tag:blogger.com,1999:blog-8807287158334608095.post5851687668035555646..comments2024-03-28T11:22:03.762-04:00Comments on The Virtuosi: I was born on WednesdayAlemihttp://www.blogger.com/profile/15394732652049740436noreply@blogger.comBlogger10125tag:blogger.com,1999:blog-8807287158334608095.post-16492759552458254042022-01-09T09:39:11.160-05:002022-01-09T09:39:11.160-05:00스포츠토토티비 Great information, thanks for sharing with...<a href="http://cse.google.com.pe/url?q=http%3A%2F%2Fsportstototv.com%2F" title="스포츠토토티비" rel="nofollow">스포츠토토티비</a> Great information, thanks for sharing with us… check out my service…<br />Champions Leaguehttps://www.blogger.com/profile/17087822094731316683noreply@blogger.comtag:blogger.com,1999:blog-8807287158334608095.post-41099577400155201492022-01-09T09:32:10.780-05:002022-01-09T09:32:10.780-05:00카지노사이트홈 Hi there, You have done a great job. I wil...<a href="https://www.google.com.mm/url?sa=t&url=https%3A%2F%2Fwww.casinositehome.com" title="카지노사이트홈" rel="nofollow">카지노사이트홈</a> Hi there, You have done a great job. I will definitely digg it and<br />personally recommend to my friends. I am confident they will be benefited from this web site.Champions Leaguehttps://www.blogger.com/profile/17087822094731316683noreply@blogger.comtag:blogger.com,1999:blog-8807287158334608095.post-71220114880793026422022-01-09T09:24:20.872-05:002022-01-09T09:24:20.872-05:00It’s perfect time to make a few plans for the long...It’s perfect time to make a few plans for the longer term and it is time to be happy. <a href="https://www.google.com.mm/url?sa=t&url=https%3A%2F%2Fwww.oncasino.site" title="온라인카지노사이트" rel="nofollow">온라인카지노사이트</a> I’ve learn this post and if I could I wish to suggest you few interesting issues or advice. Perhaps you could write subsequent articles relating to this article. I want to learn more things approximately it!<br />Champions Leaguehttps://www.blogger.com/profile/17087822094731316683noreply@blogger.comtag:blogger.com,1999:blog-8807287158334608095.post-65083053252338512412010-05-30T08:51:33.758-04:002010-05-30T08:51:33.758-04:00So essentially the *actual question* is the tricky...So essentially the *actual question* is the tricky thing. It looks like "What are the odds of a child being a boy" (50%) but it's *actually* "what are the odds of two children both being boys, and here's a headstart" (33.3..%)Geoffhttps://www.blogger.com/profile/07065848582219331473noreply@blogger.comtag:blogger.com,1999:blog-8807287158334608095.post-83639885194934063132010-05-29T08:25:21.700-04:002010-05-29T08:25:21.700-04:00It appears that the problem most people have is un...It appears that the problem most people have is understanding the difference between an ensemble average, and a sort of localized probability distribution. It's obvious to most people that every time you flip a coin, the probability of H is 1/2. So if you flip a coin 29 times and get H every time, when you flip it the 30th time you have a 1/2 chance of getting H. But if you repeat the experiment of 'flipping a coin 30 times in a row' many times, you'll get a very small probability of all-H. <br /><br />It's the same in this problem, only we have trouble defining what the 'ensemble' is here. Forgetting the Tuesday bit, you've got four choices in two flips: HH,HT,TH,TT. Now you have to specify what the ensemble is. <br /><br />If it is "the ensemble in which ONE of the coins is a H", that narrows it down to HH, HT, TH, and then the probability of HH is 1/3. <br /><br />IF, instead, you say your ensemble is "THE FIRST coin is a H", you're back to a 1/2 probability of HH. <br /><br />This Latter situation is actually the one Rob Eastway paints, because one can assign a descriptive marker to each of the coins. Coin#1 is "the coin that I see", while coin #2 is "the coin that's lost". So by not seeing the other coin, you're forcing yourself into that latter ensemble. And, in fact, if you repeat the experiment over and over, this becomes obvious, because you're not limiting yourself to coins that are "at least one girl", but actually just "a girl on the coin that I see", so if it so happened that there was no girl on the coin that you see, but a girl on the other coin, you'd discard it, despite it satisfying the condition 'at least one girl'. <br /><br />Essentially the confusion is in "what number to divide by". Of course anyone can count the number of HH that happened in N trials. The question is, do I divide by N (total number of trials), or the number of (N?+?N) (given A coin is a girl), or (N?) (given the coin that I SEE is a girl). All answers are equally valid, provided you define your ensemble rigorously.Nnoreply@blogger.comtag:blogger.com,1999:blog-8807287158334608095.post-39359708819480737742010-05-26T21:26:25.716-04:002010-05-26T21:26:25.716-04:00@Rob Eastway: Yeah, it's really counter intui...@Rob Eastway: Yeah, it's really counter intuitive. The maths is certainly sound, but it does seem contrary to experience. Think of it in terms of a multiverse, the universe branching with each possibility:<br /><br />1: There are two children. You now have four possible universes in the future, one each for BB, BG, GB, GG. 0.25 probability of each. <br /><br />2: You are told one boy is a girl. You know know you're in one of three universes, BB, BG or GB. Now what are the odds of two boys? 1 in 3. It's weird, the extra information narrows down the odds. <br /><br />It's all very quantum ;) In fact, the existence of probability in mathematics is probably some weird philosophical proof of a multiverse of some description. Probably...I mean what are the odds? ;)Geoffhttps://www.blogger.com/profile/07065848582219331473noreply@blogger.comtag:blogger.com,1999:blog-8807287158334608095.post-63593461369111516222010-05-26T10:31:31.667-04:002010-05-26T10:31:31.667-04:00NeatNeatdlonoreply@blogger.comtag:blogger.com,1999:blog-8807287158334608095.post-87163254871144363662010-05-26T10:26:20.617-04:002010-05-26T10:26:20.617-04:00Colin: All three of the statements you made are t...Colin: All three of the statements you made are true. But it is also true that "I have two boys. At least one of them is a boy, what is the probability that the other one is a boy" is 1/3 and it is also true that "I have two boys. One of them is a boy born on Tuesday, what is the probability that my other child is a boy" is 13/27. It is also true that card draws are not independent but births mostly are. But is also true that for the purpose of the article I treated card draws as mostly independent.<br /><br />Rob Eastway: Yeah, I've been racking my brain to try and figure how how this applies to the real world. The question statement is rather artificial. I mean its a little harsh to say there isn't a right answer, in the limit that I pose a well phrased mathematical question there is always an answer, the only possible contention is that my mathematical model actually doesn't replicate the kind of situation I claim it replicates in the real world. <br /><br />Now, I think you are right in that if you are in the real world, meet someone, and get introduced to their son, now the probability that their other child is a son is 1/2 effectively. Getting the probability to drop to 1/3 requires the very unrealistic scenario that you know they have exactly two children, and that least one of them is a boy, but have NO OTHER INFORMATION whatsoever about the families. This is not a situation that arises in circumstance, but it is one that I can recreate. If I parsed the US Census data for all families that have two children. Then I parsed that subset for all families that had at least one boy. Then I ask what fraction of the subset have two boys, then the answer is definitely 1/3. Similarly, parsing the Census for the boy-tuesday problem would create a circumstance where you got 48.1%, not 50 and not 33. But you're right in that its rather artificial.Alemihttps://www.blogger.com/profile/15394732652049740436noreply@blogger.comtag:blogger.com,1999:blog-8807287158334608095.post-40932174640240178162010-05-26T08:47:49.890-04:002010-05-26T08:47:49.890-04:00Hey guys, I've got two British coins, a 10p an...Hey guys, I've got two British coins, a 10p and a 50p and I've flipped them both. At least one came up as a head (The Queen, so let's call it a girl). <br /><br />What's the chance that the other coin came up as a head (i.e. a girl)? <br /><br />The answer is 1/3, right? - after all, you've seen the argument in the blog above, and my statement is exactly analogous to the one about having two children, at least one of which is a boy (let's ignore the Tuesday element, which just confuses matters).<br /><br />But wait a minute, you're making a huge assumption here. You know nothing about how I came to make my statement to you. I did indeed flip two coins. Stupidly, when they landed I couldn't find them, but a helpful friend informed me he'd found one of them and it was a head, though he didn't tell me which coin it was (10p or 50p, the younger or the older 'child'). We haven't found the second coin yet.<br /><br />So it is entirely true and fair for me to state that:<br />(a) I have two 'children'<br />(b) At least one of them is a 'girl'<br />(c) GB BG and GG are still all possible as outcomes, where GB means older child G, younger B<br /><br />However, in this case, the chance that the other coin, when it eventually is found, is a girl is 50%. Test it out with coins yourself, if you don't believe me (replicating exactly the situation I described).<br /><br />The Tuesday-boy problem makes an assumption about the circumstances in which the statement is made which I believe is actually rather implausible. I claim the coin situation I described is a more plausible analogy. So I think that 1/2 is a better answer than 13/27, but that there is not actually a 'right' answer to the Tuesday-boy problem as posed.<br /><br />Comments welcome twitter.com/robeastawayRob Eastawayhttp://www.robeastaway.comnoreply@blogger.comtag:blogger.com,1999:blog-8807287158334608095.post-19871580808204415872010-05-26T08:33:43.051-04:002010-05-26T08:33:43.051-04:00The gender of children are independent events so t...The gender of children are independent events so there are only three options: 0 boys, 1 boy, and 2 boys. The gender of one child does not affect the other. The date a child is born is independent of the gender.<br /><br />The probability of having a boy is P(B) = 0.5.<br /><br />The probability of having two boys is P(B)*P(B)=0.5*0.5 = 0.25.<br /><br />The probability of having a boy given you have one boy: P(B|B) = P(B AND B)/P(B) = P(B)*P(B)/P(B) = P(B) = 0.5.<br /><br />The flaw in using cards is that drawing a second card from the deck changes the probabilities: they are dependent. At the micro level sperm have some dependence upon each other since 1 spermatogonium yields 4 spermatids from the same DNA strand, but otherwise spermatogonium are independent. Especially if you're talking about children >~1 year apart then they are so far independent its not worth noting.Colin ML Burnetthttps://www.blogger.com/profile/00090750541507788084noreply@blogger.com