><strong>b)<\/strong> Calcule P(aucune co\u00efncidence d’anniversaire) pour ces 15 \u00e9l\u00e8ves. On donnera le r\u00e9sultat sous forme d\u00e9cimale arrondie \u00e0 0,001.<\/p>\n <p><strong>c)<\/strong> En d\u00e9duire P(au moins une co\u00efncidence). Est-ce surprenant ?<\/p>\n <p><strong>d)<\/strong> Si on rajoute 8 \u00e9l\u00e8ves suppl\u00e9mentaires (soit 23 au total), que devient cette probabilit\u00e9 ? Conclure.<\/p>\n\n <div class=\"correction\">\n <div class=\"correction-titre\">\u25b8 Correction<\/div>\n <p><strong>a)<\/strong> Nombre de paires = C(15,2) = 15 \u00d7 14 \/ 2 = <strong>105 paires<\/strong><\/p>\n <p><strong>b)<\/strong> P(aucune co\u00efncidence) = (365\/365) \u00d7 (364\/365) \u00d7 (363\/365) \u00d7 … \u00d7 (351\/365)<\/p>\n <p>= (365 \u00d7 364 \u00d7 363 \u00d7 … \u00d7 351) \/ 365\u00b9\u2075<\/p>\n <p>\u2248 <strong>0,747<\/strong><\/p>\n <p><strong>c)<\/strong> P(au moins une co\u00efncidence) = 1 – 0,747 = <strong>0,253 soit 25,3%<\/strong><\/p>\n <p>Oui, c’est d\u00e9j\u00e0 surprenant : 1 chance sur 4 avec seulement 15 personnes !<\/p>\n <p><strong>d)<\/strong> Pour 23 personnes : P(co\u00efncidence) = 1 – (365\u00d7364\u00d7…\u00d7343)\/365\u00b2\u00b3 \u2248 <strong>50,7%<\/strong><\/p>\n <p>En ajoutant 8 personnes, on passe de 25% \u00e0 plus de 50% \u2014 la probabilit\u00e9 a <strong>doubl\u00e9<\/strong> ! C’est l’effet explosif des combinaisons. \u2705<\/p>\n <\/div>\n <\/div>\n\n <div class=\"conclusion\">\n <h2>Ce qu’on retient<\/h2>\n <p>\n Le paradoxe des anniversaires nous enseigne trois choses essentielles. D’abord, que notre intuition probabiliste est structurellement d\u00e9faillante \u2014 et qu’il faut le savoir pour ne pas se laisser pi\u00e9ger. Ensuite, que la cl\u00e9 est souvent de compter les <em>paires<\/em>, pas les individus. Enfin, que les probabilit\u00e9s peuvent \u00eatre un outil p\u00e9dagogique puissant : rien ne capte l’attention d’un \u00e9l\u00e8ve comme un r\u00e9sultat qui contredit ce qu’il croyait \u00e9vident.\n <\/p>\n <p>\n La prochaine fois que tu entres dans une salle de 23 personnes ou plus, tu sais maintenant qu’il y a plus de 50% de chances que deux d’entre elles soufflent leurs bougies le m\u00eame jour. Les math\u00e9matiques sont partout \u2014 m\u00eame dans les g\u00e2teaux d’anniversaire.\n <\/p>\n <\/div>\n\n<\/article>\n\n<\/body>\n<\/html>\n","protected":false},"excerpt":{"rendered":"<p>Le paradoxe des anniversaires \u2014 MathsVivantes Probabilit\u00e9s \u00b7 Article #6 Le paradoxe des anniversaires :la surprise math\u00e9matique Dans une classe […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-4)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-4)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-4)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[8],"tags":[],"class_list":["post-85","post","type-post","status-publish","format-standard","hentry","category-fondements"],"yoast_head":"\n<title>Le paradoxe des anniversaires - AfrikIntelligent<\/title>\n<meta name=\"description\" content=\"Dans une classe de 30 \u00e9leves , deux ont le m\u00eame anniversaire avec 70% de probabilit\u00e9 . 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Le paradoxe des anniversaires expliqu\u00e9 simplement .","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/vmi3217629.contaboserver.net\/?p=85","og_locale":"fr_FR","og_type":"article","og_title":"Le paradoxe des anniversaires - AfrikIntelligent","og_description":"Dans une classe de 30 \u00e9leves , deux ont le m\u00eame anniversaire avec 70% de probabilit\u00e9 . Le paradoxe des anniversaires expliqu\u00e9 simplement .","og_url":"https:\/\/vmi3217629.contaboserver.net\/?p=85","og_site_name":"AfrikIntelligent","article_published_time":"2026-04-19T22:31:19+00:00","article_modified_time":"2026-05-02T07:26:00+00:00","author":"Odilon","twitter_card":"summary_large_image","twitter_misc":{"\u00c9crit par":"Odilon","Dur\u00e9e de lecture estim\u00e9e":"7 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/vmi3217629.contaboserver.net\/?p=85#article","isPartOf":{"@id":"https:\/\/vmi3217629.contaboserver.net\/?p=85"},"author":{"name":"Odilon","@id":"https:\/\/vmi3217629.contaboserver.net\/#\/schema\/person\/a3da39574d729b6b3ff7514ccc055923"},"headline":"Le paradoxe des anniversaires","datePublished":"2026-04-19T22:31:19+00:00","dateModified":"2026-05-02T07:26:00+00:00","mainEntityOfPage":{"@id":"https:\/\/vmi3217629.contaboserver.net\/?p=85"},"wordCount":1328,"commentCount":0,"articleSection":["Fondements"],"inLanguage":"fr-FR","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/vmi3217629.contaboserver.net\/?p=85#respond"]}]},{"@type":"WebPage","@id":"https:\/\/vmi3217629.contaboserver.net\/?p=85","url":"https:\/\/vmi3217629.contaboserver.net\/?p=85","name":"Le paradoxe des anniversaires - AfrikIntelligent","isPartOf":{"@id":"https:\/\/vmi3217629.contaboserver.net\/#website"},"datePublished":"2026-04-19T22:31:19+00:00","dateModified":"2026-05-02T07:26:00+00:00","author":{"@id":"https:\/\/vmi3217629.contaboserver.net\/#\/schema\/person\/a3da39574d729b6b3ff7514ccc055923"},"description":"Dans une classe de 30 \u00e9leves , deux ont le m\u00eame anniversaire avec 70% de probabilit\u00e9 . Le paradoxe des anniversaires expliqu\u00e9 simplement .","breadcrumb":{"@id":"https:\/\/vmi3217629.contaboserver.net\/?p=85#breadcrumb"},"inLanguage":"fr-FR","potentialAction":[{"@type":"ReadAction","target":["https:\/\/vmi3217629.contaboserver.net\/?p=85"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/vmi3217629.contaboserver.net\/?p=85#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Accueil","item":"https:\/\/vmi3217629.contaboserver.net\/"},{"@type":"ListItem","position":2,"name":"Le paradoxe des anniversaires"}]},{"@type":"WebSite","@id":"https:\/\/vmi3217629.contaboserver.net\/#website","url":"https:\/\/vmi3217629.contaboserver.net\/","name":"AfrikIntelligent","description":"Append autrement","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/vmi3217629.contaboserver.net\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"fr-FR"},{"@type":"Person","@id":"https:\/\/vmi3217629.contaboserver.net\/#\/schema\/person\/a3da39574d729b6b3ff7514ccc055923","name":"Odilon","image":{"@type":"ImageObject","inLanguage":"fr-FR","@id":"https:\/\/secure.gravatar.com\/avatar\/57bcd6a76fa395ae5e994943b66ccd3dd7e71cd0b1b77b351986e88c27801523?s=96&d=mm&r=g","url":"https:\/\/secure.gravatar.com\/avatar\/57bcd6a76fa395ae5e994943b66ccd3dd7e71cd0b1b77b351986e88c27801523?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/57bcd6a76fa395ae5e994943b66ccd3dd7e71cd0b1b77b351986e88c27801523?s=96&d=mm&r=g","caption":"Odilon"},"sameAs":["https:\/\/vmi3217629.contaboserver.net"],"url":"https:\/\/vmi3217629.contaboserver.net\/?author=1"}]}},"uagb_featured_image_src":{"full":false,"thumbnail":false,"medium":false,"medium_large":false,"large":false,"1536x1536":false,"2048x2048":false},"uagb_author_info":{"display_name":"Odilon","author_link":"https:\/\/vmi3217629.contaboserver.net\/?author=1"},"uagb_comment_info":0,"uagb_excerpt":"Le paradoxe des anniversaires \u2014 MathsVivantes Probabilit\u00e9s \u00b7 Article #6 Le paradoxe des anniversaires :la surprise math\u00e9matique Dans une classe […]","_links":{"self":[{"href":"https:\/\/vmi3217629.contaboserver.net\/index.php?rest_route=\/wp\/v2\/posts\/85","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/vmi3217629.contaboserver.net\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/vmi3217629.contaboserver.net\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/vmi3217629.contaboserver.net\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/vmi3217629.contaboserver.net\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=85"}],"version-history":[{"count":1,"href":"https:\/\/vmi3217629.contaboserver.net\/index.php?rest_route=\/wp\/v2\/posts\/85\/revisions"}],"predecessor-version":[{"id":86,"href":"https:\/\/vmi3217629.contaboserver.net\/index.php?rest_route=\/wp\/v2\/posts\/85\/revisions\/86"}],"wp:attachment":[{"href":"https:\/\/vmi3217629.contaboserver.net\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=85"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/vmi3217629.contaboserver.net\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=85"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/vmi3217629.contaboserver.net\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=85"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

{"id":85,"date":"2026-04-19T23:31:19","date_gmt":"2026-04-19T22:31:19","guid":{"rendered":"https:\/\/vmi3217629.contaboserver.net\/?p=85"},"modified":"2026-05-02T08:26:00","modified_gmt":"2026-05-02T07:26:00","slug":"paradoxe-anniversaires-probabilites","status":"publish","type":"post","link":"https:\/\/vmi3217629.contaboserver.net\/?p=85","title":{"rendered":"Le paradoxe des anniversaires"},"content":{"rendered":"\n\n\n\n\n\nLe paradoxe des anniversaires \u2014 MathsVivantes<\/title>\n<link href=\"https:\/\/fonts.googleapis.com\/css2?family=Playfair+Display:ital,wght@0,400;0,700;1,400&family=Source+Serif+4:ital,wght@0,300;0,400;0,600;1,300;1,400&display=swap\" rel=\"stylesheet\">\n<style>\n :root {\n --noir: #0f0e0b;\n --creme: #f5f0e8;\n --rose: #a0254a;\n --rose-clair: #e8547a;\n --rose-pale: #fceef3;\n --ocre: #c8850a;\n --ocre-clair: #f0c060;\n --gris: #6b6560;\n --gris-clair: #e8e2d8;\n }\n\n * { margin: 0; padding: 0; box-sizing: border-box; }\n\n body {\n background-color: var(--creme);\n color: var(--noir);\n font-family: 'Source Serif 4', Georgia, serif;\n font-weight: 300;\n line-height: 1.8;\n }\n\n .hero {\n background: #1f0a16;\n color: var(--creme);\n padding: 80px 40px 70px;\n position: relative;\n overflow: hidden;\n }\n .hero::before {\n content: \"\ud83c\udf82 ? ! \ud83c\udf89\";\n position: absolute;\n top: 20px; 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top: -10px;\n font-size: 3.5rem;\n color: var(--rose-clair);\n font-family: 'Playfair Display', serif;\n line-height: 1;\n }\n blockquote cite { display: block; font-size: 0.82rem; font-style: normal; letter-spacing: 0.1em; text-transform: uppercase; color: var(--gris); margin-top: 8px; }\n\n .conclusion { border-top: 2px solid var(--noir); margin-top: 3em; padding-top: 2em; }\n\n @media (max-width: 600px) {\n .hero { padding: 50px 20px 40px; }\n .article-body { padding: 40px 20px 60px; }\n .exercice { padding: 24px 20px; }\n .code-block { padding: 16px; font-size: 0.78rem; }\n }\n<\/style>\n<\/head>\n<body>\n\n<div class=\"hero\">\n <div class=\"hero-inner\">\n <div class=\"article-rubrique\">Probabilit\u00e9s \u00b7 Article #6<\/div>\n <h1>Le paradoxe des anniversaires :<br><em>la surprise math\u00e9matique<\/em><\/h1>\n <p class=\"hero-intro\">Dans une classe de 30 \u00e9l\u00e8ves, quelle est la probabilit\u00e9 que deux d’entre eux aient le m\u00eame anniversaire ? La plupart des gens r\u00e9pondent : \u00ab\u00a0tr\u00e8s faible\u00a0\u00bb. La r\u00e9alit\u00e9 : plus de 70%. Et dans une classe de 23 \u00e9l\u00e8ves seulement, c’est d\u00e9j\u00e0 50%. Comment est-ce possible ?<\/p>\n <div class=\"meta\">Par Odilon AKOWANOU \u00b7 MathsVivantes \u00b7 Lecture : 10 min<\/div>\n <\/div>\n<\/div>\n\n<article class=\"article-body\">\n\n <p>\n Imaginons la sc\u00e8ne. Tu entres dans ta classe de 3\u00e8me au CS Flambeau d’Afrique. 30 \u00e9l\u00e8ves sont assis devant toi. Tu lances un d\u00e9fi : <em>\u00ab\u00a0Je parie que deux \u00e9l\u00e8ves dans cette salle ont le m\u00eame jour d’anniversaire.\u00a0\u00bb<\/em> Les \u00e9l\u00e8ves sourient, incr\u00e9dules. Il y a 365 jours dans l’ann\u00e9e. Seulement 30 \u00e9l\u00e8ves. Les chances semblent minimes.\n <\/p>\n\n <p>\n Tu gagnes ton pari. Presque \u00e0 coup s\u00fbr.\n <\/p>\n\n <p>\n Ce r\u00e9sultat \u2014 qui d\u00e9fie toute intuition \u2014 s’appelle le <strong>paradoxe des anniversaires<\/strong>. Et comprendre pourquoi il est vrai, c’est comprendre l’une des erreurs les plus profondes que fait notre cerveau quand il raisonne sur les probabilit\u00e9s.\n <\/p>\n\n <h2>L’erreur de notre intuition<\/h2>\n\n <p>\n Quand on entend \u00ab\u00a0deux personnes sur 30 ont le m\u00eame anniversaire parmi 365 jours possibles\u00a0\u00bb, notre cerveau fait inconsciemment ce calcul : 30 \/ 365 \u2248 8%. Tr\u00e8s peu. Donc la probabilit\u00e9 semble faible.\n <\/p>\n\n <p>\n Mais c’est la mauvaise question ! On ne demande pas si <em>quelqu’un en particulier<\/em> a le m\u00eame anniversaire que toi. On demande si <em>n’importe quelle paire<\/em> dans le groupe partage un anniversaire. Et les paires possibles dans un groupe, \u00e7a se multipie tr\u00e8s vite.\n <\/p>\n\n <div class=\"encadre\">\n <span class=\"encadre-titre\">\ud83d\udd22 Combien de paires dans un groupe ?<\/span>\n <p>Dans un groupe de <strong>n<\/strong> personnes, le nombre de paires possibles est :<\/p>\n <p style=\"text-align:center; font-family:'Playfair Display',serif; font-size:1.2rem;\">C(n,2) = n \u00d7 (n-1) \/ 2<\/p>\n <p><strong>Pour 30 \u00e9l\u00e8ves :<\/strong> 30 \u00d7 29 \/ 2 = <strong>435 paires !<\/strong><\/p>\n <p>Ce ne sont pas 30 chances de co\u00efncidence \u2014 c’est <strong>435 chances<\/strong>. C’est \u00e7a l’intuition cl\u00e9.<\/p>\n <\/div>\n\n <h2>Le calcul : par le compl\u00e9ment<\/h2>\n\n <p>\n Le calcul direct serait complexe. On utilise une astuce classique en probabilit\u00e9s : calculer la probabilit\u00e9 de l’\u00e9v\u00e9nement <strong>contraire<\/strong> \u2014 que personne ne partage d’anniversaire \u2014 puis soustraire de 1.\n <\/p>\n\n <div class=\"definition\">\n <strong>R\u00e8gle du compl\u00e9ment :<\/strong> P(au moins une co\u00efncidence) = 1 – P(aucune co\u00efncidence). Quand l’\u00e9v\u00e9nement direct est compliqu\u00e9, on calcule son contraire, souvent plus simple.\n <\/div>\n\n <p>Calculons P(aucune co\u00efncidence) pour un groupe de n personnes :<\/p>\n\n <div class=\"etape\"><span class=\"etape-numero\">\u2460<\/span> Personne 1 : son anniversaire peut \u00eatre n’importe quel jour \u2192 365\/365 = 1<\/div>\n <div class=\"etape\"><span class=\"etape-numero\">\u2461<\/span> Personne 2 : doit \u00e9viter le jour de la personne 1 \u2192 364\/365<\/div>\n <div class=\"etape\"><span class=\"etape-numero\">\u2462<\/span> Personne 3 : doit \u00e9viter les 2 jours d\u00e9j\u00e0 pris \u2192 363\/365<\/div>\n <div class=\"etape\"><span class=\"etape-numero\">\u2463<\/span> Personne n : doit \u00e9viter les (n-1) jours d\u00e9j\u00e0 pris \u2192 (365-n+1)\/365<\/div>\n\n <p>Donc :<\/p>\n\n <div class=\"formule\">\n P(aucune co\u00efncidence) = (365\/365) \u00d7 (364\/365) \u00d7 (363\/365) \u00d7 … \u00d7 (365-n+1)\/365\n <\/div>\n\n <p>Et la probabilit\u00e9 d’au moins une co\u00efncidence vaut :<\/p>\n\n <div class=\"formule\">\n P(co\u00efncidence) = 1 – P(aucune co\u00efncidence)\n <\/div>\n\n <h2>Les r\u00e9sultats qui stup\u00e9fient<\/h2>\n\n <div class=\"table-container\">\n <table>\n <tr>\n <th>Nombre de personnes<\/th>\n <th>P(au moins une co\u00efncidence)<\/th>\n <th>Niveau de surprise<\/th>\n <\/tr>\n <tr><td>5<\/td><td>2,7%<\/td><td>\ud83d\ude10 Faible<\/td><\/tr>\n <tr><td>10<\/td><td>11,7%<\/td><td>\ud83d\ude10 Faible<\/td><\/tr>\n <tr><td>15<\/td><td>25,3%<\/td><td>\ud83e\udd14 Surprenant<\/td><\/tr>\n <tr><td>20<\/td><td>41,1%<\/td><td>\ud83d\ude32 Tr\u00e8s surprenant<\/td><\/tr>\n <tr class=\"highlight\"><td><strong>23<\/strong><\/td><td><strong>50,7%<\/strong><\/td><td>\ud83c\udfaf <strong>Seuil 50% !<\/strong><\/td><\/tr>\n <tr><td>30<\/td><td>70,6%<\/td><td>\ud83d\ude31 Choquant<\/td><\/tr>\n <tr><td>40<\/td><td>89,1%<\/td><td>\ud83d\ude31 Presque certain<\/td><\/tr>\n <tr><td>50<\/td><td>97,0%<\/td><td>\ud83d\udd25 Quasi certain<\/td><\/tr>\n <tr><td>60<\/td><td>99,4%<\/td><td>\u2705 Pratiquement s\u00fbr<\/td><\/tr>\n <\/table>\n <\/div>\n\n <p>Voici une visualisation de cette progression :<\/p>\n\n <div class=\"schema-container\">\n <div class=\"schema-titre\">\ud83d\udcca Probabilit\u00e9 de co\u00efncidence selon la taille du groupe<\/div>\n <svg width=\"100%\" viewBox=\"0 0 520 300\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\">\n <rect width=\"520\" height=\"300\" fill=\"white\"\/>\n \n <line x1=\"60\" y1=\"260\" x2=\"490\" y2=\"260\" stroke=\"#0f0e0b\" stroke-width=\"2\"\/>\n <line x1=\"60\" y1=\"260\" x2=\"60\" y2=\"20\" stroke=\"#0f0e0b\" stroke-width=\"2\"\/>\n \n <text x=\"260\" y=\"290\" font-size=\"11\" fill=\"#6b6560\" text-anchor=\"middle\">Nombre de personnes<\/text>\n <text x=\"15\" y=\"145\" font-size=\"11\" fill=\"#6b6560\" text-anchor=\"middle\" transform=\"rotate(-90,15,145)\">Probabilit\u00e9 (%)<\/text>\n \n <text x=\"40\" y=\"264\" font-size=\"9\" fill=\"#6b6560\">0%<\/text>\n <text x=\"35\" y=\"214\" font-size=\"9\" fill=\"#6b6560\">25%<\/text>\n <text x=\"35\" y=\"164\" font-size=\"9\" fill=\"#6b6560\">50%<\/text>\n <text x=\"35\" y=\"114\" font-size=\"9\" fill=\"#6b6560\">75%<\/text>\n <text x=\"35\" y=\"64\" font-size=\"9\" fill=\"#6b6560\">100%<\/text>\n \n <line x1=\"60\" y1=\"214\" x2=\"490\" y2=\"214\" stroke=\"#f0ebe3\" stroke-width=\"1\"\/>\n <line x1=\"60\" y1=\"164\" x2=\"490\" y2=\"164\" stroke=\"#e8547a\" stroke-width=\"1\" stroke-dasharray=\"4,3\" opacity=\"0.5\"\/>\n <line x1=\"60\" y1=\"114\" x2=\"490\" y2=\"114\" stroke=\"#f0ebe3\" stroke-width=\"1\"\/>\n <line x1=\"60\" y1=\"64\" x2=\"490\" y2=\"64\" stroke=\"#f0ebe3\" stroke-width=\"1\"\/>\n \n <text x=\"395\" y=\"160\" font-size=\"10\" fill=\"#e8547a\">\u2190 Seuil 50%<\/text>\n \n <polyline\n points=\"60,259 100,249 140,234 180,214 207,196 220,186 250,172 280,154 310,130 340,108 370,88 400,72 430,60 460,54 490,50\"\n fill=\"none\" stroke=\"#a0254a\" stroke-width=\"3\" stroke-linejoin=\"round\"\/>\n \n <polygon\n points=\"60,260 60,259 100,249 140,234 180,214 207,196 220,186 250,172 280,154 310,130 340,108 370,88 400,72 430,60 460,54 490,50 490,260\"\n fill=\"#a0254a\" opacity=\"0.08\"\/>\n \n <circle cx=\"207\" cy=\"196\" r=\"6\" fill=\"#a0254a\"\/>\n <line x1=\"207\" y1=\"196\" x2=\"207\" y2=\"260\" stroke=\"#a0254a\" stroke-width=\"1.5\" stroke-dasharray=\"4,3\"\/>\n <text x=\"195\" y=\"275\" font-size=\"9\" fill=\"#a0254a\" font-weight=\"bold\">23<\/text>\n \n <circle cx=\"250\" cy=\"172\" r=\"6\" fill=\"#e8547a\"\/>\n <text x=\"238\" y=\"168\" font-size=\"9\" fill=\"#e8547a\">30\u219270%<\/text>\n \n <text x=\"100\" y=\"274\" font-size=\"9\" fill=\"#6b6560\">10<\/text>\n <text x=\"180\" y=\"274\" font-size=\"9\" fill=\"#6b6560\">20<\/text>\n <text x=\"260\" y=\"274\" font-size=\"9\" fill=\"#6b6560\">30<\/text>\n <text x=\"340\" y=\"274\" font-size=\"9\" fill=\"#6b6560\">40<\/text>\n <text x=\"420\" y=\"274\" font-size=\"9\" fill=\"#6b6560\">50<\/text>\n <\/svg>\n <\/div>\n\n <h2>Calculons ensemble pour 23 personnes<\/h2>\n\n <p>\n V\u00e9rifions que pour 23 personnes, la probabilit\u00e9 d\u00e9passe effectivement 50% :\n <\/p>\n\n <div class=\"etape\"><span class=\"etape-numero\">\u2460<\/span> P(aucune co\u00efncidence, 23 personnes) = (365 \u00d7 364 \u00d7 363 \u00d7 … \u00d7 343) \/ 365\u00b2\u00b3<\/div>\n <div class=\"etape\"><span class=\"etape-numero\">\u2461<\/span> = (365!\/342!) \/ 365\u00b2\u00b3<\/div>\n <div class=\"etape\"><span class=\"etape-numero\">\u2462<\/span> \u2248 0,493<\/div>\n <div class=\"etape\"><span class=\"etape-numero\">\u2463<\/span> P(co\u00efncidence) = 1 – 0,493 = <strong>0,507 = 50,7%<\/strong> \u2713<\/div>\n\n <div class=\"formule\">Avec seulement 23 personnes \u2192 P(co\u00efncidence) > 50% !<\/div>\n\n <h2>Le code Python pour visualiser<\/h2>\n\n <div class=\"code-block\">\n <span class=\"code-label\">Python<\/span>\n<span class=\"code-comment\"># Paradoxe des anniversaires – Simulation et calcul exact<\/span>\n<span class=\"code-keyword\">import<\/span> numpy <span class=\"code-keyword\">as<\/span> np\n<span class=\"code-keyword\">import<\/span> matplotlib.pyplot <span class=\"code-keyword\">as<\/span> plt\n\n<span class=\"code-comment\"># — M\u00e9thode 1 : Calcul exact —<\/span>\n<span class=\"code-keyword\">def<\/span> <span class=\"code-function\">proba_anniversaire<\/span>(n):\n <span class=\"code-string\">\u00ab\u00a0\u00a0\u00bb\u00a0\u00bbCalcule la probabilit\u00e9 exacte de co\u00efncidence pour n personnes\u00a0\u00bb\u00a0\u00bb\u00a0\u00bb<\/span>\n p_aucune = <span class=\"code-number\">1.0<\/span>\n <span class=\"code-keyword\">for<\/span> i <span class=\"code-keyword\">in<\/span> <span class=\"code-function\">range<\/span>(n):\n p_aucune *= (<span class=\"code-number\">365<\/span> – i) \/ <span class=\"code-number\">365<\/span>\n <span class=\"code-keyword\">return<\/span> <span class=\"code-number\">1<\/span> – p_aucune\n\n<span class=\"code-comment\"># Afficher les r\u00e9sultats cl\u00e9s<\/span>\n<span class=\"code-keyword\">for<\/span> n <span class=\"code-keyword\">in<\/span> [<span class=\"code-number\">10<\/span>, <span class=\"code-number\">20<\/span>, <span class=\"code-number\">23<\/span>, <span class=\"code-number\">30<\/span>, <span class=\"code-number\">50<\/span>]:\n p = <span class=\"code-function\">proba_anniversaire<\/span>(n)\n <span class=\"code-function\">print<\/span>(<span class=\"code-string\">f\u00a0\u00bbn={n:2d} \u2192 P(co\u00efncidence) = {p:.1%}\u00a0\u00bb<\/span>)\n\n<span class=\"code-comment\"># — M\u00e9thode 2 : Simulation Monte Carlo —<\/span>\n<span class=\"code-keyword\">def<\/span> <span class=\"code-function\">simulation_anniversaire<\/span>(n, nb_simulations=<span class=\"code-number\">100000<\/span>):\n <span class=\"code-string\">\u00ab\u00a0\u00a0\u00bb\u00a0\u00bbSimule le paradoxe par tirage al\u00e9atoire\u00a0\u00bb\u00a0\u00bb\u00a0\u00bb<\/span>\n coincidences = <span class=\"code-number\">0<\/span>\n <span class=\"code-keyword\">for<\/span> _ <span class=\"code-keyword\">in<\/span> <span class=\"code-function\">range<\/span>(nb_simulations):\n anniversaires = np.random.<span class=\"code-function\">randint<\/span>(<span class=\"code-number\">1<\/span>, <span class=\"code-number\">366<\/span>, n)\n <span class=\"code-keyword\">if<\/span> <span class=\"code-function\">len<\/span>(<span class=\"code-function\">set<\/span>(anniversaires)) < n:\n coincidences += <span class=\"code-number\">1<\/span>\n <span class=\"code-keyword\">return<\/span> coincidences \/ nb_simulations\n\n<span class=\"code-comment\"># V\u00e9rification par simulation pour 23 personnes<\/span>\np_sim = <span class=\"code-function\">simulation_anniversaire<\/span>(<span class=\"code-number\">23<\/span>)\n<span class=\"code-function\">print<\/span>(<span class=\"code-string\">f\u00a0\u00bb\\nSimulation pour 23 personnes : {p_sim:.1%}\u00a0\u00bb<\/span>)\n<span class=\"code-function\">print<\/span>(<span class=\"code-string\">f\u00a0\u00bbCalcul exact pour 23 personnes : {proba_anniversaire(23):.1%}\u00a0\u00bb<\/span>)\n\n<span class=\"code-comment\"># — Graphique —<\/span>\nn_valeurs = <span class=\"code-function\">range<\/span>(<span class=\"code-number\">1<\/span>, <span class=\"code-number\">61<\/span>)\nprobas = [<span class=\"code-function\">proba_anniversaire<\/span>(n) <span class=\"code-keyword\">for<\/span> n <span class=\"code-keyword\">in<\/span> n_valeurs]\n\nplt.<span class=\"code-function\">figure<\/span>(figsize=(<span class=\"code-number\">10<\/span>, <span class=\"code-number\">6<\/span>))\nplt.<span class=\"code-function\">plot<\/span>(n_valeurs, probas, color=<span class=\"code-string\">‘#a0254a’<\/span>, linewidth=<span class=\"code-number\">2.5<\/span>)\nplt.<span class=\"code-function\">axhline<\/span>(y=<span class=\"code-number\">0.5<\/span>, color=<span class=\"code-string\">‘#e8547a’<\/span>, linestyle=<span class=\"code-string\">‘–‘<\/span>, label=<span class=\"code-string\">‘Seuil 50%’<\/span>)\nplt.<span class=\"code-function\">axvline<\/span>(x=<span class=\"code-number\">23<\/span>, color=<span class=\"code-string\">‘#c8850a’<\/span>, linestyle=<span class=\"code-string\">‘–‘<\/span>, label=<span class=\"code-string\">’23 personnes’<\/span>)\nplt.<span class=\"code-function\">fill_between<\/span>(n_valeurs, probas, alpha=<span class=\"code-number\">0.1<\/span>, color=<span class=\"code-string\">‘#a0254a’<\/span>)\nplt.<span class=\"code-function\">xlabel<\/span>(<span class=\"code-string\">‘Nombre de personnes dans le groupe’<\/span>)\nplt.<span class=\"code-function\">ylabel<\/span>(<span class=\"code-string\">‘Probabilit\u00e9 d\\’au moins une co\u00efncidence’<\/span>)\nplt.<span class=\"code-function\">title<\/span>(<span class=\"code-string\">‘Paradoxe des anniversaires’<\/span>)\nplt.<span class=\"code-function\">legend<\/span>()\nplt.<span class=\"code-function\">grid<\/span>(alpha=<span class=\"code-number\">0.3<\/span>)\nplt.<span class=\"code-function\">show<\/span>()\n <\/div>\n\n <div class=\"encadre\">\n <span class=\"encadre-titre\">\ud83d\udcbb R\u00e9sultat attendu<\/span>\n <p>n=10 \u2192 P(co\u00efncidence) = 11.7%<\/p>\n <p>n=20 \u2192 P(co\u00efncidence) = 41.1%<\/p>\n <p><strong>n=23 \u2192 P(co\u00efncidence) = 50.7%<\/strong><\/p>\n <p>n=30 \u2192 P(co\u00efncidence) = 70.6%<\/p>\n <p>n=50 \u2192 P(co\u00efncidence) = 97.0%<\/p>\n <p style=\"margin-top:8px; color:#a0254a\"><strong>Simulation pour 23 personnes : ~50.7% \u2713<\/strong><\/p>\n <\/div>\n\n <h2>Tester en classe \u2014 une exp\u00e9rience vivante<\/h2>\n\n <div class=\"encadre-histoire encadre\">\n <span class=\"encadre-titre\">\ud83c\udfeb Activit\u00e9 p\u00e9dagogique pour ta classe<\/span>\n <p>\n Voici comment utiliser ce paradoxe directement en classe, que ce soit en 3\u00e8me, en Terminale ou m\u00eame en premi\u00e8re ann\u00e9e de licence :\n <\/p>\n <p>\n <strong>\u00c9tape 1 :<\/strong> Pose la question avant d’expliquer \u2014 <em>\u00ab\u00a0Qui pense que deux \u00e9l\u00e8ves dans cette salle ont le m\u00eame anniversaire ?\u00a0\u00bb<\/em> Note les r\u00e9ponses. La majorit\u00e9 dira non.\n <\/p>\n <p>\n <strong>\u00c9tape 2 :<\/strong> Fais le tour de la classe, chaque \u00e9l\u00e8ve dit son mois et jour d’anniversaire. Arr\u00eate-toi d\u00e8s qu’une co\u00efncidence appara\u00eet. Dans 70% des classes de 30, \u00e7a arrivera.\n <\/p>\n <p>\n <strong>\u00c9tape 3 :<\/strong> Explique le calcul. Regarde les visages changer. C’est ce moment de surprise \u2014 cette rupture entre intuition et r\u00e9alit\u00e9 \u2014 qui cr\u00e9e l’envie d’apprendre.\n <\/p>\n <p>\n <strong>\u00c9tape 4 :<\/strong> Introduis la r\u00e8gle du compl\u00e9ment et le concept de paires. Tu viens d’enseigner les probabilit\u00e9s de fa\u00e7on inoubliable.\n <\/p>\n <\/div>\n\n <h2>Pourquoi notre cerveau se trompe ?<\/h2>\n\n <p>\n Ce paradoxe r\u00e9v\u00e8le un biais cognitif fondamental : notre cerveau est mauvais pour \u00e9valuer les probabilit\u00e9s qui impliquent des <strong>combinaisons<\/strong>. Nous pensons de fa\u00e7on lin\u00e9aire (30 personnes = 30 chances) alors que la r\u00e9alit\u00e9 est combinatoire (30 personnes = 435 paires).\n <\/p>\n\n <p>\n Ce m\u00eame biais nous fait sous-estimer la probabilit\u00e9 d’accidents rares qui se r\u00e9p\u00e8tent, surestimer nos chances \u00e0 la loterie, et mal \u00e9valuer les risques en m\u00e9decine. Comprendre les probabilit\u00e9s, c’est apprendre \u00e0 corriger son propre cerveau.\n <\/p>\n\n <blockquote>\n L’intuition est un outil formidable \u2014 sauf en probabilit\u00e9s, o\u00f9 elle nous trahit presque syst\u00e9matiquement.\n <cite>\u2014 Le\u00e7on des probabilit\u00e9s<\/cite>\n <\/blockquote>\n\n \n <div class=\"exercice\">\n <h3>\u270f\ufe0f Exercice corrig\u00e9<\/h3>\n <p>\n Dans un lyc\u00e9e de Cotonou, une salle de r\u00e9union accueille les d\u00e9l\u00e9gu\u00e9s de classe : <strong>15 \u00e9l\u00e8ves<\/strong> en tout.\n <\/p>\n <p><strong>a)<\/strong> Combien de paires d’\u00e9l\u00e8ves peut-on former dans ce groupe ?<\/p>\n <p><strong>b)<\/strong> Calcule P(aucune co\u00efncidence d’anniversaire) pour ces 15 \u00e9l\u00e8ves. On donnera le r\u00e9sultat sous forme d\u00e9cimale arrondie \u00e0 0,001.<\/p>\n <p><strong>c)<\/strong> En d\u00e9duire P(au moins une co\u00efncidence). Est-ce surprenant ?<\/p>\n <p><strong>d)<\/strong> Si on rajoute 8 \u00e9l\u00e8ves suppl\u00e9mentaires (soit 23 au total), que devient cette probabilit\u00e9 ? Conclure.<\/p>\n\n <div class=\"correction\">\n <div class=\"correction-titre\">\u25b8 Correction<\/div>\n <p><strong>a)<\/strong> Nombre de paires = C(15,2) = 15 \u00d7 14 \/ 2 = <strong>105 paires<\/strong><\/p>\n <p><strong>b)<\/strong> P(aucune co\u00efncidence) = (365\/365) \u00d7 (364\/365) \u00d7 (363\/365) \u00d7 … \u00d7 (351\/365)<\/p>\n <p>= (365 \u00d7 364 \u00d7 363 \u00d7 … \u00d7 351) \/ 365\u00b9\u2075<\/p>\n <p>\u2248 <strong>0,747<\/strong><\/p>\n <p><strong>c)<\/strong> P(au moins une co\u00efncidence) = 1 – 0,747 = <strong>0,253 soit 25,3%<\/strong><\/p>\n <p>Oui, c’est d\u00e9j\u00e0 surprenant : 1 chance sur 4 avec seulement 15 personnes !<\/p>\n <p><strong>d)<\/strong> Pour 23 personnes : P(co\u00efncidence) = 1 – (365\u00d7364\u00d7…\u00d7343)\/365\u00b2\u00b3 \u2248 <strong>50,7%<\/strong><\/p>\n <p>En ajoutant 8 personnes, on passe de 25% \u00e0 plus de 50% \u2014 la probabilit\u00e9 a <strong>doubl\u00e9<\/strong> ! C’est l’effet explosif des combinaisons. \u2705<\/p>\n <\/div>\n <\/div>\n\n <div class=\"conclusion\">\n <h2>Ce qu’on retient<\/h2>\n <p>\n Le paradoxe des anniversaires nous enseigne trois choses essentielles. D’abord, que notre intuition probabiliste est structurellement d\u00e9faillante \u2014 et qu’il faut le savoir pour ne pas se laisser pi\u00e9ger. Ensuite, que la cl\u00e9 est souvent de compter les <em>paires<\/em>, pas les individus. Enfin, que les probabilit\u00e9s peuvent \u00eatre un outil p\u00e9dagogique puissant : rien ne capte l’attention d’un \u00e9l\u00e8ve comme un r\u00e9sultat qui contredit ce qu’il croyait \u00e9vident.\n <\/p>\n <p>\n La prochaine fois que tu entres dans une salle de 23 personnes ou plus, tu sais maintenant qu’il y a plus de 50% de chances que deux d’entre elles soufflent leurs bougies le m\u00eame jour. Les math\u00e9matiques sont partout \u2014 m\u00eame dans les g\u00e2teaux d’anniversaire.\n <\/p>\n <\/div>\n\n<\/article>\n\n<\/body>\n<\/html>\n","protected":false},"excerpt":{"rendered":"<p>Le paradoxe des anniversaires \u2014 MathsVivantes Probabilit\u00e9s \u00b7 Article #6 Le paradoxe des anniversaires :la surprise math\u00e9matique Dans une classe […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_uag_custom_page_level_css":"","site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-4)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-4)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-4)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[8],"tags":[],"class_list":["post-85","post","type-post","status-publish","format-standard","hentry","category-fondements"],"yoast_head":"\n<title>Le paradoxe des anniversaires - AfrikIntelligent<\/title>\n<meta name=\"description\" content=\"Dans une classe de 30 \u00e9leves , deux ont le m\u00eame anniversaire avec 70% de probabilit\u00e9 . 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