{"id":1715,"date":"2026-01-31T15:34:39","date_gmt":"2026-01-31T15:34:39","guid":{"rendered":"https:\/\/www.allendowney.com\/blog\/?p=1715"},"modified":"2026-01-31T18:18:45","modified_gmt":"2026-01-31T18:18:45","slug":"the-girl-born-on-tuesday","status":"publish","type":"post","link":"https:\/\/www.allendowney.com\/blog\/2026\/01\/31\/the-girl-born-on-tuesday\/","title":{"rendered":"The Girl Born on Tuesday"},"content":{"rendered":"\n<p>Some people have strong opinions about this question:<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p>In a family with two children, if at least one of the children is a girl born on Tuesday, what are the chances that both children are girls?<\/p>\n<\/blockquote>\n\n\n\n<p>In this article, I hope to offer<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>A solution to one interpretation of this question,<\/li>\n\n\n\n<li>An explanation of <em>why<\/em> the solution seems so counterintuitive,<\/li>\n\n\n\n<li>A discussion of other interpretations, and<\/li>\n\n\n\n<li>An implication of this problem for teaching and learning probability.<\/li>\n<\/ol>\n\n\n\n<p>Let\u2019s get started.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">One interpretation<\/h2>\n\n\n\n<p>One reason this problem is contentious is that it is open to multiple interpretations. I\u2019ll start by presenting just one \u2013 then we\u2019ll get back to the ambiguity.<\/p>\n\n\n\n<p>First, to avoid real-world complications, let\u2019s assume an imaginary world where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Every family has two children.<\/li>\n\n\n\n<li>50% of children are boys and 50% are girls.<\/li>\n\n\n\n<li>All days of the week are equally likely birth days.<\/li>\n\n\n\n<li>Genders and birth days are independent.<\/li>\n<\/ul>\n\n\n\n<p>Second, we will interpret the question in terms of conditional probability; that is, we\u2019ll compute <code>P(B|A)<\/code>, where<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><code>A<\/code> is \u201cat least one of the children is a girl born on Tuesday\u201d, and<\/li>\n\n\n\n<li><code>B<\/code> is \u201cboth children are girls\u201d.<\/li>\n<\/ul>\n\n\n\n<p>Under these assumptions and this interpretation, the answer is unambiguous \u2013 and it turns out to be <code>13\/27<\/code> (about 48.1%).<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">But why?<\/h2>\n\n\n\n<p>This problem is counterintuitive because it elicits confusion between causation and evidence.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If a family has a girl born on a Tuesday, that does not <em>cause<\/em> the other child to be a girl.<\/li>\n\n\n\n<li>But the fact that a family has a girl born on Tuesday is <em>evidence<\/em> that the other child is a girl.<\/li>\n<\/ul>\n\n\n\n<p>To see why, imagine two families: the first has one girl and the other has <strong>ten<\/strong> girls. Suppose I choose one of the families at random, check to see whether they have a girl born on Tuesday, and find that they do.<\/p>\n\n\n\n<p>Which family do you think I chose?<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If I chose the family with one girl, the chance is only <code>1\/7<\/code> (about 14%) that she was born on Tuesday.<\/li>\n\n\n\n<li>If I chose the family with ten girls, the chance is about 79% that at least one of them was born on a Tuesday.<\/li>\n<\/ul>\n\n\n\n<p>And that\u2019s the key to understanding the problem:<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p>A family with more than one girl is more likely to have one born on Tuesday. Therefore, if a family has a girl born on a Tuesday, it is more likely that they have more than one girl.<\/p>\n<\/blockquote>\n\n\n\n<p>That\u2019s the qualitative argument. Now we\u2019ll make it quantitative \u2013 with Bayes\u2019s Theorem.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Bayes\u2019s Theorem<\/h2>\n\n\n\n<p>Let\u2019s start with four kinds of two-child families.<\/p>\n\n\n\n<pre id=\"codecell3\" class=\"wp-block-preformatted\">kinds = ['Boy Boy', 'Boy Girl', 'Girl Boy', 'Girl Girl']<br><\/pre>\n\n\n\n<p>Under our simplifying assumptions, these combinations are equally likely, so their prior probabilities are equal. <\/p>\n\n\n\n<pre id=\"codecell5\" class=\"wp-block-preformatted\">from fractions import Fraction\n\nprior = pd.Series(Fraction(1, 4), kinds)\ndisplay(prior, 'prior')\n<\/pre>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th> <\/th><th>prior<\/th><\/tr><\/thead><tbody><tr><th>Boy Boy<\/th><td>1\/4<\/td><\/tr><tr><th>Boy Girl<\/th><td>1\/4<\/td><\/tr><tr><th>Girl Boy<\/th><td>1\/4<\/td><\/tr><tr><th>Girl Girl<\/th><td>1\/4<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><\/p>\n\n\n\n<p>Now for each kind of family, let\u2019s compute the likelihood of a girl born on Tuesday:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If there are two boys, the probability of a girl born on Tuesday is <code>0<\/code>.<\/li>\n\n\n\n<li>If there is one girl, the probability she is born on Tuesday is <code>1\/7<\/code>.<\/li>\n\n\n\n<li>If there are two girls, the probability at least one is born on Tuesday is <code>1 - (6\/7)**2<\/code>.<\/li>\n<\/ul>\n\n\n\n<p>Let\u2019s put those values in a list.<\/p>\n\n\n\n<pre id=\"codecell6\" class=\"wp-block-preformatted\">p = Fraction(1, 7)\nlikelihood = [0, p, p, 1 - (1-p)**2]\nlikelihood\n<\/pre>\n\n\n\n<pre id=\"codecell7\" class=\"wp-block-preformatted\">[0, Fraction(1, 7), Fraction(1, 7), Fraction(13, 49)]\n<\/pre>\n\n\n\n<p>To compute the posterior probabilities, we multiply the prior and likelihood, then normalize so the results add up to 1.<\/p>\n\n\n\n<pre id=\"codecell8\" class=\"wp-block-preformatted\">posterior = prior * likelihood\nposterior \/= posterior.sum()\ndisplay(posterior, 'posterior')\n<\/pre>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th> <\/th><th>posterior<\/th><\/tr><\/thead><tbody><tr><th>Boy Boy<\/th><td>0<\/td><\/tr><tr><th>Boy Girl<\/th><td>7\/27<\/td><\/tr><tr><th>Girl Boy<\/th><td>7\/27<\/td><\/tr><tr><th>Girl Girl<\/th><td>13\/27<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><\/p>\n\n\n\n<p>The posterior probability of two girls is <code>13\/27<\/code>. As always, Bayes\u2019s Theorem is the chainsaw that cuts through the knottiest problems in probability.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Other versions<\/h2>\n\n\n\n<p>Everything so far is based on the interpretation of the question as a conditional probability. But many people have pointed out that the question is ambiguous because it does not specify <em>how we learn<\/em> that the family has a girl born on a Tuesday.<\/p>\n\n\n\n<p>This objection is valid:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>The answer depends on how we get the information, and<\/li>\n\n\n\n<li>The statement of the problem does not say how.<\/li>\n<\/ol>\n\n\n\n<p>There are many versions of this problem that specify different ways you might learn that a family has a girl born on a Tuesday, and you might enjoy the challenge of solving them.<\/p>\n\n\n\n<p>In general, if we specify the process that generates the data, we can use simulation, enumeration, or Bayes\u2019s Theorem to compute the conditional probability given the data.<\/p>\n\n\n\n<p>But what should we do if the data-generating process is not uniquely specified?<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>One option is to say that the question has no answer because it is ambiguous.<\/li>\n\n\n\n<li>Another option is to specify a prior distribution of possible data-generating processes, compute the answer under each process, and apply the law of total probability.<\/li>\n<\/ul>\n\n\n\n<p>Some of the people who choose the second option also choose a prior distribution so that the answer turns out to be <code>1\/2<\/code>. In my view, that is a correct answer to one interpretation, but that interpretation seems arbitrary \u2013 by choosing different priors, we can make the answer almost anything.<\/p>\n\n\n\n<p>I prefer the interpretation I presented, because<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>I believe it is what was intended by the people who posed the problem,<\/li>\n\n\n\n<li>It is consistent with the conventional interpretation of conditional probability,<\/li>\n\n\n\n<li>It yields an answer that seems paradoxical at first, so it is an interesting problem,<\/li>\n\n\n\n<li>The apparent paradox can be resolved in a way that sheds light on conditional probability and the idea of independent events.<\/li>\n<\/ol>\n\n\n\n<p>So I think it\u2019s a perfectly good problem \u2013 it\u2019s just hard to express it unambiguously in natural language (as opposed to math notation).<\/p>\n\n\n\n<p>But you don\u2019t have to agree with me. If you prefer a different interpretation of the question, and it leads to a different answer, feel free to write a blog post about it.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">What about independence?<\/h2>\n\n\n\n<p>I think the girl born on Tuesday carries a lesson about how we teach. In introductory probability, students often learn two ways to compute the probability of a conjunction. First they learn the easy way:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><code>P(A and B) =\u00a0 P(A) P(B)<\/code><\/li>\n<\/ul>\n\n\n\n<p>But they are warned that this only applies if <code>A<\/code> and <code>B<\/code> are <em>independent<\/em>. Otherwise, they have to do it the hard way:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><code>P(A and B) =\u00a0 P(A) P(B|A)<\/code><\/li>\n<\/ul>\n\n\n\n<p>But how to we know whether <code>A<\/code> and <code>B<\/code> are independent? Formally, they are independent if<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><code>P(B|A) = P(B)<\/code><\/li>\n<\/ul>\n\n\n\n<p>So, in order to know which formula to use, you have to know <code>P(B|A)<\/code>. But if you know <code>P(B|A)<\/code>, you might as well use the second formula.<\/p>\n\n\n\n<p>Rather than check independence by conditional probability, it is more common to assert independence by intuition. For example, if we flip two coins, we have a strong intuition that the outcomes are independent. And if the coins are known to fair, this intuition is correct. But if there is any uncertainty about the probability of heads, it is not.<\/p>\n\n\n\n<p>The coin example \u2013 and <a href=\"https:\/\/allendowney.github.io\/ThinkBayes2\/chap02.html#the-monty-hall-problem\">Monty Hall<\/a>, and <a href=\"https:\/\/allendowney.substack.com\/p\/bertrands-boxes\">Bertrand\u2019s Boxes<\/a>, and many more \u2013 demonstrate the real lesson of the girl born on Tuesday \u2013 <strong>our intuition for independence is wildly unreliable<\/strong>.<\/p>\n\n\n\n<p>Which means we might want to rethink the way we teach it.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">In general<\/h2>\n\n\n\n<p>Previously I wrote about <a href=\"https:\/\/allendowney.substack.com\/p\/the-lost-chapter\">a version of this problem where the girl is named Florida<\/a>. In general, if we are given that a family has at least one girl with a particular property, and the prevalence of the property is <code>p<\/code>, we can use Bayes\u2019s Theorem to compute the probability of two girls.<\/p>\n\n\n\n<p>I\u2019ll use SymPy to represent the priors and the probability <code>p<\/code>.<\/p>\n\n\n\n<pre id=\"codecell9\" class=\"wp-block-preformatted\">from sympy import Rational\n\nprior = pd.Series(Rational(1, 4), kinds)\ndisplay(prior, 'prior')\n<\/pre>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>prior<\/th><\/tr><\/thead><tbody><tr><th>Boy Boy<\/th><td>1\/4<\/td><\/tr><tr><th>Boy Girl<\/th><td>1\/4<\/td><\/tr><tr><th>Girl Boy<\/th><td>1\/4<\/td><\/tr><tr><th>Girl Girl<\/th><td>1\/4<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><\/p>\n\n\n\n<p>Here are the likelihoods in terms of <code>p<\/code>.<\/p>\n\n\n\n<pre id=\"codecell10\" class=\"wp-block-preformatted\">from sympy import symbols\n\np = symbols('p')\n\nlikelihood = [0, p, p, 1 - (1-p)**2]\nlikelihood\n<\/pre>\n\n\n\n<pre id=\"codecell11\" class=\"wp-block-preformatted\">[0, p, p, 1 - (1 - p)**2]\n<\/pre>\n\n\n\n<p>And here are the posteriors.<\/p>\n\n\n\n<pre id=\"codecell12\" class=\"wp-block-preformatted\">posterior = prior * likelihood\nposterior \/= posterior.sum()\n\nfor kind, prob in posterior.items():\n    print(kind, prob.simplify())\n<\/pre>\n\n\n\n<pre id=\"codecell13\" class=\"wp-block-preformatted\">Boy Boy 0\nBoy Girl -1\/(p - 4)\nGirl Boy -1\/(p - 4)\nGirl Girl (p - 2)\/(p - 4)\n<\/pre>\n\n\n\n<p>So the general answer is <code>(p-2) \/ (p-4)<\/code>.<\/p>\n\n\n\n<p>If we plug in <code>p = 1\/7<\/code>, we get <code>13\/27<\/code> again.<\/p>\n\n\n\n<pre id=\"codecell14\" class=\"wp-block-preformatted\">prob = posterior['Girl Girl'].subs({p: Rational(1, 7)})\nprob\n<\/pre>\n\n\n\n<p>Or for the girl named Florida, let\u2019s assume one girl out of 1000 is named Florida.<\/p>\n\n\n\n<pre id=\"codecell15\" class=\"wp-block-preformatted\">prob = posterior['Girl Girl'].subs({p: Rational(1, 1000)})\nprob\n<\/pre>\n\n\n\n<p>The following figure shows the probability of two girls as a function of the prevalence of the property.<\/p>\n\n\n\n<pre id=\"codecell16\" class=\"wp-block-preformatted\">xs = np.linspace(0, 1)<br>ys = (xs-2) \/ (xs-4)<br><br>plt.plot(xs, ys)<br>plt.xlabel('Prevalence of the property')<br>plt.ylabel('Conditional probability of two girls')<br><\/pre>\n\n\n\n<figure class=\"wp-block-image\"><img loading=\"lazy\" decoding=\"async\" width=\"584\" height=\"432\" src=\"https:\/\/www.allendowney.com\/blog\/wp-content\/uploads\/2026\/01\/c81aa262e67d9b56ecabe5664c2397cdd0375ce23e2d2c683d8d281e36c47726.png\" alt=\"_images\/c81aa262e67d9b56ecabe5664c2397cdd0375ce23e2d2c683d8d281e36c47726.png\" class=\"wp-image-1718\" srcset=\"https:\/\/www.allendowney.com\/blog\/wp-content\/uploads\/2026\/01\/c81aa262e67d9b56ecabe5664c2397cdd0375ce23e2d2c683d8d281e36c47726.png 584w, https:\/\/www.allendowney.com\/blog\/wp-content\/uploads\/2026\/01\/c81aa262e67d9b56ecabe5664c2397cdd0375ce23e2d2c683d8d281e36c47726-300x222.png 300w, https:\/\/www.allendowney.com\/blog\/wp-content\/uploads\/2026\/01\/c81aa262e67d9b56ecabe5664c2397cdd0375ce23e2d2c683d8d281e36c47726-365x270.png 365w\" sizes=\"auto, (max-width: 584px) 100vw, 584px\" \/><\/figure>\n\n\n\n<p>If the property is rare \u2013 like the name Florida \u2013 the conditional probability is close to <code>1\/2<\/code>. If the property is common \u2013 like having a name \u2013 the conditional probability is close to <code>1\/3<\/code>.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Objections<\/h2>\n\n\n\n<p>Here are some objections to the \u201cgirl born on Tuesday\u201d problem along with my responses.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">You have to model the message, not just the event<\/h3>\n\n\n\n<p><strong>Objection.<\/strong><br>The statement \u201cat least one child is a girl born on Tuesday\u201d should not be treated as a bare event in a probability space. It should be treated as the outcome of a random process that generates <em>messages<\/em> or <em>facts we learn<\/em>. Therefore, the probability space must include not only family composition, but also the mechanism by which that information is produced. Any solution that conditions only on the family outcomes is incomplete.<\/p>\n\n\n\n<p><strong>Response.<\/strong><br>I agree that if the problem is interpreted as conditioning on a <em>message<\/em> (something that is said, reported, or chosen from among several true statements), then the reporting mechanism matters and must be modeled explicitly. However, I don\u2019t think such a mechanism is required in all cases. It is standard and meaningful to interpret a question as conditioning on an <em>event<\/em> \u2013 an extensional property of outcomes \u2013 without introducing an additional random variable for how the information was obtained. That is the interpretation I adopt here.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Without a specified selection rule, symmetry forces the answer to 1\/2<\/h3>\n\n\n\n<p><strong>Objection.<\/strong><br>If the problem does not specify how the information was obtained, then we must assume a symmetric rule for selecting which true statement is revealed. Under that assumption, conditioning on \u201cat least one boy\u201d or \u201cat least one girl\u201d must give the same answer, and applying the law of total probability forces the posterior probability to equal the prior. Therefore, the correct answer must be 1\/2.<\/p>\n\n\n\n<p><strong>Response.<\/strong><br>This conclusion follows only if we assume that the conditioning is on a <em>message<\/em> chosen from a symmetric set of alternatives. Under that interpretation, the result does depend on the selection rule, and <code>1\/2<\/code> is a valid answer for one particular choice of rule. But if the conditioning is on an <em>event<\/em> rather than a message, there is no requirement that different events form a symmetric partition or that the law of total probability be applied across them in this way. Under the event-based interpretation, the argument forcing <code>1\/2<\/code> does not apply.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">The problem is ambiguous and therefore has no answer<\/h3>\n\n\n\n<p><strong>Objection.<\/strong><br>Because the problem does not specify how we learn that there is a girl born on Tuesday, it is fundamentally ambiguous. Since different interpretations lead to different answers, the question has no single correct solution.<\/p>\n\n\n\n<p><strong>Response.<\/strong><br>It\u2019s true that the problem is ambiguous as stated in natural language. One option is to declare it unanswerable. Another is to resolve the ambiguity by adopting a conventional default interpretation. I choose the latter: I interpret the question as a conditional probability defined on an explicit probability model and make that interpretation clear by enumerating the sample space. Under that interpretation, the answer is unambiguous and, in my view, interesting and instructive \u2013 even if other interpretations lead to different answers.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">You are changing the sampling procedure<\/h3>\n\n\n\n<p><strong>Objection.<\/strong><br>Some people object that the <code>13\/27<\/code> result comes from changing how families are selected. Conditioning on \u201cat least one child is a girl born on Tuesday\u201d oversamples families with more girls, so the conditional distribution no longer represents the original population of two-child families. From this perspective, the result feels like an artifact of biased sampling rather than a genuine probability update.<\/p>\n\n\n\n<p><strong>Response.<\/strong><br>That description is accurate, but it is not a flaw. Conditioning <em>is<\/em> biased sampling: evidence changes the distribution of outcomes. Families with more girls really are more likely to satisfy the condition, and the conditional probability reflects that fact.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">The day of the week seems irrelevant<\/h3>\n\n\n\n<p><strong>Objection.<\/strong><br>Tuesday has nothing to do with gender, so it feels wrong that adding this detail should change the probability. Since the day of the week does not <em>cause<\/em> a child to be a girl, it seems irrelevant to the question.<\/p>\n\n\n\n<p><strong>Response.<\/strong><br>This objection reflects a common confusion between causal independence and evidential relevance. While the day of the week does not cause the other child\u2019s gender, it provides evidence about the number of girls in the family. Evidence can change probabilities even when there is no causal connection.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">The result depends on unrealistic independence assumptions<\/h3>\n\n\n\n<p><strong>Objection.<\/strong><br>The solution assumes that genders and days of the week are independent and uniformly distributed, which is not true in the real world. If those assumptions are relaxed, the answer changes.<\/p>\n\n\n\n<p><strong>Response.<\/strong><br>That is correct, but those assumptions are not the source of the puzzle. Relaxing them changes the numerical value of the answer, but not the underlying logic. The same kind of reasoning applies under more realistic models.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">The problem is artificial or pathological<\/h3>\n\n\n\n<p><strong>Objection.<\/strong><br>Some readers reject the problem not because the calculation is wrong, but because the setup feels artificial or unlike how information is learned in real life. From this view, the problem is a trick rather than a meaningful probability question.<\/p>\n\n\n\n<p><strong>Response.<\/strong><br>Whether this is a flaw or a feature depends on the goal. The problem is artificial, but it is intended to expose how unreliable our intuitions about conditional probability and independence can be. In that sense, its artificiality is what makes it pedagogically useful. The underlying issue \u2013 determining how evidence bears on hypotheses \u2013 comes up in real-world problems all the time. And getting it wrong has real-world consequences.<\/p>\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Some people have strong opinions about this question: In a family with two children, if at least one of the children is a girl born on Tuesday, what are the chances that both children are girls? In this article, I hope to offer Let\u2019s get started. One interpretation One reason this problem is contentious is that it is open to multiple interpretations. I\u2019ll start by presenting just one \u2013 then we\u2019ll get back to the ambiguity. First, to avoid real-world&#8230;<\/p>\n<p class=\"read-more\"><a class=\"btn btn-default\" href=\"https:\/\/www.allendowney.com\/blog\/2026\/01\/31\/the-girl-born-on-tuesday\/\"> Read More<span class=\"screen-reader-text\">  Read More<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_publicize_message":"","jetpack_publicize_feature_enabled":true,"jetpack_social_post_already_shared":true,"jetpack_social_options":{"image_generator_settings":{"template":"highway","default_image_id":0,"font":"","enabled":false},"version":2}},"categories":[1],"tags":[63,65],"class_list":["post-1715","post","type-post","status-publish","format-standard","hentry","category-uncategorized","tag-bayess-theorem","tag-probability"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.5 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>The Girl Born on Tuesday - Probably Overthinking It<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.allendowney.com\/blog\/2026\/01\/31\/the-girl-born-on-tuesday\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The Girl Born on Tuesday - Probably Overthinking It\" \/>\n<meta property=\"og:description\" content=\"Some people have strong opinions about this question: In a family with two children, if at least one of the children is a girl born on Tuesday, what are the chances that both children are girls? In this article, I hope to offer Let\u2019s get started. One interpretation One reason this problem is contentious is that it is open to multiple interpretations. I\u2019ll start by presenting just one \u2013 then we\u2019ll get back to the ambiguity. First, to avoid real-world... 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