{"id":1828,"date":"2026-06-04T19:13:19","date_gmt":"2026-06-04T19:13:19","guid":{"rendered":"https:\/\/www.allendowney.com\/blog\/?p=1828"},"modified":"2026-06-04T19:15:26","modified_gmt":"2026-06-04T19:15:26","slug":"the-frog-puzzle","status":"publish","type":"post","link":"https:\/\/www.allendowney.com\/blog\/2026\/06\/04\/the-frog-puzzle\/","title":{"rendered":"The Frog Puzzle"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Here\u2019s a probability puzzle from a TED-Ed video called <a href=\"https:\/\/www.youtube.com\/watch?v=cpwSGsb-rTs\">Can you solve the frog riddle?<\/a> by Derek Abbott. It came up recently in this <a href=\"https:\/\/www.reddit.com\/r\/probabilitytheory\/comments\/1ttzcjt\/frog_riddle_again\/\">Reddit thread<\/a>:<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">You\u2019re stranded in a rainforest after accidentally eating a poisonous mushroom. To survive the poison, you need to lick a certain species of frog. Only female frogs produce the antidote. Male and female frogs occur in equal numbers and look identical, but male frogs have a distinctive croak.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">You see one frog alone on a tree stump. In another direction, you hear the croak of a male frog coming from a clearing with two frogs. You can\u2019t tell which one made the sound.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">You only have time to go to one place. What are your chances of survival if you go to the clearing and lick both frogs? What if you go to the lone frog?<\/p>\n<\/blockquote>\n\n\n\n<p class=\"wp-block-paragraph\">The second question is relatively easy: if we assume that you are equally likely to see a male or female frog, the probability is 50% that the lone frog is female.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The first question depends on how we interpret the puzzle. In particular, it hinges on the word \u201cdistinctive\u201d \u2013 does that mean:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Only male frogs croak, and the sound is distinguishable from background noises, or<\/li>\n\n\n\n<li>Both male and female frogs croak, but the male croak is distinguishable from the female croak.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Based on the answer presented in the video, the first meaning is intended. So we\u2019ll start by solving that version.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">But the second meaning makes the problem a little harder, so we\u2019ll solve that one, too.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Only Male Frogs Croak<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">To solve the intended version of the puzzle, we\u2019ll assume<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Only male frogs croak, and<\/li>\n\n\n\n<li>When two frogs appear together, their sexes are independent.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">So we\u2019ll start with a prior where all two-frog combinations are equally likely.<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">from sympy import Rational\n\nhypo = ['FF', 'FM', 'MF', 'MM']\nprior = Rational(1)\n<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Now let\u2019s think about the likelihood of the data under each scenario. In the video, the solution is based on these assumptions:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If both frogs are female, the probability of hearing the male croak is 0.<\/li>\n\n\n\n<li>If either frog is male, the probability that one of them croaks is 1.<\/li>\n<\/ul>\n\n\n\n<pre class=\"wp-block-preformatted\">likelihood = [0, 1, 1, 1]\n<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">I\u2019ll use a <code>BayesTable<\/code> to compute the posterior probability for each scenario.<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">import pandas as pd\nimport numpy as np\n\nclass BayesTable(pd.DataFrame):\n    def __init__(self, hypo, prior=1, **options):\n        columns = ['prior', 'likelihood', 'unnorm', 'posterior']\n        super().__init__(index=hypo, columns=columns, **options)\n        self.prior = prior\n    \n    def update(self, likelihood):\n        self.likelihood = likelihood\n        self.unnorm = self.prior * self.likelihood\n        nc = self.unnorm.sum()\n        self.posterior = self.unnorm \/ nc\n<\/pre>\n\n\n\n<pre class=\"wp-block-preformatted\">table = BayesTable(hypo, prior)\ntable.update(likelihood)\ntable\n<\/pre>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>prior<\/th><th>likelihood<\/th><th>unnorm<\/th><th>posterior<\/th><\/tr><\/thead><tbody><tr><th>FF<\/th><td>1<\/td><td>0<\/td><td>0<\/td><td>0<\/td><\/tr><tr><th>FM<\/th><td>1<\/td><td>1<\/td><td>1<\/td><td>1\/3<\/td><\/tr><tr><th>MF<\/th><td>1<\/td><td>1<\/td><td>1<\/td><td>1\/3<\/td><\/tr><tr><th>MM<\/th><td>1<\/td><td>1<\/td><td>1<\/td><td>1\/3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">From the table, we can extract the posterior probability that both frogs are male.<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">from sympy import init_printing\ninit_printing(use_latex=False)\n<\/pre>\n\n\n\n<pre class=\"wp-block-preformatted\">table.posterior['MM']\n<\/pre>\n\n\n\n<pre class=\"wp-block-preformatted\">1\/3\n<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">With these assumptions, the probability 1\/3 that both frogs are male (and you die), so the probability is 2\/3 that at least one is female (and you live).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">And that\u2019s the answer in the video.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Poisson (not Poison) Frogs<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">But is that the right likelihood? Suppose frogs are equally likely to croak at any instant in time, so their croaks follow a Poisson process. If we assume that these croaking processes are independent, two frogs would be more likely to croak, during a given interval, than one.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">If the interval is much longer than the average time between croaks, the probability that either frog croaks approaches 1, which is consistent with the previous solution.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">But if the interval is short \u2013 as it might be if you were deciding whether to approach the first frog \u2013 the probability of hearing a croak would be double if there are two male frogs rather than one.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In that case, the likelihood of the data would be:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">half = Rational(1, 2)\nlikelihood = [0, half, half, 1]\n<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">And here are the posterior probabilities:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">table = BayesTable(hypo, prior)\ntable.update(likelihood)\ntable\n<\/pre>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>prior<\/th><th>likelihood<\/th><th>unnorm<\/th><th>posterior<\/th><\/tr><\/thead><tbody><tr><th>FF<\/th><td>1<\/td><td>0<\/td><td>0<\/td><td>0<\/td><\/tr><tr><th>FM<\/th><td>1<\/td><td>1\/2<\/td><td>1\/2<\/td><td>1\/4<\/td><\/tr><tr><th>MF<\/th><td>1<\/td><td>1\/2<\/td><td>1\/2<\/td><td>1\/4<\/td><\/tr><tr><th>MM<\/th><td>1<\/td><td>1<\/td><td>1<\/td><td>1\/2<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">With Poisson frogs and a short interval, the probability of two male frogs is 1\/2, so it doesn\u2019t matter whether you approach the lone frog or the pair of frogs.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Female Frogs Croak, Too<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Now let\u2019s think about the other interpretation of the puzzle: suppose both male and female frogs croak, but we can distinguish one from the other. And suppose male and female frogs croak at different rates, but they are still independent.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Assume that male frogs croak at a rate of 1 per time unit, and female frogs at a rate of <code>r<\/code> per time unit. In that case, if we start listening at a random time, the probability that we hear a male frog first is <code>1 \/ (r+1)<\/code> if there\u2019s only one male frog, and <code>1<\/code> if there are two male frogs.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So the likelihood in this case is:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">from sympy import symbols\n\nr = symbols('r')\nlikelihood = [0, 1 \/ (r+1), 1 \/ (r+1), 1]\n<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">And here are the posteriors<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">table = BayesTable(hypo, prior)\ntable.update(likelihood)\ntable\n<\/pre>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>prior<\/th><th>likelihood<\/th><th>unnorm<\/th><th>posterior<\/th><\/tr><\/thead><tbody><tr><th>FF<\/th><td>1<\/td><td>0<\/td><td>0<\/td><td>0<\/td><\/tr><tr><th>FM<\/th><td>1<\/td><td>1\/(r + 1)<\/td><td>1\/(r + 1)<\/td><td>1\/((1 + 2\/(r + 1))*(r + 1))<\/td><\/tr><tr><th>MF<\/th><td>1<\/td><td>1\/(r + 1)<\/td><td>1\/(r + 1)<\/td><td>1\/((1 + 2\/(r + 1))*(r + 1))<\/td><\/tr><tr><th>MM<\/th><td>1<\/td><td>1<\/td><td>1<\/td><td>1\/(1 + 2\/(r + 1))<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">In this scenario, here\u2019s the probability you die.<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">prob_die = table.posterior['MM']\nprob_die.simplify()\n<\/pre>\n\n\n\n<pre class=\"wp-block-preformatted\">r + 1\n\u2500\u2500\u2500\u2500\u2500\nr + 3\n<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">If female frogs don\u2019t croak, we get the same answer as in the first scenario.<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">prob_die.subs({r: 0})\n<\/pre>\n\n\n\n<pre class=\"wp-block-preformatted\">1\/3\n<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">If male and female frogs croak at the same rate, the probability that both frogs are male is 1\/2.<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">prob_die.subs({r: 1})\n<\/pre>\n\n\n\n<pre class=\"wp-block-preformatted\">1\/2\n<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">But if female frogs croak much more often, the fact that a male croaked first is strong evidence that both are male, so the posterior probability is close to 1.<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">prob_die.subs({r: 1000}).evalf()\n<\/pre>\n\n\n\n<pre class=\"wp-block-preformatted\">0.998005982053839\n<\/pre>\n\n\n\n<h2 class=\"wp-block-heading\">Assortative Mating<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Now suppose that when we see two frogs together, their sexes are not independent; specifically, let\u2019s assume that the probability of a same-sex pair is <code>p<\/code>, so the probability of a mixed-sex pair is <code>1-p<\/code>. In this scenario, the priors (before we hear the croak) are not equal.<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">p = symbols('p')\nprior = [p, 1-p, 1-p, p]\n<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Here are the posterior probabilities, assuming again that both male and female frogs, possibly at different rates.<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">likelihood = [0, 1 \/ (r+1), 1 \/ (r+1), 1]\ntable = BayesTable(hypo, prior)\ntable.update(likelihood)\ntable\n<\/pre>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>prior<\/th><th>likelihood<\/th><th>unnorm<\/th><th>posterior<\/th><\/tr><\/thead><tbody><tr><th>FF<\/th><td>p<\/td><td>0<\/td><td>0<\/td><td>0<\/td><\/tr><tr><th>FM<\/th><td>1 &#8211; p<\/td><td>1\/(r + 1)<\/td><td>(1 &#8211; p)\/(r + 1)<\/td><td>(1 &#8211; p)\/((p + 2*(1 &#8211; p)\/(r + 1))*(r + 1))<\/td><\/tr><tr><th>MF<\/th><td>1 &#8211; p<\/td><td>1\/(r + 1)<\/td><td>(1 &#8211; p)\/(r + 1)<\/td><td>(1 &#8211; p)\/((p + 2*(1 &#8211; p)\/(r + 1))*(r + 1))<\/td><\/tr><tr><th>MM<\/th><td>p<\/td><td>1<\/td><td>p<\/td><td>p\/(p + 2*(1 &#8211; p)\/(r + 1))<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<pre class=\"wp-block-preformatted\">table.posterior['MM'].simplify()\n<\/pre>\n\n\n\n<pre class=\"wp-block-preformatted\"> p\u22c5(r + 1) \n\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\u2500\np\u22c5r - p + 2\n<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">If <code>p=1\/2<\/code>, this simplifies to the previous scenario.<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">table.posterior['MM'].subs({p: half}).simplify()\n<\/pre>\n\n\n\n<pre class=\"wp-block-preformatted\">r + 1\n\u2500\u2500\u2500\u2500\u2500\nr + 3\n<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">And if <code>r=0<\/code> (female frogs don\u2019t croak), we get the answer presented in the video.<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">table.posterior['MM'].subs({p: half, r: 0}).simplify()\n<\/pre>\n\n\n\n<pre class=\"wp-block-preformatted\">1\/3\n<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">But depending on the assumptions, the probability can be as low as <code>0<\/code><\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">table.posterior['MM'].subs({p: 0, r: 1}).simplify()\n<\/pre>\n\n\n\n<pre class=\"wp-block-preformatted\">0\n<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Or as high as <code>1<\/code>.<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">table.posterior['MM'].subs({p: 1, r: 0}).simplify()\n<\/pre>\n\n\n\n<pre class=\"wp-block-preformatted\">1\n<\/pre>\n\n\n\n<p class=\"wp-block-paragraph\">Or anything in between. As is often the case with problems like these, the answer depends on a precise specification of the data-generating process.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Discussion<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">If all of this seems like more trouble than it\u2019s worth, let me suggest a metacognitive shortcut for solving puzzles like this.<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Notice that in all probability puzzles, the answer is either 1\/2 or 1\/3.<\/li>\n\n\n\n<li>Also, the answer is always counterintuitive; otherwise it wouldn\u2019t be a puzzle.<\/li>\n\n\n\n<li>Therefore, if your intuition says the answer is 1\/2, it\u2019s actually 1\/3, and vice versa.<\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\">That might save you some time.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This notebook uses methods and materials from <em>Think Bayes<\/em>, second edition. If you like this sort of thing, you can read the whole book, and more examples, at <a href=\"https:\/\/allendowney.github.io\/ThinkBayes2\/\">allendowney.github.io\/ThinkBayes2\/<\/a>.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><br><br><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Here\u2019s a probability puzzle from a TED-Ed video called Can you solve the frog riddle? by Derek Abbott. It came up recently in this Reddit thread: You\u2019re stranded in a rainforest after accidentally eating a poisonous mushroom. To survive the poison, you need to lick a certain species of frog. Only female frogs produce the antidote. Male and female frogs occur in equal numbers and look identical, but male frogs have a distinctive croak. You see one frog alone on&#8230;<\/p>\n<p class=\"read-more\"><a class=\"btn btn-default\" href=\"https:\/\/www.allendowney.com\/blog\/2026\/06\/04\/the-frog-puzzle\/\"> 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_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_feature_clip_id":0,"_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},"jetpack_post_was_ever_published":false},"categories":[1],"tags":[63,131,65],"class_list":["post-1828","post","type-post","status-publish","format-standard","hentry","category-uncategorized","tag-bayess-theorem","tag-frog-puzzle","tag-probability"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v28.3 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>The Frog Puzzle - 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\/06\/04\/the-frog-puzzle\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"The Frog Puzzle - Probably Overthinking It\" \/>\n<meta property=\"og:description\" content=\"Here\u2019s a probability puzzle from a TED-Ed video called Can you solve the frog riddle? by Derek Abbott. It came up recently in this Reddit thread: You\u2019re stranded in a rainforest after accidentally eating a poisonous mushroom. To survive the poison, you need to lick a certain species of frog. Only female frogs produce the antidote. Male and female frogs occur in equal numbers and look identical, but male frogs have a distinctive croak. You see one frog alone on... 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