tag:blogger.com,1999:blog-3500036.post8440480555401411616..comments2024-01-02T04:49:16.658-05:00Comments on He Lives: The Tuesday Child Puzzle via Monte Carlo (repost)Davidhttp://www.blogger.com/profile/08688240424047203541noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-3500036.post-81832395755056934272017-12-02T09:57:26.744-05:002017-12-02T09:57:26.744-05:00Let me try to emphasize the issue here, with the f...Let me try to emphasize the issue here, with the following generic problem: You meet a man who tells you that the has two child. He then provides additional information that is partially dependent on the gender combination of his two children, but was freely offered. What is the probability that he has a boy and a girl?<br /><br />First, notice that I reversed the question to be about two different genders, not two of the same. A lot of people have difficulty accepting that "are both the same?" and "are both boys?" is the same question if what you were told is that there is a boy. But the reversed question is not so difficult, and its answer is always one minus the answer for the more popular version.<br /><br />Let K represent the knowledge you have at this point. And use B0, B1, and B2 to represent the cases where the man has zero, one, or two boys, respectively. Note that without the information, Pr(B0)=Pr(B2)=1/4, and Pr(B1)=1/2. The solution is a fairly trivial application of Bayes' Theorem:<br /><br />Pr(B1)|K) = Pr(K|B1)*Pr(B1)/Pr(K) = Pr(K|B1)/[2*Pr(K)]<br /><br />And note that the law of Total Probability says:<br /><br />Pr(K) = Pr(K|B0)*Pr(B0)+Pr(K|B1)*Pr(B1)+Pr(K|B2)*Pr(B2)<br />Pr(K) = [Pr(K|B0)+2*Pr(K|B1)+Pr(K|B2)]/4.<br /><br />So<br /><br />Pr(B1)|K) = 2*Pr(K|B1)/[Pr(K|B0)+2*Pr(K|B1)+Pr(K|B2)]<br /><br />So the answer is entirely dependent on the event K, and how it can occur in the various family types. The confusing part, is that K represents the event where you acquire this information, not the event where it is true. The difference is that you can acquire different information even when it is true.<br /><br />The importance of this difference is evident in the Monty Hall Problem. If the knowledge K that you have there is the event where a goat is behind the door Monty Hall opened? Then the answer is 1/2. Because there is a goat behind that door in two out of every three games. In one out of those two, the car is behind your original door, and in the other one it is behind the door you'd switch to.<br /><br />But the actual information you have is that there is a goat behind that door AND Monty Hall choose it. Say he chooses a door by flipping a coin; he uses it if he needs to, but still flips it if he doesn't so that you don;t know he doesn't need it. So now there are six cases, not three. In two of them, your door has a goat and he opens the door he did. In one of them, your door has the car and he open the door he did. In the other three, he opens a different door. This way, we see that the chances are 1/3 if you stay, and 2/3 if you switch.<br /><br />My point is that "K is true" is not necessarily the same event as "you know K." If there are equivalent alternative options - like Monty Hall opening Door #2 instead of Door #3 - you have to assume both could have happened with the same probability, so "you know K" is more restrictive that "K is true."<br /><br />In my problem, if K is "I have at least one boy," then the conditional probabilities that K is true are Pr(K|B0)=0, Pr(K|B1)=Pr(K|B2)=1, and the answer is 2/3 (remember, I reversed the question). And if K is "I have at least one girl," the answer is also 2/3. But then, if K is "I'm going to tell you the gender of at least one tomorrow," and the answer is 2/3 regardless of what that gender turns out to be, the answer changes to 2/3 today. This is Bertrand's Box Paradox.<br /><br />But there is an alternative to "I have at least one boy" is case B1, namely "I have at least one girl." If we use Pr(K|B1)=1/2, the the answer is 1/2, and the paradox goes away.<br /><br />+++++<br /><br />You can object to Pr(K|B1)=1/2 in your version #1, because the version is ambiguous. We have no idea how we learned K. But that means no answer is possible, which is what Martin Gardner concluded for that very question.<br /><br />In versions #2 and #3, the parent clearly chose information from the set available. Just like Monty Hall clearly choose between whatever goat doors were available. You have to assume Pr(K|B1)=1/2.<br /><br />It is only in version #4 that Pr(K|B1)=1.JeffJohttps://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-3500036.post-15714399349197631522017-11-26T08:34:47.823-05:002017-11-26T08:34:47.823-05:00There are a couple of horrible wording choices in ...There are a couple of horrible wording choices in these versions. I'll get there, but I want to add a fourth one:<br /><br />4) You meet a man on the street and he says, “I have two children." You ask "Is at least one of them a son born on a Tuesday?” He says yes. What is the probability that both are boys?<br /><br />The answer to #2 and #3 is 1/2. I have explained why: You can't conclude that a person with a Tuesday Boy, and another child who is not a Tuesday Boy, would always tell you about the Tuesday Boy and never tell you about the other child. You can conclude this in #4, which is why its answer is 13/27.<br /><br />As Martin Gardner pointed out, #1 is ambiguous and could be 13/27 or 1/2. But the **ONLY** scenario that would make it 13/27 is version #4, which is completely unreasonable from the wording of #1.<br /><br />+++++<br /><br />Horrible wording choice #1: If "one is XXX" can be interpreted as "at least one, but maybe two, are XXX", then "I have two" can also be interpreted as "I have at least two, but maybe more." Every version of the problem should say "at least one." Note that Gardner's (And Tanya's, if I recall correctly, after she once asked me to maybe publish some analysis with her) does, but Rosenhouse's does not.<br /><br />Horrible wording choice #2: Asking about "the other child" implies that you have identified which child is the one you are talking about. The answer is the same as the versions where you know about the older child. The point is that how identification is accomplished is irrelevant, just the fact that one is. "I have two children, and the one who sits to Mother's right at the dinner table is a boy" makes it a 50% chance that the other child is a boy.<br /><br />Merely counting cases is never the correct way to solve a probability problem. It can get the right answer when every case has the same chance, but getting the right answer does not mean the solution is correct. And assuming it is correct can get the wrong answer.<br /><br />The Monty Hall Problem is a good example: there are three doors where the car could be, one is eliminated, so the answer is that each door has the same chance. Since they must add up to 100%, each is 50%, right? Wrong. There are effect8ively six cases: Two where the car is behind Door #2 and Monty opens Door #3, two where the car is behind Door #3 and Monty opens Door #2, one where the car is behind door #1 (contestant's door) and Monty opens Door #3, and one where the car is behind door #1 and Monty opens Door #2. Eliminate the ones where Monty opens #2 (since we saw him open #3), and the answer is 2/3.<br /><br />This solution is more properly ex[pressed by using a 0% probability he would open #3 if the car is behind #3, a 100% chance if it is behind #2, and a 50% chance if it is behind #1. So the odds are 100%:50%, or 2:1, or a 2/3 probability.<br /><br />Any version of the Two Child Problem can only be solved by assessing the probability that we would know about a boy (with property XXX, like "born on Tuesday") in a family with a boy having that property and another child who is not a boy with that property. By merely counting cases, you are assuming that probability is 100%, and you cannot assume that.<br /><br />Look up Tanya Khovanova's blog at http://blog.tanyakhovanova.com/2011/05/a-son-named-luigi/JeffJohttps://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-3500036.post-81266030857289039352017-11-25T12:28:51.101-05:002017-11-25T12:28:51.101-05:00Are you saying that these three formulations of th...Are you saying that these three formulations of the problem are no equivalent?<br /><br />1) Mr. Smith has two children, and one of them is a boy, but also that the boy was born on Tuesday (Wikipedia)<br /><br />2) You meet a man on the street and he says, “I have two children and one is a son born on a Tuesday.” What is the probability that the other child is also a son? (Jason Rosenhouse)<br /><br />3) I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?<br /><br />All three claim the same answer of 13/27. I assume you think at least Jason is correct, and probably wikipedia, so how do you view my wording as significant? Is not my wording the same as Jason's, except I am the man on the street?<br /><br />This is interesting, I actually hope I am wrong because the problem will be even more fascinating.<br />Davidhttps://www.blogger.com/profile/08688240424047203541noreply@blogger.comtag:blogger.com,1999:blog-3500036.post-71522981147695712782017-11-25T10:01:35.549-05:002017-11-25T10:01:35.549-05:00As proved by the Bertrand Box Paradox - and btw, I...As proved by the Bertrand Box Paradox - and btw, I got my interpretation of it from Jason Rosenhouse's book starting on page 24 - having two children including a boy who was born on a Tuesday, and telling somebody that you have two children including a boy who was born on a Tuesday, ARE NOT THE SAME EVENT.<br /><br />I can't stress that enough. You are equating them in your solution, but your problem statement does not.<br /><br />As Rosenhouse explains, you are making the exact same mistake as the people who think the answer to the Monty Hall Problem is 1/2. They think that because Door #3 has a goat, that they should "eliminate" all possibilities where it has the car, and "keep" all where it has a goat. In EXACTLY THE SAME MANNER, you suggest that we should "eliminate" all possibilities where you don't have a Tuesday boy, and "keep" all where you do.<br /><br />Your table, and your simulation, are wrong because they simply count the cases, "keeping" those with a Tuesday Boy and "eliminating" those without one. Make a similar table, or simulation, for Monty Hall, and the answer you get will be 1/2.<br /><br />If I had asked you "does either of your two children happen to be a boy who was born on a Tuesday?", it would be correct to "keep" all cases where you have a Tuesday Boy. Just like, on the game show, if we had asked Monty Hall to open door #3 for us, 1/2 would be correct.<br /><br />But when you have the freedom to mention a Girl born on a Thursday even if you her brother was born on a Tuesday, then we can't simply count cases. We can only consider both descriptions as being equally likely to be passed along.n Just like the game show contestant must consider the possibility that door #2 could have been opened.JeffJohttps://www.blogger.com/profile/09110352332876400907noreply@blogger.comtag:blogger.com,1999:blog-3500036.post-11153089156322966802017-11-25T06:23:05.538-05:002017-11-25T06:23:05.538-05:00JeffJo,
Thanks for the comment.
I disagree, I t...JeffJo, <br /><br />Thanks for the comment.<br /><br />I disagree, I think the problem as stated is unambiguous. For a simple analysis, see Jason Rosenhouse (a mathematician who wrote a book on the Monty Hall problem). <br /><br />http://scienceblogs.com/evolutionblog/2011/11/08/the-tuesday-birthday-problem/<br /><br />I have update the post with a tabular based discussion.<br /><br />Out of curiosity, where do think the simulation, which confirms the 13/27 answer, went wrong?Davidhttps://www.blogger.com/profile/08688240424047203541noreply@blogger.comtag:blogger.com,1999:blog-3500036.post-16821037752530669072017-11-24T17:48:05.579-05:002017-11-24T17:48:05.579-05:00In 1889, Joseph Bertrand published a cautionary ta...In 1889, Joseph Bertrand published a cautionary tale about how *NOT* to solve a probability problem like this. There are three boxes, each with two coins: one has two of gold, one has two of silver, and the last box has one of each kind. Pick a box at random. What is the probability that it has two of the same kind of coin? (This is supposed to be easy: 2/3.)<br /><br />Now suppose I open the chosen box, pull out a coin, and show you that it is gold. Does this affect the answer? Many people - including you from your answer above - would say it does. The box can't be the one with two silver coins, but could be either of the other two. Since only one of these boxes has two of the same kind of coin, the answer appears to change to 1/2.<br /><br />Bertrand then changed the second problem slightly. Suppose I pull a coin out, but don't show it to you. Does the answer still change? If the coin I pulled out is gold, the problem is the same as the one where it seemed to change to 1/2. If it is silver, similar logic says it is also changes to 1/2. Since these are the only two possibilities, and the answer is 1/2 regardless of which, the answer to this question is also 1/2. (Note: a more formal solution uses conditional probability and the Law of Total Probability to get this result.)<br /><br />But that is a paradox: You have no information that would allow you to update the probability from 2/3 to 1/2. (Note: the name "Bertrand's Box Paradox" does not mean the problem itself, it means this paradox). So what seems to be true about the probability change, can't be.<br /><br />Bertrand's point was that you cannot assume you would get the information you have in every case where it is true. The correct solution to recognizes that the probability I would pull a gold coin from a box with two gold coins is 100%, but it is 50% if the box has one of each. Using these values, the answer to the second question is 2/3, just like the first question, and there is no paradox.<br /><br />In 1959, Martin Gardner published the predecessor to your problem in Scientific American: "Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?" He originally said the answer was 1/3, BUT HE RETRACTED THAT ANSWER for the same reason found in Bertrand's Box Paradox. The answer could be 1/3 or 1/2, depending on how you learned the information. Since, unlike Bertrand, he gave no information about this, he said the problem was ambiguous. In my opinion, he is wrong - the ambiguity means you can only assume it was randomly obtained, and the answer is 1/2. But that isn't really important.<br /><br />In 2010, at a puzzle convention held in honor of Martin Gardner, Gary Foshee presented the problem you give above. It is less ambiguous than Gardner's earlier one, since Foshee told us that he picked one description from the (most likely) two that describe his children. And he DIShonored Gardner by overlooking how Gardner changed his answer.<br /><br />The answer to your question is 1/2. There is a 1/196 chance that you have two Tuesday Boys, and a 100% chance you would tell us about a Tuesday Boy if you do. There is a 12/196 chance that you have a Tuesday Boy and a non-Tuesday Boy, and a 14/196 chance that you have a Tuesday Boy and a girl; but in either case, only a 50% chance you would tell us about the Tuesday Boy. Ignoring the "196" since it divides out of all terms, the answer is (1+12/2)/(1+12/2+14/2)=(1+6)/(1+6+7)=1/2.JeffJohttps://www.blogger.com/profile/09110352332876400907noreply@blogger.com