## Probability Puzzles

Today Greg Egan mailed me two puzzles in probability theory: a “simple” one, and a more complicated one that compares Bayesian and frequentist interpretations of probability theory.

Try your hand at the simple one first. Egan wrote:

A few months ago I read about a very simple but fun probability puzzle. Someone tells you:

“I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?”

Please give it a try before moving on. Or at least figure out what this is:

Of course, your first reaction should be “it’s irrelevant the boy was born on a Tuesday“. At least that was my first reaction. So I said:

I’d intuitively assume that the day Tuesday is not relevant, so I’d ignore that information – or else look at some hospital statistics to see if it is relevant. I’d also assume that boy/girl births act just like independently distributed fair coin flips — which is surely false, but I’m guessing the puzzle wants us to assume it’s true. And then I’d say there are 4 equally likely options: BB, BG, GB and GG.

If you tell me “one is a boy”, it’s very different from “the first one is a boy”. If one is a boy, we’re down to 3 equally likely options: BB, BG, and GB. So, the probability of two boys is 1/3.

But that’s not the answer Egan gives:

The usual answer to this puzzle — after people get over an initial intuitive sense that the “Tuesday” can’t possibly be relevant — is that the probability of having two sons is 13/27. If someone has two children, for each there are 14 possibilities as to boy/girl and weekday of birth, so if at least one child is a son born on a Tuesday there are 14 + 14 – 1 = 27 possibilities (subtracting 1 for the doubly-counted intersection, where both children are sons born on a Tuesday), of which 7 + 7 – 1 = 13 involve two sons.

If you find that answer unbelievable, read his essay! He does a good job of making it more intuitive:

• Greg Egan, Some thoughts on Tuesday’s child.

But then comes his deeper puzzle, or question:

That’s fine, but as a frequentist if someone asks me to take this probability seriously and start making bets, I will only do so if I can imagine some repetition of the experiment. Suppose someone offered me $81 if the parent had two sons, but I had to pay$54 if they had a son and a daughter. The expected gain from that bet for P(two sons)=13/27 would be $11. If I took up that bet, I would then resolve that in the future I’d only take the same bet again if the person each time had two children and at least one son born specifically on a TUESDAY. In fact, I’d insist on asking the parent myself “Do you have at least one son born on a Tuesday?” rather than having them volunteer the information (since someone with two sons born on different days might not mention the one born on a Tuesday). That way, I’d be sampling a subset of parents all meeting exactly the same conditions, and I’d be satisfied that my long-term expectation of gain really would be$11 per bet.

But I’m curious as to how a Bayesian, who is happier to think of a probability applying to a single event in isolation, would respond to the same situation. It seems to me (perhaps naively) that a Bayesian ought to be happy to take this bet any time, and then forget about what they did in the past — which ought to make them willing to take the bet on future offers even when the day of the week when the son was born changes. After all, P(two sons)=13/27 whatever day is substituted for Tuesday.

However, anyone who agreed to keep taking the bet regardless of the day of the week would lose money! Without pinning down the day to a particular choice, you’re betting on a sample of parents who simply have two children, at least one of whom is a son. That gives P(two sons)=1/3, and the expectation for the $81/$54 bet becomes a $9 loss. Now, I understand how the difference between P(two sons)=13/27 and P(two sons)=1/3 arises, despite the perfect symmetry between the weekdays; the subsets with “at least one son born on day X” are not disjoint, so even though they are isomorphic, their union will have a different proportion of two-son families than the individual subsets. What’s puzzling me is this: how does a Bayesian reason about the thought experiment I’ve described, in such a way that they don’t end up taking the bet every time and losing money? ### 105 Responses to Probability Puzzles 1. Robert Smart says: The hypotheses are GG, GB, BG and BB, each has prior probability 1/4. In the initial question the probability of result given the hypotheses are 0, 7/49, 7/49 and 13/49 [1/7 converted to 7/49]. Bayes theorem then gives the 13/(0+7+7+13). To take a bet on the probability of a result maybe you need to define the result before looking at the evidence (though Bayes is very flexible if you can think clearly about hypotheses and prior probabilities and probability of result given hypothesis). Obviously I don’t know much, but by some coincidence the author of a great book about saving the world (SEWTHA: Sustainable Energy: Without the Hot Air) that is freely available on the Internet is also the author of a great book covering Bayesian statistics and other related stuff: “Information Theory, Inference, and Learning Algorithms” that is also freely available: http://www.inference.phy.cam.ac.uk/mackay/itila/. 2. bane says: This response isn’t anything about Bayesian inference so much as the “protocol” aspects of the problem. Bringing in hopefully useful explicitness, Greg appears to be asking 1. Given forall d in days . P(2 sons | query reveals has son born on d)=13/27 2. wouldn’t you then think P (2 sons | exists d in days . query reveals has a son born on d) = 13/27 ? (The odd form of 2 is a way of saying that someone tells you they have a son born on day d without d being specified in advance.) You’d only take the bet on the non-day-prespecified protocol if you think 2 is correct. Greg is suggesting that knowledge of 1 implies 2, but I don’t see it. The question is what valid logical manipulation on the quantifier in 1 would make you think it was equivalent to 2? • bane says: To clear up an unfortunate turn of phrase, it should be “Greg is suggesting that someone applying non-frequentist reasoning would think knowledge of 1 implies 2″; clearly Greg himself is aware this isn’t so. 3. Philip Gibbs says: These kind of conclusions are wrong because you are making unwarranted assumptions about what else the person may or may not have said depending on what children he has. To see this consider the simple case of a parent with two children in two controlled experiments. For the first experiment instruct the parent to approach someone and tell them “I have a boy” if and only if it is true, otherwise just say hello. In this experiment the person can conclude that the probability that the parent has two boys is 1/3 assuming he knows the instructions. For the second experiment instruct the parent to say either “I have a boy” or “I have a girl”. He can say either provided it is true and should toss a coin to decide which to say if both are true. In this case the conclusion is different. If the person spoken to knows the rules then when he is told that the parent has a boy the probability of him having two boys is now 1/2. Similar arguments apply to the case of “I have a boy born on a Tuesday” Unless you know what the parent may or may not have said in various circumstances you cannot evaluate the probabilities. It is partly because you are throwing in other assumptions that would not be justified in real life that you get counterintuitive answers. • Philip Gibbs says: A search shows that this puzzle has been circulated widely by Gary Foshee, and points similar to mine have of course been made before. e.g. at http://news.bbc.co.uk/1/hi/programmes/more_or_less/8735812.stm The points Greg Egan makes about the different probablity models may be more novel and worthy of further discussion, but the hidden assumptions need to be understood first I think. • John Suppa says: In the second experiment if we are always assuming the parent has 2 children as the puzzle stated to begin with then the probability is not 1/2 because if we assume the parent can have any of the following: b/b b/g or g/g then the parent will have flipped a coin only 1/3 and of those 1/3 if we can in this case assume the coin flip is 50/50 then in a sample size of 300 (to make this clean) with an equal amount of b/b b/g g/g then 100 parents will flip a coin first and 50 times the parent will say “I have a boy” and 50 times will say “I have a girl” and if we can assume that 100 parents have b/b and 100 parents have g/g then we can know that 150 out of 300 parents will say “I have a boy” because clearly we can discount g/g which leaves 200 parents, 100 with b/b and 100 with b/g. 100 of the 100 b/b parents will say “I have a boy” while 50 of the b/g parents will say “I have a boy” Therefor of 200 parents, 150 will say “I have a boy” while only 50 will say “I have a girl” Now since the puzzle began with this person having 2 children we’ll keep it that way and not account for other possibilities (1 boy, 1 girl or 0 children) so if this parent in your scenario says “I have a boy” then we’re left with only these possibilities. He has 2 boys or he has a boy and a girl and he flipped a coin to determine what he would say. Under these guidelines we now know that this parent can only have 2 boys or a boy and a girl. You can keep the remaining possibilities at a sample size of 200 parents or we can expand it to 2 million, either way under the scenario that you put forth the probability of the parent’s second child also being a boy is 2/3 Here’s why, if the parent (P) had 2 girls P would have under your scenario said “I have a girl” if P had a boy and a girl then P would have flipped a coin, if we just assume that 50 out of 100 times P would get a coin flip result of “I have a boy” and we use a sample size of 300 P (100 b/b 100 b/g 100 g/g) then we know that “I have a boy” means 1 of 2 scenarios, the first is that P has b/b in which case no coin flip is necessary and the chances of the second child being a boy is 100/100 the other scenario is that P flipped a coin and the result was to say “I have a boy” so the chances of the second child being a boy in that scenario is 0/100. Sounds like 1/2 however of those 150 P who would say “I have a boy” only 50 or 1/3 will actually have b/g and 100 or 2/3 will have b/b. So the answer to scenario 2 is that if the person spoken to knows the rules as you put them then when he is told that the parent has a boy the probability of him having 2 boys is now 2/3 • PalmerEldritch says: In a sample size of 300 families 150 families (not 100) will have a boy and girl. Redo your figures and the answer comes out as 1/2 4. David Pollard says: It’s an Orwellian world. Looking at Greg Egan’s drawings it’s easy to see that one of the 13/27 squares is double-plus-uncounted. Simply treat the vertical and horizontal groups, first- and second-born, as distinct with equal probability and the sanity or 14/28 returns. Meanwhile, there appears to be a ghastly real-world problem in Fallujah, where birth sex-ratios are way from normal. Fairly certainly the cause is something other than ionising radiation, for which widespread claims are being made. However the likely cause, environmental toxins, is no less alarming. http://www.independent.co.uk/news/world/middle-east/toxic-legacy-of-us-assault-on-fallujah-worse-than-hiroshima-2034065.html http://forums.theregister.co.uk/forum/1/2010/08/19/iran_defacers_hit_genetics_website/ • John Baez says: So you think the ‘sane’ answer to the original puzzle is 1/2, not 1/3 or 13/27? Interesting. • David Pollard says: Rather than saying the answer is definitely 1/2, I was hoping to point out the danger of following a red herring. Nevertheless, when someone says, “One is a boy …” this, in everyday use of language, can be taken to imply that the other isn’t a boy. Otherwise the sentence would have begun, “One of the boys is …” The implication isn’t quite as strong as the implication that “I have two children” means *only* two children and not, say, two children (who have with red hair) but it’s still there. 5. Greg Egan says: bane: I think what you’re saying is right, and in fact in the essay I conclude that 13/27 is the correct probability only in a context where you can pre-specify a particular day. But it’s still not clear to me how a Bayesian would avoid the conclusion: P(2 sons | parent volunteers “I have two children, one a son born on a Tuesday”) = 13/27 even when this is just a chance encounter between people out in the world, and no one has pre-specified anything. Using the usual probability update rules: Let hypothesis H=”parent has two sons”. Prior probability P(H)=1/4 [which should be uncontroversial when you only know “parent has two children”]. New data D=”parent has a son born on a Tuesday”. Marginal probability P(D)=27/196 [from the 27 red-bordered squares in the 3rd diagram in the essay]. P(D|H) =13/49 [from the 13 red-bordered squares within the “two sons” blue square in the same diagram]. P(H|D) = P(D|H) P(H) / P(D) = (13/49) (1/4) (196/27) = 13/27 In the essay I do suggest that in most contexts it’s appropriate to ignore the “Tuesday” part and conclude that P(H)=1/3, but I still can’t see how to justify that without invoking a kind of frequentist embedding of the event in a series of repetitions. So I suppose my question now is: when exactly are you allowed to ignore data when performing a Bayesian update of the likelihood of a hypothesis? Or have I misapplied the rules somehow, and would the correct application detect the irrelevance of “Tuesday” automatically? • Dmitry says: May I ask for clarification (about your Bayesian scenario)? Imagine a line of thousands of parents each having exactly 2 children. You ask every one of them one by one: “Do you have at least one son born on an X?” where X is some particular weekday. You choose this weekday for every parent separately by a 7-sided dice roll. If the answer is “Yes” then you bet according to P=13/27 probability. Are you saying that this would lead to a money loss? • Dmitry says: See below, I elaborated on my point there. • John F says: Philip Gibbs is right that if a Bayesian can know the assumptions, then he can calculate the probabilities correctly. These will then be the exact same probabilities that a frequentist will calculate. There is provably always exactly zero difference between a correct Bayesian and a correct frequentist calculation. The only time there can be difference between the two interpretations is that, for some situations, there is a larger probability that a Bayesian will report an incorrect answer (because the formula permits calculation based on wrong assumptions, so it can encourage “filling in the blanks”) while a frequentist will not report any answer (he refused to count because he didn’t have complete information). It is perhaps of some psychological interest that a frequentist, while counting all possibilities, doesn’t think he is being Bayesian, but the forms of self-deception are many. And not limited to Bayesians. 6. Guy Srinivasan says: His essay gets it wrong, forgetting that Eve asked Anna about Robert in particular. Instead of 13/27, we want P(two boys|Robert is first)*P(Robert is first) + P(two boys|Robert is second)*P(Robert is second), which assuming Robert is equally likely to be first or second is 7/14*1/2+7/14*1/2=1/2. The question he sent you is a little different, and notice how much it matters exactly what the context is. Suppose an oracle presented you with all parents with exactly two children where at least one of which is a boy born on a Tuesday. Then you should make the bet over and over. Suppose then an oracle presented you with all parents with exactly two children where at least one of which is a boy born on a Wednesday. Make the bets, they’re great! etc. But notice that every distinct-day set of boys will come up *twice*, while every duplicate-day set of boys or boy-girl set will come up *once*. Does this translate to the real world? Will Eve* encounter Anna twice if she has boys born on different days, forgetting the first time? Hardly. So if you only have a Tuesday oracle then make the bets, if you have (say) a SMTWThFS oracle that presents you with all parents with exactly two children where at least one is a boy born on a day and if both are boys picks randomly which day to tell you then don’t make the bets, etc. etc. How do we avoid losing money? One way is to do it Egan’s way, precommit to having nothing but the Tuesday oracle. You don’t need to commit to Tuesday, though, you could also use this algorithm when meeting someone new: 1. Ask if they have exactly two children. If yes, 2. Ask if at least one is a boy. If yes, 3. Pick a random day and ask if at least one was born on that day. If yes, 4. Bet your coworker. If any were no, don’t bet, but you don’t get to sample that family again. Of course if parents are coming up to you and volunteering that they have exactly two children at least one of which is a boy born on a Tuesday and asking you to bet? You should probably run. • Greg Egan says: You’re right about the Robert thing. Thanks, I’ve fixed it now. Attaching a name to the son born on a Tuesday is useful, but the original way I did this altered the problem, as you point out. 7. While I don’t have a solution to the Bayesian problem, I propose the following simplification that gets rid of the days and only asks about boys and girls. (The day thing is just an additional layer of the same paradox.) Namely, the procedure is the following: we have a person X with two children. hypothesis H = the two children have the same sex data Db = person X says that one person is a boy data Dg = person X says that one person is a girl Given a sequence of persons that tell you either Db or Dg, should you make a bet on hypothesis H? 8. It’s a lot of gibberish. The question states that one child is a boy and asks what the probability is that both children are boys. In the answer, it’s stated that “And then I’d say there are 4 equally likely options: BB, BG, GB and GG” This is false. One child is a boy, therefore GG is *not* an option. Moreover, it’s *either* BG or GB but not both, therefore the correct options are BB or BG/GB. The probability is therefore 50%. The day of birth is irrelevant. • John Baez says: So you too think the answer is 1/2. Interesting. Let’s see if I understand your reasoning. We’ll ignore the bit about ‘Tuesday’ since it’s irrelevant (most people here disagree, but never mind). Since GG is ruled out, that leaves three equally likely options: BB, BG, and GB. So, the chance that it’s BB is 1/2. • OK, there are two children, each of whom may be either a girl or a boy. We know that one of the children is a boy, and we are asked what the probability is that both children are boys is. If we put down options, we have: Child One Child Two B B and or B G The remaining child may be either a boy or a girl, and all things being equal, it makes sense that the probability of the child being a boy is 50% and the probability of the child being a girl is 50%. The options, having ruled out GG, since one of the children is a boy, are BB or BG, not BB or BG or GB. “GB” is a false choice, it states a fact which is not in evidence, that Child One is a girl, G, when Child One has been previously identified as a boy, B. The only question remaining is the gender of Child Two, and the chances of B or G are equal, thus P(Child One|B)*P(Child Two|B) + P(Child One|B)*P(Child Two|G) = 1.0 and given that P(Child One|B) = 1.0 and that P(Child One|B) = P(Child Two|B), then the only choice for P(Child One|B)*P(Child Two|B) is 0.5, or 50%, or 1/2. That’s my reasoning. And it is irrelevant that Child One was born on a Tuesday, or under a full moon, or on a day not February 29 of a leap year. • John Baez says: streamfortyseven wrote: Child One has been previously identified as a boy, B. Okay, you’re interpreting the puzzle differently from how Greg intended it. It’s not your fault: the wording is ambiguous. You’re solving this puzzle: “I have two children, Child One and Child Two, and Child One is a boy. What is the probability that both are boys?” The answer to this is indeed 1/2, as you say. And adding the information that Child One was born on a Tuesday would not change the answer. Greg was starting from a different puzzle: “I have two children, and at least one is a boy. What is the probability that both are boys?” Now the answer is 1/3. And now adding the information that at least one is a boy born on a Tuesday changes the answer to 13/27. When you earlier wrote the correct options are BB or BG/GB. The probability is therefore 50%. I thought you were trying to solve Greg’s intended version of the puzzle, where BB, BG and GB are all options that must be considered. So, I had trouble seeing how you got the answer 1/2 instead of 1/3. 9. John F says: I’ve tried to figure out what “this” is. It’s a picture of 400 x 5310 pixels, probably (Bayesian?) a portion of a larger image originally in 12 bit gray. That leads me to believe (ahem) that it was some real data. This data has been resampled and lightly compressed, now in jpg format. It looks like a topographical map of something squished flat in the general direction to the right, but then again so do a lot of things. My gut feeling is it is either a portion of a scan of the surface of a real object (scale I can’t choose, maybe radar of Mars or electron microscope of paint), or generated data with 2d correlations. • John Baez says: Excellent — you tried to solve the third puzzle! It’s indeed part of Mars, but seen in visible light, not radar. Despite what the real estate agents say, not all of Mars is red! This is a region near the south pole, which is covered with ‘dry ice’ — frozen carbon dioxide. For some mysterious reason this dry ice forms a wide variety of baroque patterns. This is just one of many beautiful pictures of Mars taken by HiRISE (the High Resolution Imaging Science Experiments). For example: • South polar residual cap monitoring: rare stratigraphic contacts, http://hirise.lpl.arizona.edu/ESP_014379_0925 • Evolution of the south polar residual cap, http://hirise.lpl.arizona.edu/PSP_004687_0930 • South polar carbon dioxide ice cap, http://hirise.lpl.arizona.edu/ESP_014261_0930 The picture in my blog entry is the last of these. Patrick Rusell wrote this description of it: This HiRISE image is of a portion of Mars’ south polar residual ice cap. Like Earth, Mars has concentrations of water ice at both poles. Because Mars is so much colder, however, the seasonal ice that gets deposited at high latitudes in the winter and is removed in the spring (generally analogous to winter-time snow on Earth) is actually carbon dioxide ice. Around the south pole there are areas of this carbon dioxide ice that do not disappear every spring, but rather survive winter after winter. This persistent carbon dioxide ice is called the south polar residual cap, and is what we are looking at in this HiRISE image. Relatively high-standing smooth material is broken up by semi-circular depressions and linear, branching troughs that make a pattern resembling those of your fingerprints. The high-standing areas are thicknesses of several meters of carbon dioxide ice. The depressions and troughs are thought to be caused by the removal of carbon dioxide ice by sublimation (the change of a material from solid directly to gas). HiRISE is observing this carbon dioxide terrain to try to determine how these patterns develop and how fast the depressions and troughs grow. While the south polar residual cap as a whole is present every year, there are certainly changes taking place within it. With the high resolution of HiRISE, we intend to measure the amount of expansion of the depressions over multiple Mars years. Knowing the amount of carbon dioxide removed can give us an idea of the atmospheric, weather, and climate conditions over the course of a year. In addition, looking for where carbon dioxide ice might be being deposited on top of this terrain may help us understand if there is any net loss or accumulation of the carbon-dioxide ice over time, which would be a good indicator of whether Mars’ climate is in the process of changing or not. 10. Greg Egan says: Ah, thanks to Philip Gibbs and bane, I think I’ve finally resolved everything that was puzzling me! If we’re starting from a pool of equal numbers of BB,BG,GB, and we specify that a parent with two boys will toss a coin to decide which one’s birth day to spontaneously disclose (with the others always disclosing the birth day of their sole son) then the data: D=Parent says “I have a son born on a Tuesday” has the conditional probabilities: P(D|BG) = P(D|GB) = 1/7 for the sole son P(D|BB) = (1/2)x(1/7) + (1/2)x(1/7) = 1/7 where the (1/2)s are for the coin landing heads or tails. This makes it pretty obvious that a spontaneous disclosure made in this way will leave P(BB) at 1/3; if we work through the mechanics of a formal Bayesian update based on the disclosure, P(D) and P(D|BB) are equal, so P(BB|D)=P(BB)=1/3. In contrast to this, answering “yes” to the predetermined question “Do you have at least one son born on a Tuesday?” gives P(BB|yes) of 13/27, as previously calculated. Thanks for all the helpful comments. • John F says: Greg, here is another sort of puzzle. My son came home a few years ago with his first probability homework in elementary school. Anyway they were doing Venn diagrams, and since independence is a fundamental concept, a couple of problems involved Venn diagrams of indepedent events. They had to move around like a red square and a blue circle to precisely overlap. Of course that is the wrong way to do it. The Venn diagram of independent red events in a red circle and blue events in a blue square is the color purple. The puzzle is – what is the shape? • bane says: I’m going to be very nitpicky: Greg has definitely figured out the kernel of the problem, and I think is completely correct. However, it’s always tricky to reason about probability when an “observation” is used twice (in this case initially “I have a son” to exclude GG, then “I have a son who was born on a Tuesday”). So here’s a slightly rejigged argument (using the same spontaneous disclosure protocol): Writing the complement of a set A as A^c, an equivalent form of Bayes theorem (that I’m more familiar with) is P(D|S) = P(D|S)P(S) / [P(D|S)P(S) + P(D|S^c)P(S^c) Then having S be “I have 2 sons” (with obviously P(S)=1/4, P(S^c)=3/4) and D be disclosure “I have a son who was born on a Tuesday”, then utilising the P(BB|D) = [P(D|BB)*1/4] / [P(D|BB)*1/4 + P(D| BB^c)*3/4)] =[1/7*1/4] / [1/7*1/4 + 2/3*1/7*3/4)] = 1/3 The fact that P(D|BB^C)=P(D|one of BG,GB,GG)=2/3*1/7 is just saying that in two of the three cases there’s a 1/3 probability, in GG case there’s a zero probability. Again, this is exactly Greg’s argument, just rejigged to treat proposition D as “atomic”. • bane says: D’oh: the LHS of first eqn should be P(S|D). • Scott says: Indeed, the key is to keep the gender of the child independent of the birth day. If you ask the boy’s day of birth after you already know that there is a boy then the random variables are not independent. 11. Tom Leinster says: I see a spectrum of possible questions: 1. “I have two children. One is a boy. What is the probability that both are boys?” 2. “I have two children. One is a boy born on a Tuesday. What is the probability that both are boys?” n. “I have two children. One is a boy born on Tuesday 4 February 1947 in Indianapolis, who grew up in Arizona, and who is married to a woman called Marilyn. What is the probability that both are boys?” The answer to the first question is 1/3. The last question is equivalent, I suspect, to: “What is the probability that Dan Quayle’s unique sibling is male?” If this means anything at all, the answer is 1/2. So, I’d intuitively expect the answers to the intermediate questions 2, 3, …, (n – 1) to be between 1/3 and 1/2, increasing with each question. From that point of view, it doesn’t surprise me that the answer to question 2 is between 1/3 and 1/2. (Disclaimer: I have no idea how many siblings Dan Quayle has. But he was born on a Tuesday.) • Philip Gibbs says: If Dan Quayle had a unique brother the probability would be 1. If he had a unique sister the probability would be 0. Actually he has a sister and a brother. :) • Tom Leinster says: OK :-) But let’s ignore your knowledge that he has two siblings. I tried to pick someone not too famous, so that no one would know off that information off the tops of their heads. I should have known that someone would either know or find out. We could alter the question by talking about Dan Quayle’s oldest sibling. But let’s keep things simple and pretend he has only one. I wrote “If this means anything at all” for exactly the reason that you mention in your first two setences. From a frequentist perspective, the question is certainly meaningless, as meaningless as the question “what is the probability that so-and-so wins the next election?” But there is a straightforward Bayesian interpretation — the interpretation that comes to everyone’s minds — and the answer then is 1/2. • Vladimir says: Tom Leinster: I see a spectrum of possible questions: The entire spectrum can be stated as: “I have two children. One is a boy with a property X. Each boy in the entire population has the property X with the probability p, independent of others.” Then the probability that the other sibling is a boy will be (call it P): P = 1 – 2p/(4p – p2) If X = “born on Tuesday,” then p=1/7 and P=13/27. The extreme cases 1 and N from your comment correspond to p=1 (every boy has the property X), in which case P=1/3, and the limit for p → 0, so that p2 << p and thus P ~ 1/2. (To be fully precise, the case where you state the exact identity of the boy, rather than just some rare property that's unlikely to be shared by any other member of the population, doesn't fit this model since the property "being the particular individual A" is not independent across samples with multiple individuals — if the first sibling has it, it's impossible for the other one to have it too.) 12. bane says: It seems to be agreed in this thread that the “disclosure protocol” matters. With that in mind, what’s the answer to the question if the parent makes a spontaneous declaration? 1. If the parent will always tell you about a boy (if they have one), and will flip a coin to decide which one to tell you about if they have two boys, if they say “One is a boy born on a Tuesday” the probability they have two boys comes out at 1/3. 2. If the parent picks a child by a coin flip, then tells you about that child’s gender and birth day then, if they say “One is a boy born on a Tuesday” the probability they have two boys comes out at 1/2. (These have been calculated via Bayes theorem but more reliably verified by a computer doing exhaustive elimination.) Obviously 1 and 2 are different protocols, but arguably 2 is a marginally more “natural” assumption in treating both gender and birth day equally, although that’s highly debatable. [I’m sure this point has been made in other internet discussions of Forshee’s question; I haven’t bothered to look at them.] 13. John Baez says: Bane wrote: It seems to be agreed in this thread that the “disclosure protocol” matters. Yes, it seems that like all really confounding puzzles, there’s a tricky interaction of subtleties. (Think of Newcomb’s paradox, for example.) Let me say some stuff that’s obvious to everyone who has thought about this seriously, but still deserves saying. When I first read the puzzle I didn’t think at all about the fact that someone was saying “I have a boy born on a Tuesday”. I assumed that was just the puzzle-teller’s way of introducing the assumption that we know this person has a boy born on a Tuesday. So I thought we should compute P(two boys | two children, at least one being a boy born on a Tuesday) And then Greg’s answer, 13/27, is clearly correct, though initially shocking if you know P(two boys | two children, at least one being a boy) = 2/3 especially after you’ve painstakingly disabused yourself of the common misimpression that this equals 1/2, which seems to arise from the fact that P(two boys | two children, the elder being a boy) = 1/2 and also P(two boys | two children, the younger being a boy) = 1/2 But you’re right, the puzzle invites us to dig into another layer of subtlety: taking the puzzle literally, all we know is that someone says “I have a boy born on a Tuesday”. If we take this seriously, we have to worry about things like P(they say “I have a boy born on a Tuesday” | they have a boy born on a Tuesday) and P(they say “I have a boy born on a Wednesday” | they have a boy born on a Wednesday) and P(they say “I have a girl born on a Tuesday” | they have a girl born on a Tuesday) and so on. Question abound. Does the speaker only pipe up if they have a boy born on Tuesday, or do they spontaneously reveal the gender and/or birth day of one child to anyone they meet, regardless of the values of those quantities? Are they perhaps liars, or do they come from a tribe where it’s deeply shameful to have a daughter born on a Friday? — nobody here seems to have considered these possibilities! And so on. So yes, the “disclosure protocol” matters a lot, if you want to dig into this layer of subtlety. • John Baez says: But if you don’t want to get into the issue of the “disclosure protocol”, you might want to use this stripped-down version of the puzzle: You flip two coins on randomly chosen days of the week. At least one lands heads-up and is thrown on a Tuesday. What is the probability that both coins land heads-up? I think this has a certain minimalist charm. • bane says: I was thinking along the lines of the Monty Hall problem where, although “switch” is always an optimal answer, the player infers different probabilities depending on the protocol the gameshow host uses (along with being a bit worried about whether using the information “has a son” twice was being done validly). Incidentally, I think changing the “I have …” to refer to the reasoner’s state of knowledge just moves the necessity to make assumptions elsewhere. In this case, given an equal probability of BB, BG, GB and GG to go to the case where BB, BG, GB are equally likely you need the slightly stronger assumption “I always know when someone has a son, and this parent has a son”. • Mark Meckes says: Yes, I’ve always felt that this is the most crucial point that is rarely made about the Monty Hall problem. I’m pleased to see that the Wikipedia page says of the version from vos Savant’s famous column: “Although not explicitly stated in this version, solutions are almost always based on the additional assumptions that the car is initially equally likely to be behind each door and that the host must open a door showing a goat, must randomly choose which door to open if both hide goats, and must make the offer to switch.” 14. John Baez says: Here’s another puzzle, which I urge everyone to try: Anna is an extremely avid mountaineer and photographer, famous for her illustrated book The World’s Hundred Tallest Peaks. She has two children. One of them, Thor, is quite well-known in mountaineering circles. He was born shortly after Anna summited Mount Kilimanjaro. As he was being delivered, the sun set through a declivity between two peaks, Kibo and Mawenzi — a once-in-a-year event that Anna had wanted to photograph, despite the obvious risks. A freak thunderstorm created an amazing rainbow at the moment Thor was born. Anna’s husband took a picture of Anna cradling the newborn in her arms with the rainbow behind her — you can see it in her book. What is the probability that both Anna’s children are boys? • Tom Leinster says: Isn’t Thor Dan Quayle? • John Baez says: The odds are one in a million. I realized after posting my comment that you had discussed the same idea. On this thread I’m making it easier to post comments without me having to approve them all, which is nice — but the problem is that I’m not absolutely forced to read all of them before saying anything myself. However, I’m hoping there’s still something novel in my idea, namely how it’s linked to issues involving ‘identical particles’ in physics, and how identical particles can seem ‘non-identical’ when you’re told enough information about their individual states — enough information to effectively ‘name’ them. Instead of trying to explain my thoughts about this, I thought it would be easier and more fun to pose a series of further puzzles, starting with this and this. So, I hope people take a crack at these puzzles! • Tom Leinster says: I did Google to see if there really was an Anna answering to your description. The fact that there isn’t makes your way of putting the point more convincing: it gets rid of the problem (well, both problems) that Philip Gibbs mentioned. • John F says: 1/2. See Vladimir’s comment. 15. John Baez says: Here’s yet another puzzle. I’m curious about how it’s related to the puzzle I just posted. Suppose you have two identical particles. Suppose they can each be in one of two states. Just for fun, let’s call these the ‘boy’ state and the ‘girl’ state. Suppose at least one particle is a boy. What is the probability that both particles are boys? • Philip Gibbs says: I think there are lots of ways to interpret this quesion. If they are fermions and there is no other components to the phase space (e.g. they cant be in different positions) then the answer is zero. If they are bosons or fermions that can also have different positions, you could answer that the probablity is the mod squared amplitude of the component of the state vector that corresponds to both particles being boys, but that is perhaps not the answer you are looking for :) Perhaps you have to think of the particles as being initially in some sort of unbiased mixed state, but you have to be precise about what that means before you can answer the question. What do you mean by “at least one particle is a boy”? Have some measurements been made to provide this knowledge or is it just a statement about how the state was prepared? There may be some optimal answers to these questions that make this a nice generalisation of the classical problem. I’m not sure. • John Baez says: Phil wrote: I think there are lots of ways to interpret this quesion. True, but I had mainly two in mind. If they are fermions and there is no other components to the phase space (e.g. they can’t be in different positions) then the answer is zero. That’s one of the interpretations I had in mind. Indeed, when I said the particles can each be in one of two states, I meant only two states: the Hilbert space for a single particle of this sort is just $\mathbb{C}^2$. (For example, you can imagine that the particles’ position degrees of freedom are nailed down by having them lie in the ground state of a potential well. This is how people create spin qubits in an ion trap. But I don’t want to assume that the ‘boy’ and ‘girl’ states are necessarily spin up and spin down states. Just two states of some sort. None of this parenthetical remark really matters.) Now surely you can guess the other interpretation I had in mind, and say the answer with that interpretation! What do you mean by “at least one particle is a boy”? Have some measurements been made to provide this knowledge or is it just a statement about how the state was prepared? Personally I don’t feel any need to get into those questions when answering questions about quantum probability theory: I just pick the maximum-entropy mixed state consistent with the known facts, and use that to compute the probabilities. (I also do that in classical probability theory — just using a different definition of ‘mixed state’, and ‘entropy’. I explained this a little in week27.) • Philip Gibbs says: So for bosons the state space is three dimensional with basis |bb>, |gg>, (|bg> + |gb>)/sqrt(2) The “maximum-entropy mixed state consistent” with at least one particle is a boy will have a density matrix with (1/2, 0, 1/2) down the diagonal, so the probability for both are boys is 1/2, if I have not missed some subtlety. If the particles were not identical the answer would be 1/3. • John F says: I am confused. I had thought there was a need to use a reduced density matrix like Zhou http://arxiv.org/PS_cache/arxiv/pdf/0805/0805.3867v2.pdf but that doesn’t take into account that you’ve already measured one of the particles. So the density matrix is diagonal and two-dimensional. But I think the boy-boy probability is not specified by assuming that that the two one-particle states are equally probable. Sigh. • John Baez says: I’m not sure what’s making you sigh. I’m sure Phil got the right answer for identical bosons, namely 1/2, but it might help if you said what answer said alternative method would give. • John F says: The sigh indicates frustrated confusion, which tends to reign over my attempts to do quantum mechanics. I’m not sure the answer to the puzzle is always 1/2 without additional caveats about mutual independence or something – what if these were the kinds of identical particles who happened to prefer to be of the same gender when taken in couples? I don’t think (that’s probably my problem right there …) that preference necessarily implies an interaction energy. For example you could arrange they were always created that way when taken in couples, or alternatively create the measuring system that way. Such gender preference might imply that gender couldn’t be spin. Anyway, how I was approaching the answer with the reduced density matrix is undoubtedly wrong. Sorry about the lack of formatting. We suppose one particle (call him 1 for now) is a boy, and that the other particle (call it 2) could be a boy with some probability. First, forget that we know the gender of 1! The expectation value of any operator A operating on just 2 is <A> = Tr(ρ A), expanded in a basis as Equation 1: <A> = sum over states of 2,2′,1 {(C*21 C2’1) } The eigenvalues of ρ are the probabilities of its eigenstates. Now suppose the basis is gender i.e. |b> and |g>. If <A> is to be the probability that 2 is a boy, then with A the boy gender projection, i.e. diagonal 1 0, we have Equation 2: <A>= (C*bb Cbb + C*bg Cbg) Identity imposes sign constraints on the expansion coefficents C, but we don’t have to invoke here spin or space or anything. Of course there are still normalization constraints. I think the reasoning is correct so far, even if the math typing isn’t. Now what to do about the fact that we know that the gender of 1 is b? I think the answer using Equation 2 is simply that Cbg = 0. (Cgg would be zero too if it mattered). Then <a> = C*bb Cbb It looks to me like in the end the sign constraints can’t matter, so I presume there is major error. • John F says: Insert bra A ket everywhere it should be. Sigh. • John Baez says: The problem with your typing was that this blog allows html, and to begin a link in html you use <A HREF = So, each time you tried to write the expectation value of A: <A> you fooled the poor thing into thinking you were starting a link! That’s why the whole last half of your comment came out in color: it was a big fat link to nowhere. I fixed it. The solution, which I am using here: don’t type < and >, type &lt; and &gt;, which are html character entity references. Any good blogger needs to know lots of these. So, to get <A> you need to type &lt;A&gt; Either that, or use LaTeX, which is also allowed here, but requires the first dollar sign to be followed by the word ‘latex’, thus:$latex

Puzzles for html whizzes:

1) what did I need to type for you to see &lt;A&gt;?

2) what did I need to type for you to see \$latex?

• John Baez says:

John F wrote:

The sigh indicates frustrated confusion, which tends to reign over my attempts to do quantum mechanics.

Yeah, it’s tough stuff. I wouldn’t recommend trying to learn it from articles like Zhou’s — too fancy.

I started with the third volume of Feynman’s Lectures on Physics, spent a couple years on that, took a few classes on, taught a few classes… and by now, after thirty-two years of work, I think I understand the stuff pretty well. But I don’t understand the calculation you’re trying to do. What are those “C” things? Matrix elements of the operator A?

I’m not sure the answer to the puzzle is always 1/2 without additional caveats about mutual independence or something – what if these were the kinds of identical particles who happened to prefer to be of the same gender when taken in couples?

Interesting notion! You could dream up such particles, but the main particles that actually exist come in two kinds: bosons and fermions. So, when someone asks a puzzle like mine you’re supposed to try it for these two options before trying anything fancy.

I don’t think (that’s probably my problem right there …) that preference necessarily implies an interaction energy.

Identical fermions absolutely refuse to be in the same state. Identical bosons effectively prefer to be in the same state, through a subtle mechanism that this puzzle was supposed to illustrate. And non-identical particles don’t give a darn: strictly speaking, because they’re non-identical, they can’t really be in the ‘same’ state, just analogous states.

More later. Time to make breakfast!

• John F says:

Breakfast? I suppose that means you think you pass the Turing test. I don’t recall Feynman Lecture’s Vol. III discussing density matrices. I think I recall hearing him say “nobody understands quantum mechanics”.

The C coefficients were intended to represent elements of the density matrix. There were supposed to be extra factors on the rhs like bra 2 A 2′ ket. The square magnitudes of the C coefficients are like probabilities, and are probabilities when in the appropriate basis. Anyway I come up one short on counting constraints with this toy quantum puzzle as originally given. Hence the boy-boy probability is underdetermined.

One problem I have with the statement of this particular puzzle is that you claimed the “Hilbert space for a single particle of this sort is just C^2″, but then you changed the universe of discussion to *reality* (note the lack of parsing of formerly unobsolete emphasis formatting). A second problem with the given answer is that for nonidentical particles you count gg as an acceptable state (indeed, equally probable) even though it is actually ruled out.

I do think it is possible to constrain the universe of discussion to couples of entangled particles. What answer do you get then?

• John Sidles says:

Hmmm … isn’t it (arguably) inefficient for students to spend time being puzzled about quantum dynamics, until they have spent an adequate amount of time being puzzled about classical dynamics?

For example, there’s a large body of 20th century physics literature that argues (or even axiomatically asserts) that quantum dynamics is described by a linear unitary transformation on a Hilbert state-space … so much so, that almost all 20th century quantum textbooks teach this point-of-view uncritically … with results that have proved to be highly unfortunate for STEM students in general.

Some of the physics-related and information-theory-related reasons for this have been discussed over on Dick Lipton’s blog, under the topic Quantum Algorithms A Different View—Again … here we will instead take a mathematical and historical point-of-view.

In historical retrospect, two key events were Paul Dirac’s 1930 textbook Principles of Quantum Mechanics, which gave rise to today’s vast physics literature on quantum dynamics. Shortly after Dirac;’s book came mathematician Erich K¨hler’s 1933 article Uber eine bemerkenswerte Hermitesche metrik which gave rise to today’s vast mathematical literature on Kählerian geometry.

Today we appreciate that the quantum and Kählerian literatures are comparably vast because they are (in essence) the same topic … which traditionally is discussed by mathematicians from a geometric and holomorphic point of view (with an admixture of dynamics) … and by physicists from a dynamical and algebraic point of view (with an admixture of geometry).

These contrasting-but-related viewpoints are unsurprising, given that Dirac was a dynamicist with a strong interest in geometry, while Kähler was a geometer with a strong interest in dynamics.

The dominance of dynamics and algebra over geometry and flow in 20th century physics textbooks arose largely because Dirac’s 1930 textbook is clearly written, includes many practical examples, is well-suited to students … and came first. In contrast, Kähler’s 1933 article was short, obscure, and had no practical examples … and came second.

Thus Dirac’s ideas have dominated (among physicists) for the remainder of the 20th century … with regrettable consequences.

For example, Dirac’s discussion of bra and ket vectors asserts “The bra vectors, as they have here been introduced, are quite a different kind of vector from the kets, and so far there is no connection between them” … and then stops just short of explaining the geometric connection between them.

Had Dirac waited until 1935, he might have continued—with Dirac’s gift for concision and clarity—along lines similar to “The connection between bras and kets is essentially geometric, and is provided by the Kähler triple of metric, symplectic, and complex structure.”

The remainder of Dirac’s textook might then have focussed—without loss of concision, generality, or practicality—upon continuous dynamical flows on (what are today called) tensor network manifolds, instead of discrete algebraic operations on (what are today called) Hilbert spaces.

In this Kähler-antecedes-Dirac world, the whole history of 20th century physics surely would have been different … with end results that arguably might have less confusing for students.

A well-written article on this topic is Abhay Ashtekar and Troy Schilling’s arxiv preprint Geometrical formulation of quantum mechanics (arXiv:gr-qc/9706069).

• John F says:

John S,
Cohen-Tannoudji’s text also contained many practical examples with clarity of calculation. I wouldn’t praise concision over clarity for any textbook. If one views quantum mechanics as a means to an end, namely calculating probabilities, the usual linear transformation view is best.
Geometrization can add extra confusion, and does for me. It especially mucks up calculations of probabilities, which require extra care and extra intelligence to deal with properly. For instance Ashtekar and Schilling’s discussion of the geometry of wave function collapse during measurement is pasted in and, bluntly, hokey; viz “the system is more likely to collapse to a nearby state than a distant one.”

• John F says:

FWIW, Don Page has been obsessed with correctly calculating probabilities. Although he tends to think largely

http://arxiv.org/PS_cache/arxiv/pdf/0903/0903.4888v2.pdf

I think what he calls observational averaging

http://arxiv.org/PS_cache/arxiv/pdf/1003/1003.2419v1.pdf

is also correct for calculating probabilities in general, e.g. potentially entangled states.

Don Page has been my favorite physicist for several decades, not least after he destroyed any hope of doing semiclassical quantum gravity, e.g. averaging over stress-energy, by an obvious-after-you-see-it gedanken Schrodinger’s cat experiment.

• John Baez says:

Phil wrote:

So for bosons the state space is three dimensional with basis

|bb>, |gg>, (|bg> + |gb>)/√2

The “maximum-entropy mixed state consistent” with at least one particle is a boy will have a density matrix with (1/2, 0, 1/2) down the diagonal, so the probability for both are boys is 1/2, if I have not missed some subtlety.

If the particles were not identical the answer would be 1/3.

That’s right! Let me summarize:

Suppose you have two particles, each of which can be in either of two states, ‘boy’ or ‘girl’. Suppose you know that at least one is a boy. What is the probability that both are boys?

If the particles are identical bosons, the probability is 1/2, since there are two equally likely states: BB and BG = GB. (The product here is “symmetrized”, as usual for bosons. You wrote down the symmetrization explicitly.)

If the particles are identical fermions, the probability is 0, since the only allowed state is BG = -GB. (The product here is “antisymmetrized”, as usual for fermions, so BB = -BB, so BB = 0, meaning there’s no chance of two boys.)

If the particles are not identical, the probability is 1/3, since there are three equally likely possibilities: BB, BG, and GB.

This is the easiest example showing how identical bosons have an enhanced probability of being in the same state while identical fermions have a suppressed probability of being in the same state, as compared with non-identical particles.

I find it interesting that a number of people, facing the puzzle “I have two children, and at least one is a boy — what’s the probability that both are boys?” give the probability 1/2. They’re acting like children are bosons!

Of course the real answer is 1/3, since children are distinguishable — even twins.

I find it even more interesting that the fancier puzzle “I have two children, and at least one is a boy born on a Tuesday — what’s the probability that both are boys?” gives a probability closer to 1/2, namely 13/27 ∼ 0.48.

Why should the extra piece of information make the answer get closer to the answer for bosons?

Of course a look at the picture in Greg’s essay makes it clear why the answer is close to 1/2, and would get closer if there were more days in the week:

Picking a day of week almost doubles the chance of the BB state as compared to the BG or GB states. Bosons, on the other hand, halve the chance of the BG or GB state, by declaring that BG = GB. In other words, making the kids into identical bosons ‘folds over’ the square along the diagonal.

But there might be more to say…

• bane says:

John wrote “They’re acting like children are bosons!”. I’ll just note that there is a different way to get a probability of 1/2 that doesn’t involve any “bosonic”-type behaviour, namely
option 2 here. (Of course that doesn’t mean it’s actually an explanation for why people “gravitate towards” a 1/2 here. My limited knowledge of psychologists who study these things is that non-trained people try and do everything mentally and “exceed their working memory”, leading to faulty identifications between different values.)

• Greg Egan says:

I added a new diagram to a revised version of the essay, which makes it very clear how the probability reverts back to 1/3 from 13/27 when the birth day is reported whatever it might have been. In the BB region there’s a random choice of which son’s birth day to mention, so each possibility there gets a half-square corresponding to (say) heads on a flipped coin for the first son and tails for the second … and the extra probability within the BB region vanishes.

• Philip Gibbs says:

“They’re acting like children are bosons”. Very thought provoking. I sometimes think my two children are bosons :)

What happens with three children? Someone says “I have three children including a boy” — what is the probability that all three are boys? The naive answer is 1/4. The Foshee-Egan answer is 1/7. But the bosonic answer is 1/3. So the naive answer is not the same as the bosonic answer. Even if the answer is the same for two children, the thinking could be different.

• westy31 says:

But wait.

The states do not have equal probability. We should assume that the parent of a Boy-Girl might equally well have said “I have at least one Girl”: The probability the parent says “At least one Boy” is 1/2. So these states should be weighted by 1/2. A Parent of a Boy-Boy combination has probability 1 of saying “At least one Boy”, so it should be weighted by 1, twice the weight for boy-Girl.

So after applying these weights, you end up with probability 1/2.

Interesting how this is the same as for bosons.
Gerard

• westy31 says:

Basically, this is the same as what bane is saying.

16. Dmitry says:

I left a short comment above, but I’m afraid that it will be left unnoticed, so I decided to extend it here. I think that Greg Egan is wrong in his analysis of “second question”. A Bayesian won’t lose any money.

Here’s the scenario that Greg is correctly explaining. Let’s imagine a line of thousands of parents each having exactly 2 children. You ask every one of them one by one: “Do you have at least one son born on Tuesday?” If the answer is “Yes” then you bet according to P=13/27 probability that he or she has 2 sons, and eventually earn money.

Now what happens if the weekday is changed every time? Let’s say that we now ask the following question: “Do you have at least one son born on X?”, where X is a weekday chosen for every parent separately by a 7-sided dice roll. If the answer is “Yes” then you still bet according to P=13/27 probability.

If I understood Greg correctly, then he thinks that (1) a Bayesian would do it, and (2) this would lead to a money loss. I think (2) is wrong, and to be sure I wrote a simple matlab script to check. If the question is fixed then 13/27 of those who answer “Yes” have 2 boys. If the question is being randomly changed, the result is the same.

One more reason to be a Bayesian :) Seriously.

• bane says:

It’s tricky to because the discussion is spread. Greg agrees that if you get to choose and ask the question, then you can pick some day and you have an expectation of winning on average. What was unclear was how this relates to the situation where the parent says with no prompting from you to you “I have a son born on day X”, where X is some particular day. In the course of the thread what approaches can be taken to this question have been clarified.

Incidentally, I think the consensus is that upon reflection this issue doesn’t relate to anything about frequentist versus Bayesianism interpretations, it’s really about how one decides what mathematical structure the verbal problem corresponds to and then the calculation is the same in both interpretations.

• Dmitry says:

To this I fully agree. It’s just in the original Greg’s question he writes: “It seems to me (perhaps naively) that a Bayesian ought to be happy to take this bet any time, and then forget about what they did in the past — which ought to make them willing to take the bet on future offers even when the day of the week when the son was born changes” — and I’m saying that even if the day of the week changes it doesn’t make a difference.

What makes a difference, is that you have to ask a question yourself, here I agree with you. But there is no need to be specifically constrained on only one weekday (as Greg was suggesting: “If I took up that bet, I would then resolve that in the future I’d only take the same bet again if the person each time had two children and at least one son born specifically on a TUESDAY”).

I just wanted to point out that given the correct “protocol”, weekday can be changed from parent to parent.

17. elelias says:

There are a lot of comments, so I’ll make mine brief. I think a bayesian would aknowledge that it all comes down to formulating the question with a specific day of the week. The fact that every question has a different day is of course irrelevant, the only thing that matters is that the mentioning of a specific day modifies the sampling. Right?

18. Daniel Lewis says:

It’s true that the choice of Tuesday is arbitrary to a Bayesian, and that they would accept the bet for any day. These questions all give a Bayesian the same information about the answerer’s children, assuming we already know one is a boy:

(1) “Is one a boy born on Monday?”
(2) “Is one a boy born on Tuesday?”

(7) “Is one a boy born on Sunday?”
or even (Rand) “Pick a day of the week at random. Is one a boy born on that day?”

A “yes” answer to any of these questions is more likely when there are two boys than when there is only one, since having two boys gives more chances for one of them to match the criteria. If we work out the math as in the post, we find that our posterior is 13/27 as expected.

This question, on the other hand, is completely uninformative: (Any) “Is there a day of the week that you have a son who was born on?” Everyone with at least one boy answers “yes” to this question, which means its answer doesn’t give us new information if we know they have a boy. If a Bayesian asks this question, his posterior equals his prior.

Where Greg goes wrong is in supposing that if a Bayesian accepts (1)-(7) all as useful, equivalent tests then he must accept (Any) as well. But it’s just not true (in general) that if P(BB | (1)) = P(BB |(2)) = … = P(BB | (7)) then P(BB | (1) or (2) or … or (7)) must also be the same quantity.

(Rand) on the other hand does follow from (1)-(7), and as Dmitry showed by simulation it results in the same winnings as picking a single day to ask about.

19. John Baez says:

Let’s try another puzzle. This is a followup to the earlier ‘let boys be bosons’ puzzle.

Suppose we have two particles, each of which can be in any of 14 states. Just for fun, we’ll call 7 of these states ‘boys’, and label them by the days of the week. We call the other 7 ‘girls’ and also label those by the days of the week: for example, ‘girl Friday’.

Now suppose at least one of the particles is a boy Tuesday. What is the probability that both particles are boys?

There are three answers: one for identical bosons, one for identical fermions, and one for non-identical particles. Anyone who understands how to solve the earlier puzzle can solve this one. It’s really just a matter of counting the options.

• There are actually *four* answers:

There are two sets of 7 possible states: (B/Mon), (G/Mon); (B/Tues),(G/Tues); and so on.

One possible statement is: “We know the state of Child One, which is boy/Tuesday. The question is: what is the probability of Child Two being a boy?”

For identical bosons as well as non-identical particles, the day of the week is irrelevant, hence the probability reduces to 1/2, since it is possible for both of the particles to be in a “boy Tuesday” state, 7 boy states being available and 7 girl states being available, out of 14 possible states for the Child Two particle.

For identical fermions, since it is not possible for both of the particles to be in a “boy Tuesday” state, but still possible for at least one particle, Child One, to be in a boy Tuesday state while the other particle, Child Two, is in one of the other six boy/not Tuesday states or in one of the seven girl/day states, the probability of both being boys seems to be 6/13 (six possible boy/day allowed states out of 13 possible particle allowed states for Child Two, the state of boy/Tuesday being forbidden).

For non-identical particles, the day of the week is irrelevant once again, and so the probability reduces to 1/3, the choices being BB, BG, or GB.

===============

Another possible statement is: “We know the state of Child One or Child Two, which is boy/Tuesday. The question is: what is the probability of the other child, Child Two or Child One, respectively, being a boy?”

For identical bosons, the day of the week is irrelevant, hence the probability reduces to 1/2, since it is possible for both of the particles to be in a “boy Tuesday” state, 7 boy states being available and 7 girl states being available, out of 14 possible states for the Child Two or Child One particle. (BB, BG = GB)

For identical fermions, since it is not possible for both of the particles to be in a “boy Tuesday” state, but still possible for at least one particle, Child One or Child Two, to be in a boy Tuesday state while the other particle, Child Two or Child One, respectively, is in one of the other six boy/not Tuesday states or in one of the seven girl/day states, the probability of both being boys seems to be 12/26 (six possible boy/day allowed states out of 13 possible particle allowed states for Child Two, six possible boy/day allowed states out of 13 possible particle allowed states for Child One, the state of boy/Tuesday being allowed for only one child).

For non-identical particles, the day of the week is irrelevant once again, and so the probability reduces to 1/3, the choices being BB, BG, or GB.

• John Baez says:

streamfortyseven wrote:

One possible statement is: “We know the state of Child One, which is boy/Tuesday. The question is: what is the probability of Child Two being a boy?”

Thanks for being the first to take a crack at this!

Personally I am very scared of talking about ‘Child One’ and ‘Child Two’ when we are dealing with identical bosons or identical fermions. After all, the whole point is that in quantum mechanics, it makes no sense to ‘name’ identical particles — to treat them as having individual identities.

So, while it’s possible that you are doing something quite reasonable in these cases, I will skip them, and only tackle your question above in the case of non-identical particles.

Let me try it. We have two non-identical particles called Child One and Child Two, each of which can be in 14 states: ‘boy Monday’, ‘boy Tuesday’, …, ‘boy Sunday’, ‘girl Monday’, ‘girl Tuesday’, … ‘girl Sunday’.

But in fact, we have been told that Child One is in the boy Tuesday state. So we only need to consider the 14 possible states of Child Two. We treat these as equally probable (again, thanks to the principle of maximum entropy). So, the chance that Child Two is a boy is 7/14. So, the chance that both particles are boys is 7/14 = 0.500.

Another possible statement is: “We know the state of Child One or Child Two, which is boy/Tuesday. The question is: what is the probability of the other child, Child Two or Child One, respectively, being a boy?”

I would like to see answer to the question I originally asked, and this is still not that… at least, not word for word. Maybe it’s equivalent, but I’m too tired to think about that. My original question was:

At least one of the particles is a boy Tuesday. What is the probability that both are boys?

so this is the question I’ll answer. I’ll do it in three cases:

1) identical bosons,
2) identical fermions,
3) non-identical particles.

1) Identical Bosons. Before receiving any extra information we count the allowed states by counting the little squares in the diagonal of this big square, plus half the number of little squares that aren’t on the diagonal, since we can’t tell which particle is which — the kids are bosons, so it doesn’t make sense to distinguish between the first child and the second child.

If you work out the answer, you’ll get a triangular number:

14×15/2 = 105

states.

But that was just for fun. We’ve been told that at least one particle is a boy Tuesday, so we only need to consider the squares marked in red — with the proviso that now there’s no difference between, say ‘first child boy Tuesday, second child girl Friday’ and ‘second child girl Friday, first child boy Tuesday’. Remember, they’re bosons.

If we count the red squares with this all-important proviso, we get 14. So, there are 14 states, all equally probable (by maximum entropy) — and both particles are boys in 7 of these states. So the answer is 7/14 = 0.5000.

2) Identical Fermions. Before receiving any extra information we count the allowed states by working out half the number of little squares that aren’t on the diagonal, since we can’t tell which particle is which, and states on the diagonal are forbidden:

If you work out the answer, you’ll get a triangular number:

14×13/2 = 91

states.

But that was just for fun. We’ve been told that at least one particle is a boy Tuesday, so we only need to consider the squares marked in red — with the proviso that now there’s no difference between, say ‘first child boy Tuesday, second child girl Friday’ and ‘second child girl Friday, first child boy Tuesday’, and also the extra proviso that we can’t have two boy Tuesdays.

If we count the red squares with these all-important provisos, we get 13. So, there are 13 states, all equally probable — and both particles are boys in 6 of these states. So the answer is 6/13 ∼ 0.4615.

3) Non-identical particles.

Before receiving any extra information we count the allowed states simply by counting all the little squares in this big square — no funny business this time:

This time you obviously get

142 = 196

states: a square number instead of a triangle number.

Again, that was just for fun. My point is simply that these ‘just for fun’ counts of states for pairs of bosons, pairs of fermions and pairs of non-identical particles are related:

101 + 95 = 196

and that’s no coincidence: that’s always how it works!

In fact, we’ve been told that at least one particle is a boy Tuesday, so we only need to consider the squares marked in red. So, there are 27 states, all equally probable — and both particles are boys in 13 of these states. So, the answer is 13/27 ∼ 0.4814.

Case 3) was Greg’s original ‘simple question’. In case 1) the answer is slightly bigger, since bosons like to be in the same state. In case 2) the answer is slightly smaller, since fermions can’t stand being in the same state.

And again — just as in my original ‘let boys be bosons’ puzzle — the bosonic case gives an answer that matches the answer where we distinguish between the two children, assert that the first was a boy Tuesday, and ask what’s the probability that both are boys.

20. Thomas Larsson says:

I haven’t read through all the comments, so forgive me if someone else has pointed this out.

1. Some 52% of all infants are boys, perhaps because Y sperms are somewhat lighter (or more competitive) than X sperms.

2. Not all infants are XX and XY. Also X, XXX, and XYY are viable genders.

This may affect the odds.

• John Baez says:

Hi, Thomas!

In my original post I mentioned something about how in this puzzle we’d ignore subtlety 1. But it’s still interesting. I’d heard this percentage has evolved to make up for the fact that males kill themselves off more. I have no idea if that’s true.

I didn’t mention subtlety 2.

21. Rob Shaw says:

“I have two children. One is a boy born on a certain day of the week, but I’m not going to tell you which it is. What is the probability I have two boys?”

It’s clear that the second sentence adds zero information to the problem. Even if the “someone” wrote the day on a piece of paper, and showed it to you after your probability estimate, it wouldn’t change things, as you already know the probability of a given day of the week occurring is one-seventh.

Therefore the original intuition, that the day of the week of the birth is irrelevant, is correct.

22. ar18 says:

Apparently, no one here understands elementary probability.

Let’s presume that having a boy or girl is random, or 50:50 (it isn’t 50:50 as the odds are slightly higher for having a boy than a girl, but everyone here is making the unspoken assumption that it is 50:50, so I will continue using that same presumption)…

You are thinking that if I have a boy, that will change the odds of the sex of my next child. Wrong! The odds will never change because if they did change, it would no longer be a random thing since previous events would be physically influencing subsequent events. That would be magical thinking and not probability. The odds of having a boy or girl, no matter how many boys or girls you have previously had, is always 50:50. Period. Now if the problem had been stated as what are the odds of having two boys, then, and only then, would the odds be different from 50:50. But that wasn’t what the alleged puzzle presented, which also means that none of you payed attention to how the problem was presented:

I am pregnant. What are the odds of me having a boy? 50:50.

I want to have only two children. What are my odds of having two boys? Not 50:50.

I want to have only two children. What are my odds of having two boys, with one of them born on Tuesday? Not 50:50.

None of the above is what the puzzle presented. What the puzzle presented was this — I have a [three-year-old] child and I am pregnant. What are the odds of me having a boy? The odds are 50:50. The sex of the three-year-old is irrelevant, since that fact won’t magically determine the sex of babies. It will also not matter what time of the week it is, since weekdays won’t magical determine the sex of babies. Previously having a boy will also not magically determine the sex of babies. What will determine the sex of babies? Random chance, which can be mathematically calculated to be 50:50 when there are only two outcomes per event.

It is the same thing with coin flips. If I flip a coin 99 times and then ask you, “What are the odds that the next flip will land heads?”, if your answer is anything other than 50:50, you don’t understand probability. If I tell you that, “I have just flipped a coin 99 times, and every time it came up heads, so now what will the odds that the next flip will land heads?”, and you answer anything other than 50:50, you don’t understand probability. Now if I ask you what are the odds that I could get 100 heads in a row and you say 50:50, you don’t understand probability.

You guys are playing illogical games with numbers and you don’t even know what those numbers mean.

• John Baez says:

ar18 writes:

Apparently, no one here understands elementary probability.

I discourage personalizing the discussion in this way, but for what it’s worth, I don’t think that’s true.

You are thinking that if I have a boy, that will change the odds of the sex of my next child. Wrong!

I’ve read all the comments on this thread pretty carefully, and I don’t think anyone is making that assumption. We’re all treating the genders of the children as stochastically independent.

For example:

Everybody here agrees that if we say “I have two children, Child One and Child Two. Child One is a boy. What is the probability that Child Two is a boy?”, the right answer — for the purposes of this discussion — is 50%.

(It’s not exactly correct, due to the existence of identical twins and other factors. But it’s close enough, and it’s the right answer for the purposes of the calculations we’re doing.)

But consider a different question: “I have two children. At least one is a boy. What is the probability that both are boys?” Now there are 3 possibilities: BB, BG and GB, all equally likely. So, the probability is 1/3.

Some people on this thread are phrasing this question as follows. “I have two children. One is a boy. What is the probability that both are boys?” Here they are using “one” to mean “at least one”. This may be what’s annoying you: one can interpret the sentence “one is a boy” in a very different way that gives the answer 1/2.

Note that I’ve already explained this stuff to streamfortyseven, who began by saying “it’s a lot of gibberish” but wound up understanding what Egan was talking about.

23. I apologize for not having read all the above comments in detail, so perhaps this has been said already.

how does a Bayesian reason about the thought experiment I’ve described, in such a way that they don’t end up taking the bet every time and losing money?

Answer: By carefully differentiating among various protocols.

You have a population of parents, each with two children. You have the opportunity to place bets that are expected money winners given the 13/27 probability, but losers given the 1/3 probability.

Protocol One: You go down the line and ask each parent “Do you have at least one boy born on a Tuesday?” and place a bet with each one who says yes. You come out a winner. Then you go down the line and do the same thing with “Wednesday” instead of Tuesday. You come out a winner again. Etc.

Protocol Two: You go down the line and ask each parent “Do you have at least one boy born on an X?” where X is chosen randomly and place a bet with each one who says yes. You come out a loser.

The discrepancy is fully accounted for by the fact that in Experiment One, some parents get bet on twice, while in Experiment Two, each parent gets bet on at most once.

Now what about Protocol Three?

Protocol Three: Go down the street randomly asking people “Do you have at least one boy born on an X?”, where X varies randomly. This might sound a lot like Protocol Two, but it’s actually equivalent to Protocol One, since anyone with two boys born on different days is twice as likely to answer “yes” to your random question than someone with, say, a boy and a girl.

So our Bayesian will go down the street asking about random days of the week, bet when the answer is yes, and get rich.

• Dmitry says:

Protocol Two: You go down the line and ask each parent “Do you have at least one boy born on an X?” where X is chosen randomly and place a bet with each one who says yes. You come out a loser.

I think you are wrong here. Moreover, to convince myself I did a simulation experiment some time ago (see above, I left a comment here), and yes, you seem to be wrong: one would come out a winner.

The proof of it is actually contained in Greg’s original confusion. If you ask about Monday, then the probability is 13/27. If you ask about Tuesday, it’s the same. So if you ask randomly about Mondays and Tuesdays this probability can’t change, which means you can safely bet according to 13/27 probability. As you go down the line of parents you can every time randomly choose a day and ask a corresponding question; in the end you’ll still win.

• So if you ask randomly about Mondays and Tuesdays this probability can’t change, which means you can safely bet according to 13/27 probability.

You’re right of course. I’m dashing out the door but will post a corrected version of this later in the day.

24. Dmitri pointed out that in my earlier comment, I thoroughly misdescribed my intended Protocol Two. (For the record, I copy-pasted the wrong description from an earlier draft.)

Rather than list errata, here is a corrected version of the entire comment:

I apologize for not having read all the above comments in detail, so perhaps this has been said already.

how does a Bayesian reason about the thought experiment I’ve described, in such a way that they don’t end up taking the bet every time and losing money?

Answer: By carefully differentiating among various protocols.

You have a population of parents, each with two children. You have the opportunity to place bets that are expected money winners given the 13/27 probability, but losers given the 1/3 probability.

Protocol One: You go down the line and ask each parent “Do you have at least one boy born on a Tuesday?” and place a bet with each one who says yes. You come out a winner. Then you go down the line and do the same thing with “Wednesday” instead of Tuesday. You come out a winner again. Etc.

Protocol One-Prime: You go down the line and ask each parent “Do you have at least one boy born on an X?” where X is chosen randomly and place a bet with each one who says yes. Once again you come out a winner.

Protocol Two: You go down the line and ask each parent “Do you have at least one boy born on a Saturday?” You bet with those who say yes. With those who say “no”, you then ask “Do you have at least one boy born on a Sunday?”. You bet with those who say yes. This time you lose.

The discrepancy is fully accounted for by the fact that in Protocols One and One-Prime, some parents get bet on twice, while in Protocol Two, each parent gets bet on at most once. It is Protocol Two to which the 1/3 calculation is relevant.

Now what about Protocol Three?

Protocol Three: Go down the street randomly asking people “Do you have at least one boy born on an X?”, where X varies randomly. Note that any parent with two boys born on different days is twice as likely to answer “yes” as a parent with, say, a boy and a girl. Thus you are essentially in the world of Protocols One and One-Prime, not in the world of Protocol Two.

So our Bayesian will go down the street asking about random days of the week, bet when the answer is yes, and get rich.

25. John Bentin says:

Let me pose a slight variant: Instead of “born on a Tuesday” your colleague says “born on—now let me think…..ah yes: it was a Tuesday.” Being rather more fast-witted than your fumbling colleague, you have already calculated the probability of two boys to while away the time while he was thinking. Now you know the day, do you revise your estimate?

• And another slight variant: instead of “born on a Tuesday”, it’s “born a Tuesday and having red hair” where the possible hair colors are black, brown, red, and blond … so instead of 7(days) x 4(gender mixes) we now have 7(days) x 4(gender mixes) x 4(hair colors) .

I think that the more factors you add in, the series beginning with 13/27 converges to 1/2. How about it?

• John Baez says:

streamfortyseven wrote:

I think that the more factors you add in, the series beginning with 13/27 converges to 1/2. How about it?

As John Bentin pointed out, you just answered this puzzle of mine. The answer, I believe, is 1/2 – ε for some very small value of ε.

• Sure, because you go from 1/2 (or 1/3) to 13/27 or whatever.

• John Bentin says:

@stream47: Your red-hair point was made by Tom Leinster (Dan Quayle) and John Baez (Thor). I would argue, in the hesitation variant, that you should switch from 1/3 to 1/2 even before you hear the day: the hesitation is crucial while the day is irrelevant. For the original, unhesitant, “born on a Tuesday” case, 13/27 is correct.

26. […] into fuels, led to a great discussion that taught us a lot about this issue. Greg Egan’s Probability Puzzles were a fun way to sharpen our understanding of probability theory. Keep ‘em […]

27. Puzzled says:

I have a dumb question.

I get the 1/3 probability for the simple version of “one of my children is a boy”. We eliminate the GG possibility and what we’re left with is three equally likely outcomes, hence BB=1/3.

But I worked what I thought was an alternative but straightforward approach and was left scratching my head. What did I do wrong?

I interpreted the statement “one of my children is a boy” to be equivalent to “either my first or second born child is a boy” and then I broke the statement into two cases:

Case 1, the first child was a boy (p=.5)

OR

Case 2, the second child was a boy (p=.5)

Given case 1, the possibles are BB and BG. I assigned p=.5 within case 1 for p=.25 for each.

Given case 2, the possibles are BB and GB. Again I assigned p=.5 within case 2 and p=.25 for each.

So I get p(BB)= .25+.25 = .5 and not 1/3 but 1/2.

I guess I should not have counted both BB’s, when I eliminate one of them then I get p(BB)=.25/.75

But what is the reason to eliminate one of the BB’s?

And if that is not the problem with the approach, what is?

• bane says:

You aren’t doing anything wrong (at least from a quick scan).

John’s answer of 1/3 comes from applying what’s called the Maximum Entropy Principle to assign unknown probabilities (namely in a way to maximise the similarity between unknowns). In the case-by-case analysis you’re assigning probabilities to unknowns in a slightly different way based on the tree structure of responses, and hence getting a different result of 1/2. In a sense, the difference is highlighting that the problem isn’t completely well specified.

FWIW, David McKay (mixed physicist and machine learning guy) says in his book ITILA

The maximum entropy method has sometimes been recommended as a method for assigning prior distributions in Bayesian modelling. While the outcomes of the maximum entropy method are sometimes interesting and thought-provoking, I do not advocate maxent as THE approach to assigning priors.

• bane says:

I’ll modify my statement slightly. Going from the interpretation

either my first or second born child is a boy

into two cases each occurring with probability of 1/2 is, I think, not correct. If you’d used an interpretation “I’ve just picked one of my children, and it’s a boy” then the rest of the development is fine. (The key point is that under that interpretation, it was a possible outcome that they did in fact have one boy and didn’t tell you. It’s these “lost” BG and GB samples that cause the probability to rise from 1/3 to 1/2.)

• John Baez says:

Puzzled writes:

I have a dumb question.

Good. Smart questions are usually way too hard.

I hope you understood both Bane’s replies. In his first reply I think he was putting too much emphasis on foundational issues which are a bit beside the point here, though interesting. His second reply really hit the nail on the head:

Going from the interpretation

either my first or second born child is a boy

into two cases each occurring with probability of 1/2 is, I think, not correct.

But he didn’t say why it’s not correct. Here’s the problem. You are treating the two cases, ‘the first child is a boy’ and ‘the second child is a boy’ as mutually exclusive alternatives. But they’re not: they overlap!

Suppose you’ve got mutually exclusive alternatives: alternative A and alternative B. Suppose each occurs with probability 1/2, and some event has probability p in the first case and q in the second case. Then you can calculate the overall probability of that event as you do, namely:

½ p + ½ q

But if alternative A and alternative B can both happen at once, you need to do a more complicated calculation. You need to consider three cases: one where just A happens, one where just B happens, and one where they both happen. So you need a formula where we add up three terms, not just two.

In our problem, there’s a ⅓ chance that only the first child is a boy. There’s a ⅓ chance that only the second child is a boy. There’s a ⅓ that both are boys. And obviously:

• If only the first child is a boy, there’s a 0 chance that both are boys.
• If only the second child is a boy, there’s a 0 chance that both are boys.
• If both are boys, there chance is 1 that both are boys.

So the right answer, doing it case by case, is

⅓ 0 + ⅓ 0 + ⅓ 1 = ⅓

Fun question!

• bane says:

Actually, I was thinking my first answer was the one that contained the important conceptual stuff, the second one was the equivalent of pointing out an minor slip :-) . The key point I was trying to convey to Puzzled was that one shouldn’t be worried that thinking about what the question means in different ways gives rise to different numerical probabilities.

I don’t disagree with any of the technical things you wrote as such, but need to point out that it’s not uncontestably the “right” answer. I think it’s key to point out that in the step between “You need to consider three cases” and assigning them 1/3 each, you’re assigning a probability to something you have no knowledge of based on some “general principles about what you expect” (ie, a prior). In this case you’re using the Principle of Maximum Entropy, which is one interesting, useful and sometimes effective way to do it, but it’s not the only way. As the David McKay quote illustrates, it’s not the only procedure used by those doing (typically huge, complicated) probabilistic inference.

(I haven’t answered any of the questions about particles, basically because I’m unsure if in that field there’re reasons why max-ent is the only appropriate principle to use.)

• John Baez says:

Bane wrote:

I think it’s key to point out that in the step between “You need to consider three cases” and assigning them 1/3 each, you’re assigning a probability to something you have no knowledge of based on some “general principles about what you expect” (ie, a prior).

That’s true, and important, but I didn’t get the feeling that this was what Puzzled was puzzled about… so I was afraid your first comment might scare him more than enlighten him.

(I haven’t answered any of the questions about particles, basically because I’m unsure if in that field there’re reasons why max-ent is the only appropriate principle to use.)

In my opinion, the principle of maximum entropy is no more or less appropriate in particle physics than in other fields. In my opinion, everyone should use maximum entropy a whole lot, and it’s just sort of a historical coincidence that it got so entrenched in physics, especially statistical mechanics, before Jaynes jumped up and down and shouted that it’s a generally applicable method. It’s not the only method, but it’s a darn good one.

28. […] course I instantly thought of the probability puzzles we’ve discussed here. So I thought you folks might enjoy […]

29. Popular Math says:

Here is the conventional Bayes treatment of this case:

Let q = p(2B | at least 1B) = 1/3, and
1-q = p(1B | at least 1B) = 2/3

The probability of two boys given that one is born on Tues is given by Bayes Thm as
(*) p(2B | 1B Tues) = p(1B Tues | 2B) q / p(1B Tues)
(here and throughout “1B Tues” means *at least* one boy who is born on Tuesday, rather than exactly one born on Tues).

As usual, we expand the denominator as
p(1B Tues) = p(1B Tues | 2B)q + p(1B Tues | 1B)(1-q)
where p(1B Tues | 1B) = 1/7 , and
p(1B Tues | 2B) = 1- (6/7)^2 = 13/49,
(because the probability of a boy not being born on Tues is 6/7)

Inserting into (*) gives
p(2B | 1B Tues) = (13/49) (1/3) / ((13/49) (1/3) + (1/7)(2/3))
= 13 / (13 + 14) = 13/27
This makes explicit what additional info is learned by pinpointing the weekday of birth, and from the terms in the denominator moreover what’s going on in the betting protocol: i.e., there’s an enhanced probability of two boys due to the increased likelihood (from 1/7 to 13/49) of having one on Tues if there are two (or equivalently the greater difficulty of not having one on Tues if you have two).

If instead there are N species of boys, let “1B f.s.” mean at least one boy of known fixed species, and then independent of the number of species of girls:
p(1B f.s. | 1B) = 1/N
p(1B f.s. | 2B) = (1 – (1-1/N)^2) = (2N-1)/N^2
and consequently (*) now gives
p(2B | 1B f.s.) = p(1B f.s. | 2B) q / p(1B f.s.)
= (2N-1)/ ((2N-1) + 2N) = (1/2)(1 – 1/(4N-1))
which makes it clear how this ranges from 1/3 to 1/2 as the degree of specificity grows, i.e., as N ranges from 1 to \infty.

Note 1: The issues (and confusions) in this discussion thread were typically seen while teaching naive Bayes methodology to non-mathematically inclined undergraduates. (Curiously, more mathematically inclined colleagues are frequently equally befuddled by the elementary probability issues.) The rest have shown up in teaching distinguishable vs. indistinguishable bosons and fermions to physics undergraduates.

Note 2: In some circumstances (e.g., the Wason selection task), putting problems in a social context can improve performance, but in this probability case it appears disadvantageous: In successive years, I’ve assigned the problem as stated here, as probability of two boys, given that one is born on a given day of week; or as the probability of two blue balls, choosing twice at random from a set of fourteen balls, of which seven are blue and numbered 1 through 7 and similarly seven green (with replacement after the first selection). The performance is surprisingly better for the same problem phrased in terms of blue and green balls (perhaps because the answer for two blue balls given blue ball number 3 somehow doesn’t seem as intuitively obvious as for two boys given a boy born on Tues, so they had engage more mathematical brain apparatus from the outset).

Note 3: The problem is equally straightforward in terms of M colors of balls (generalizing gender) and N flavors (generalizing day of week), choosing r balls (with replacement), and determining the probability of s color collisions conditioned on knowing the flavors of t of them.

30. Popular Math says:

Finally, it is worthwhile pointing out that the Tues birth problem is as well protocol dependent.

The Bayes methodology above, based on the two equations
(*) p(2B | 1B Tues) = p(1B Tues | 2B) q / p(1B Tues)
(**) p(1B Tues) = p(1B Tues | 2B)q + p(1B Tues | 1B)(1-q)
implicitly assumed the following protocol: if the parent has two children, including at least one boy born on Tues, then the parent will always inform that he or she has a child born on Tuesday. But there are other protocols: suppose if there are two boys, but only one of whom is born on Tues, then the parent will always give the weekday of the other child’s birth. Then learning that there is a boy who was born on Tues makes it *less* likely that there are two boys. To calculate this probability, now interpret “1B Tues” as your having been informed that the parent has one child born on Tues. Then
p(1B Tues | 1B) remains 1/7, but
p(1B Tues | 2B) = 1/49
because you only get to learn about the Tues boy in this protocol when both are born on Tues. Substituting into (*) gives
(*’) p(2B | 1B Tues) = (1/49)(1/3) / ((1/49)(1/3) + (1/7)(2/3)) = 1/15 .
So hearing that there is a boy born on Tues in this protocol makes it *less* likely that there are two boys, for the obvious reason that you only hear about the Tues birth in the case of two boys when both are born on Tues. If this protocol is used, then you will lose badly if you bet on two boys based on a 13/27 probability.

There is a simple class of such protocols in which if there are two boys with only one born on Tues, the parent preferentially reports Tues over exactly m of the 6 other weekdays when they occur. An example of the m=3 case would be if the other boy is born on Wed, Thu, Fri, then you’ll be told that day, but otherwise (Sat, Sun, Mon) you’ll be told Tues. The original case posed in this thread corresponded to m=6 (always inform Tues if there is at least one Tues), and the other extreme case posed above in this message corresponds to m=0 (only inform Tues if both boys are born on Tues). For these protocols in general for m=0 through 6:
p_m(1B Tues | 2B) = (2m+1)/49 = 1/49, 3/49, 5/49, 1/7, 9/49, 11/49, 13/49
(because there are 2m cases, Tues X and X Tues, where you’ll be told Tues, plus 1 case where both are born on Tues).

There is a total of 2^6=64 such protocols, {6\choose m} of them for each value of m (represent as six binary bits, with a 1 when Tues is reported over one of the other six days, and m corresponds to the number of 1’s), and the probability of two boys
(*”) p_m(2B | 1B Tues) = p_m(1B Tues | 2B) q / p(1B Tues)
increases monotonically from 1/15 to 13/27 as m increases from 0 to 6.

Now the punchline: suppose we don’t know which of the protocols is being employed. Let’s assume maximal ignorance (Bayesian flat prior) and hence that any of the 64 possible protocols is used with equal probability. Then p(1B Tues | 2B) should be taken as the average over all the values for all the protocols weighted equally. Their values are symmetric about the m=3 value of 1/7, as is {6\choose m} (=1,6,15,20,15,6,1), so the average is
<p(1B Tues | 2B)>_{flat prior} = 1/7.
Substitution into (*) now gives
(*”’) p(2B | 1B Tues) = (1/7)(1/3) / ((1/7)(1/3) + (1/7)(2/3)) = 1/3 .

So if we assume maximal ignorance of the protocol for reporting Tues, we learn nothing from the statement that a boy is born on Tues. This is ultimately the resolution to the seeming paradox of why the probability shifts from 1/3 to 13/27, just from learning that a boy is born on Tues. (“We already knew the boy had to be born on some day, so what additional information did we receive?”). If we don’t know the protocol that determines why a Tues birth is disclosed, then indeed we have learned nothing, and the expectation of the probability of two boys given that one is born on Tues remains 1/3. For the same reason, if we’re maximally ignorant re the protocol, it only makes sense to bet on a 1/3 chance of two boys despite “learning” that one is born on Tues.

If you know the precise protocol from within the above simple class, then the probability of two boys ranges through the seven discrete values from 1/15 to 13/27, according to the number of weekdays a Tues birth is reported preferentially over other weekdays. If you have partial probabilistic knowledge of the protocol, then the expectation <p(1B Tues | 2B)> is calculated with the p_m(1B Tues | 2B)’s weighted accordingly, and p(2B | 1B Tues) can take any real value from 1/15 to 13/27.

Final Note: While I understand I’m lecturing into a vacuum at this point, with all activity in this thread having ceased a week ago, for completeness I mention that the comment in the preceding message: “and then independent of the number of species of girls” still assumed that the total probability of girls is 1/2. In this formalism, the probability of girls enters entirely through the value of q. In general if r is the boy/girl ratio, then q = r / (r+2), and of course the probability of two boys (*) increases with the value of this ratio. The Bayesian formulation gives the same result as simple counting, but frequently expresses the parameter dependencies more transparently.

31. John Bentin says:

Yes, but there are other considerations. Why would anyone tell you, unasked, that (a) he (or she) had two children and (b) one was a boy? In my view, (a) he is proud of having children, and more proud of having two than having one, and (b) he is proud of having a boy, and would have been even more proud had he fathered two boys, which he would surely have boasted about had that been the case. From this, you may conclude that there is little chance that he had two boys.

The Tuesday information won’t change this conclusion much unless we put great weight on it. If we do so, we might conclude that the person was either mad or trying to tease us or test us in some way. Then it’s all up in the air again.

32. SV says:

Q. I have two children. One is a boy. What is the probability I have two sons?
A. 1/2
There are a lot of ways to look at the problem.

1) One child is a boy. The probability of the *other* child being a son is 1/2. So the probability of two sons is 1/2.

2) The boy *being referred to* is either the first child or the second child.
The probability that the boy being referred to is the first child, let’s call it P(1B), is 1/2.
The probability that the child being referred to is the second child, let’s call it P(2B) is 1/2.
These are mutually exclusive cases whose probabilities sum to 1.
So probability of two sons = P(1B) * Prob of 2 boys given 1B + P(2B) * Prob of 2 boys given 2B = 1/2*1/2 + 1/2*1/2 = 1/2

3) For each possibility of boys and girls in 2 children, calculate the *number of boys*, let’s call it NB.
The possibilities for NB are 0(GG), 1(BG or GB) and 2(BB).
The original question can be rephrased as, “I have two children. The count of boys is at least 1. What is the probability
that the count of boys is 2?”
Our possibility space is NB = 1 and NB = 2. Prob of NB = 2 is 1/2.

4) The various combinations of boy/girl (B/G), with the order(first or second child) being important are
BG, GB and BB.
Let me label the scenarios with 1,2,3 and the first child and second child with a and b respectively.
Then the combinations are
1Ba,1Gb 2Ga,2Bb 3Ba,3Bb
The boy being referred to can be 1Ba or 2Bb or 3Ba or 3Bb = total of 4 possibilities.
Out of these, possibilities in which both children are boys = 3Ba and 3Bb = 2 possibilities
So required probability = 2/4 = 1/2

Q. I have two children. One is a boy born on a Tuesday. What is the probability I have two sons?
A. 1/2

The boy being born on a Tuesday is irrelevant. The flaw in Greg Egan’s square counting argument can be seen by
looking at point of view 4) to the question above.
I am referring to the third image from the top on this page, http://gregegan.customer.netspace.net.au/ESSAYS/TUESDAY/Tuesday.html
When both children are boys born on a Tuesday, there are *2* possibilities added to the total possibility space.
The boy being referred to can be either the first child or the second child, so we have to add both
possibilities to the total possibility space. Both these possibilities (in the case when both children are boys)
similarly also contribute to the possible outcomes with 2 boys. That’s why the required probability is 14/28 = 1/2.

33. My Name says:

The day of the week has nothing to do with the answer. 49 combinations with 7 out of 49 being both kids born on the same day. 1/7 is to high to be ignored as subtracting 1. And combination of kids gender are: BB GB GG(since the order in which they were born is irrelevant) which 33.333% or 1/3 simply. but since we know that one kid is boy than it goes to 50% BB or GB.

34. Werner says:

The Bayesian is making bets on 14 different, simultaneously running, games. It’s not that it doesn’t matter, it’s that each is done in isolation as an iteration of one of the 14 running games.

The real question is, does the probability change if he asks about the day the first child was born, or can he just think it into changing? Also … hasn’t someone run this expereiment in some way?

• Werner says:

Side comment … please tell Mr. Egan, I LOVE Diaspora.

35. Martin-2 says:

The answer to the first puzzle is 1/2. This doesn’t invalidate the second “deep” puzzle because Mr. Egan thought he was working on this similar puzzle:
Mother: “I have two children.”
You: “Is one of them a boy born on a Tuesday?”
Mother: “Yes. Now what is the probability that both are boys?”

Here the answer is 13/27, for reasons outlined above. OP’s puzzle is instead:
Mother: “I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?”

Here you need to account for the fact that P(she says “boy”|Boy&Girl) = 1/2 and P(she says “Tuesday”|Tuesday&~Tuesday) = 1/2. Surprisingly, this makes the final answer P(2 boys) = 1/2. I wish I could show you a nice proof but I arrived at this conclusion by brute force assuming a 2-day week and then generalizing. My inspiration comes from this LessWrong post.

I’ve never seen this distinction made outside of LW, though I haven’t read most of the comments here or on Prof. Landsburg’s post on TBQ. I love both versions of this problem. There’s something mischievous about how there’s an obvious answer, but then some deep reasoning gets you to a cool non-obvious answer, but then often some even more deep reasoning brings you right back to the obvious answer, so everyone from math illiterates to math professors gets it wrong about half the time.

• Martin-2 says:

Now, you may object to my step P(she says “boy”|Boy&Girl) = P(she says “Tuesday”|Tuesday&~Tuesday) = 1/2. Maybe you expect the parent to be biased toward/against singling out boys (or Tuesdays). But that doesn’t really matter for our understanding of the puzzle. The point is that Mr. Egan and Prof. Baez implicitly assumed that these probabilities are 1, and there’s simply no way that we live in such a boy-and-Tuesday-centric world.

• Martin-2 says:

I’d also like to add that this reasoning runs parallel to the Monty Hall problem. When you pick a door you have some chance of picking the winner, but Monty never picks the winning door. Likewise, when you ask “is one of them a boy?” you have some chance of hitting the jackpot if the parent says “no”, but if the parent does all the talking you don’t get this chance.