tag:blogger.com,1999:blog-38600807.post3596642133775789507..comments2018-06-02T14:19:34.554-04:00Comments on Advanced Football Analytics (formerly Advanced NFL Stats): Stories vs. StatisticsUnknownnoreply@blogger.comBlogger25125tag:blogger.com,1999:blog-38600807.post-71882501362304430492010-11-15T16:34:32.612-05:002010-11-15T16:34:32.612-05:00For the Anon poster who is having trouble understa...For the Anon poster who is having trouble understanding why the correct answer is 1/3, perhaps you should try running a simulation. If you assign equal probability of having a boy or girl, you will find that 25% of two-child families are BB, 25% are GG, and 50% are a mix (25% with the youngest being a boy, 25% with the youngest being a girl). So, let's say you were lucky enough to get exactly 25% with only 100 samples, or in other words, 100 2-child families.<br /><br />You would have:<br /><br />GG - 25 Families<br />BB - 25 Families<br />GB - 25 Families<br />BG - 25 Families<br /><br />Now let's restate the question: A set of parents have 2 children. At least one is a boy. What is the probability both children are boys?<br /><br />The "At least one is a boy clause" eliminates GG as a possibility. That leaves us with three types of families left:<br /><br />BB - 25 Families<br />GB - 25 Families<br />BG - 25 Families<br /><br />Now when we consider the OTHER child (i.e. the one that is not definitely a boy), we find that only 25/75 (or, simplified, 1/3) families have a boy as the other child - and that is the BB family.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-66730427338831170832010-11-09T10:42:24.746-05:002010-11-09T10:42:24.746-05:00We use Bayes. What is the probability of two boys,...We use Bayes. What is the probability of two boys, given there is one boy? <br /><br />The numerator: (1)(1/4) which = (1/4). <br /><br />The denominator: (1)(1/4)+(1/10)(1/4)+(1/10)(1/4)+0 which =(12/40). <br /><br />So, (1/4)/(12/40)=5/6, which is our answer. <br /><br />This makes sense. After all, if we see Smith with a son, it's very likely he doesn't have a daughter. <br /><br />Anyway, Chris, you are exactly right to think backward, that's the whole point of Bayes. <br /><br />JudoJohnAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-46093356662167968642010-11-08T22:20:22.785-05:002010-11-08T22:20:22.785-05:0090%? Here's my logic, let me know if I'm ...90%? Here's my logic, let me know if I'm doing this right. I think its a matter of thinking backwards. So, if he has a son and a daughter. There is only a 10% chance that his son happened to go with him that time. If it wasn't that 1/10 chance that is happening in this scenario, then it must be a boy/boy family. Hence 90%?Chrisnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-76722066803463602032010-11-08T21:36:12.599-05:002010-11-08T21:36:12.599-05:00Of course Brian is exactly right. Mr. Smith has tw...Of course Brian is exactly right. Mr. Smith has two children. There is 1/4 chance to have two sons. You see him with a son. You trim the sample space, eliminating two girls. You are left with boy/boy girl/boy or boy/girl....1 in 3. <br /><br />Now, there is a 90% chance that if Mr. Smith goes to the theater, it will be with his daughter (if he has one). You see him at the theater with a son. What are is the probability he has two sons? <br /><br />JudoJohnAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-76908014735376046462010-11-08T15:58:32.960-05:002010-11-08T15:58:32.960-05:00A set of parents have 2 children. At least one is ...A set of parents have 2 children. At least one is a boy. What is the probability both children are boys?<br /><br />GG (eliminated)<br />BB<br />BG<br />GB<br /><br />Answer = 1/3<br /><br />It does depend on how you ask the question.Brian Burkehttps://www.blogger.com/profile/12371470711365236987noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-50214447079326519252010-11-08T15:42:06.521-05:002010-11-08T15:42:06.521-05:00The basket problem isn't analogous to the two ...The basket problem isn't analogous to the two sons problem, but in the two sons problem, you aren't choosing between 3 choices. They say the parents have a boy. Here are the possibilities:<br /><br />Boy, Boy<br />Boy, Boy<br />Girl, Boy<br />Boy, Girl<br /><br />Half of them are BB<br /><br />Another way to look at it:<br /><br />Boy, Boy<br />Boy, Girl<br />Girl, Boy<br /><br />There are 4 boys here. If you randomly pick one, there is a 50% chance he is in a 2 two son family. They aren't asking you to pick a certain pairing. They certainly didn't ask for order. They are asking if the boy is in a two son family. 50% of the boys are in a two son family. It's as simple as that. If you don't understand, you are using bad math. You know that there are 4 possible boys and 2 of them are in a 2 two son family.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-9940731395486811402010-11-08T14:32:25.333-05:002010-11-08T14:32:25.333-05:00This is awesome. I could read the back-and-forth ...This is awesome. I could read the back-and-forth of the two sons problem all day.<br /><br />(FYI: I'm in the 33% camp)MattyPhttps://www.blogger.com/profile/09238596881583799736noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-54547538881956902082010-11-08T08:37:31.911-05:002010-11-08T08:37:31.911-05:00Anonymous #2 again. Actually, please ignore this ...Anonymous #2 again. Actually, please ignore this last comment - I think the chance that you've picked basket #2 is actually higher than 1/3.<br /><br />There is an important distinction between the basket problem and the 2-sons problem. They're not quite the same thing. Knowing that you have at least 1 boy does NOT increase the probability of being BB vs. BG or GB. It's still 1/3 chance of being BB. In the basket puzzle, having more blue balls in a basket DOES increase the chance that you've picked that basket. It's a bad analogy.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-27160976676454497742010-11-08T07:55:48.038-05:002010-11-08T07:55:48.038-05:00To the other Anonymous - In your basket example, i...To the other Anonymous - In your basket example, if you randomly pick a blue ball, it means you picked from baskets #2, #3, or #4. <br /><br />1/3 chance of basket #2 x 100% chance the other ball is blue.<br />+<br />1/3 chance of basket #3 x 0% chance the other ball is blue.<br />+<br />1/3 chance of basket #4 x 0% chance the other ball is blue.<br /><br />= 1/3 chance the other ball is blue.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-65670724073211614722010-11-07T22:13:07.045-05:002010-11-07T22:13:07.045-05:00Brian, the order is not relevant at all. The quest...Brian, the order is not relevant at all. The question didn't ask just like the Tuesday question is wrong. <br /><br />Think of it like this. Say you have a 4 baskets with red balls (girls) and blue balls (boys) and you are blindfolded. 1 has 2 red balls. 1 has 2 blue balls. 2 has 1 red ball and 1 blue. If you randomly pick a blue ball, there is a 50% chance the other ball is blue.<br /><br />Boy, Boy<br />Boy, Girl<br />Girl, Boy<br /><br />There are 4 boys here. If you pick the first boy, he has another boy with him. If you pick the second boy, he has another boy with him. That is 2 with 2 boys. If you pick the 3rd boy, he has a girl paired with him. If you pick the fourth boy, he has a girl paired with. That means 50% of the boys are in 2 boy families.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-52445064035110438852010-11-07T20:43:00.356-05:002010-11-07T20:43:00.356-05:00Anonymous: Your #1 and #3 *are the same thing* in ...Anonymous: Your #1 and #3 *are the same thing* in the "either" question. You need to stop thinking about sequence, and then you'll get it.Brian Burkehttps://www.blogger.com/profile/12371470711365236987noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-34275337613964180612010-11-07T16:13:56.635-05:002010-11-07T16:13:56.635-05:00Yes they are wrong. Mathematicians can be wrong. I...Yes they are wrong. Mathematicians can be wrong. It's not like peer reviewed work isn't false around half the time. Smart guys are just as wrong at times. <br /><br /> Here are all the possible combinations:<br /><br />1) Boy(known), Boy<br />2) Boy(known), Girl<br />3) Boy, Boy(known)<br />4) Girl, Boy(known)<br /><br />That is still 50%Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-2889949745552724812010-11-07T15:39:06.234-05:002010-11-07T15:39:06.234-05:00Wrong again, my stubborn friend. You'll figure...Wrong again, my stubborn friend. You'll figure it out eventually.<br /><br />If you remove the GG combination from the 4 possible combinations, you get:<br /><br />GB<br />BG<br />BB<br /><br />each with 1/3 probability.<br /><br />But hey, I, together with thousands of mathematicians, could be wrong. Could you be?Brian Burkehttps://www.blogger.com/profile/12371470711365236987noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-89263455465652178362010-11-07T13:35:52.264-05:002010-11-07T13:35:52.264-05:00You are still wrong Brian. Birth order is irreleva...You are still wrong Brian. Birth order is irrelevant because the question didn't ask that and even if it did, it is still 50%<br /><br />If you are not asked for an order then it is:<br /><br />A: Boy<br />B: Girl<br /><br />That is 50%<br /><br />If you are asked for an order then it is:<br />A: Boy A: Boy B<br />B: Boy A: Girl <br />C: Girl: Boy A<br />D: Boy B: Boy A<br /><br />Either way it is 50%.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-26850083253647339582010-11-06T21:00:09.985-04:002010-11-06T21:00:09.985-04:00Ian hit it on the head above.Ian hit it on the head above.Brian Burkehttps://www.blogger.com/profile/12371470711365236987noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-17967179765103025832010-11-06T20:58:56.390-04:002010-11-06T20:58:56.390-04:00Not wrong. It's in the wording of the question...Not wrong. It's in the wording of the question. <br /><br />If you are told that at least <i>either one</i> of the two children is a boy, and are asked for the probability the other is also a boy, then the answer is 1/3.<br /><br />If you are told that <i>this particular child</i> is a boy, and are asked for the probability that the other <i>particular child</i> is boy, then the answer is 1/2.<br /><br />Nice try. But I do admire your defiant certainty.Brian Burkehttps://www.blogger.com/profile/12371470711365236987noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-22578678445311590202010-11-06T20:29:01.477-04:002010-11-06T20:29:01.477-04:00Brian, you are completely wrong on the 2 sons prob...Brian, you are completely wrong on the 2 sons problem. No matter what way you calculate it, you get 50%. The only way not to get 50% is to use statistical sophistry that makes 0 logic.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-4105933917241828612010-11-05T17:42:34.500-04:002010-11-05T17:42:34.500-04:00Paulos' book Irreligion is also quite good and...Paulos' book Irreligion is also quite good and not filled with the type of inflammatory rhetoric many such books include.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-85597274036362285022010-11-05T15:51:28.599-04:002010-11-05T15:51:28.599-04:00Thought everyone here would appreciate this quote....Thought everyone here would appreciate this quote. How many times have you heard this before?<br /><br />Igor Olshanksy, Cowboys DE: "I mean, there's a lot of things that can go wrong when you pass the ball. You can get sacked, the ball can get tipped, you can get an interception. If you run the ball, you can get tackled and get a forced fumble. See, there's a lot more things [that can go wrong on pass plays]. ... You've got to run the ball."Jameshttps://www.blogger.com/profile/01838293735141324662noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-86374013868341311282010-11-05T14:31:24.838-04:002010-11-05T14:31:24.838-04:00Brian, I have a project to do in my Probability Th...Brian, I have a project to do in my Probability Theory class that is a variation on the two son problem. It's some pretty interesting stuff...and yeah, Bayes to the rescue. I'll send you a copy of the power point when I'm done if you want. <br /><br />JohnAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-89239250546803405242010-11-05T14:12:17.882-04:002010-11-05T14:12:17.882-04:00See I wonder if I'm not understanding the two-...See I wonder if I'm not understanding the two-son problem properly, because it makes sense to me. The important thing is understanding exactly how much information you are being given.<br /><br />Normally when you're told 'at least ...', that's when you need to think about it. Maybe I'm just more wary than I should be.Ian Simcoxhttps://www.blogger.com/profile/01518825067469269377noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-72474621878819223242010-11-05T12:43:16.776-04:002010-11-05T12:43:16.776-04:00Yeah, the selection process is key. There is a gr...Yeah, the selection process is key. There is a great discussion of the two-son problem (and a related variant) here:<br /><br />http://www.sciencenews.org/view/generic/id/60598/title/Math_Trek__When_intuition_and_math_probably_look_wrongDrewnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-46018708065336830122010-11-05T12:10:27.397-04:002010-11-05T12:10:27.397-04:00Yes, it seems similar to Monty Hall. The way I fin...Yes, it seems similar to Monty Hall. The way I finally accepted it was to think Bayes.<br /><br />For all two-child families, having a boy-girl combination (in either sequence) is twice as likely as a boy-boy combination. You have to start with that prior distribution.<br /><br />Alternatively, the 4 possible combinations (with probabilities) are:<br />BB (1/4) BG (1/4) GB (1/4) and GG (1/4)<br /><br />Now remove the possibility of GG. We are left with:<br />BB (1/3) BG (1/3) and GB (1/3)<br /><br />Therefore the probability of having the other child be boy is 1/3. The difference is in the sequence. If we were told the <i>younger</i> child is a boy, then the answer would be different. We would have:<br /><br />BB (1/2) and BG (1/2) <br /><br />There can be no GB combination.Brian Burkehttps://www.blogger.com/profile/12371470711365236987noreply@blogger.comtag:blogger.com,1999:blog-38600807.post-8197288620646751392010-11-05T12:00:10.520-04:002010-11-05T12:00:10.520-04:00Maybe the Monty Hall paradox?Maybe the Monty Hall paradox?JMMnoreply@blogger.comtag:blogger.com,1999:blog-38600807.post-24947250631650699792010-11-05T10:32:44.932-04:002010-11-05T10:32:44.932-04:00Is the two son problem just a restatement of "...Is the two son problem just a restatement of "the price is right paradox?"JMMnoreply@blogger.com