tag:blogger.com,1999:blog-264226589944705290.post885741176038003767..comments2024-09-25T08:51:01.854-07:00Comments on God Plays Dice: A quiz on probability from the Wall Street JournalMichael Lugohttp://www.blogger.com/profile/15671307315028242949noreply@blogger.comBlogger30125tag:blogger.com,1999:blog-264226589944705290.post-70181422207100273842008-04-12T00:14:00.000-07:002008-04-12T00:14:00.000-07:00Infact the strange name is a non sense. Not really...Infact the strange name is a non sense. Not really important for the answer. Also the age! But if you try to think plainly the answer is 1/2 = 50%. Infact: given you have seen a girl what is the probability for the other child to be a girl? Nothing else that 1/2. But, what's the strange about this reasoning? It's the first obvious answer given here.riemannhttps://www.blogger.com/profile/07014580248413882424noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-88218819121485785972008-04-11T19:43:00.000-07:002008-04-11T19:43:00.000-07:00Dr Armstrong: Touché! :DDr Armstrong: Touché! :Dhydrahttps://www.blogger.com/profile/04730791838590072890noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-4354674631844140352008-04-11T17:57:00.000-07:002008-04-11T17:57:00.000-07:00Vishal: the technical term is "red herring".Vishal: the technical term is "red herring".Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-36852341465125791782008-04-11T15:18:00.000-07:002008-04-11T15:18:00.000-07:00#3 is just crap! :) Unless someone points out some...#3 is just crap! :) Unless someone points out some obvious thing that I have been missing all along!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-66366601953931464822008-04-11T14:46:00.000-07:002008-04-11T14:46:00.000-07:00I understand! So what's your answer for #3?I understand! So what's your answer for #3?riemannhttps://www.blogger.com/profile/07014580248413882424noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-32003125586084726662008-04-11T14:32:00.000-07:002008-04-11T14:32:00.000-07:00riemann: if we take {FF, FM, MF} as our sample spa...riemann: if we take {FF, FM, MF} as our sample space, then the probability that both the children are girls is of course 1/3, as you will agree.<BR/><BR/>However, if we choose {FF, FM, MF, MM} as our sample space, then we are computing the conditional probability P({FF}/{FF, FM, MF}), which is again 1/3. So, our answers will be the same! It's just that I am calculating the probability using a different route!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-43311913710367485412008-04-11T14:04:00.000-07:002008-04-11T14:04:00.000-07:00are you sure? #2 reads:2. You know that a certain ...are you sure? #2 reads:<BR/><I>2. You know that a certain family has two children, and that at least one is a girl. But you can’t recall whether both are girls. What is the probability that the family has two girls — to the nearest percentage point?</I><BR/><BR/>So in this case isn't the sample space = {MF,FM,FF}?riemannhttps://www.blogger.com/profile/07014580248413882424noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-27618689783876035172008-04-11T13:48:00.000-07:002008-04-11T13:48:00.000-07:00riemann: I see what you are saying. But, my earlie...riemann: I see what you are saying. But, my earlier point had to do with first constructing the sample space, i.e. determining the set of of all possible outcomes before any observation is made. In my solution, the sample space I provided was {MM, MF, FF, FM}. But, I couldn't determine the sample space in your solution. If you could provide one, then I think our disagreement will be over. And, I do agree we could have different answers depending on what sample spaces we choose.<BR/><BR/>(My solution is really for problem #2. Problem #3, I admit I find it confusing.)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-16872005593869031982008-04-11T13:02:00.000-07:002008-04-11T13:02:00.000-07:00Anyway you cannot consider the possibility MM anym...Anyway you cannot consider the possibility MM anymore 'cause your observation has excluded it! I think you were talking about "a priori possibility" which is not the case.<BR/>So conditional probability takes you to the right answer...riemannhttps://www.blogger.com/profile/07014580248413882424noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-28973826039713320812008-04-11T12:46:00.000-07:002008-04-11T12:46:00.000-07:00Hmmm... my solution is indeed for problem 2. But, ...Hmmm... my solution is indeed for problem 2. But, I really wonder how the clause/condition "girl with a very unusual name" in problem 3 changes anything at all! Of course, I could be missing something, in which case, it would be great if Isabel threw some light on that strange condition!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-7219712004885058052008-04-11T12:37:00.000-07:002008-04-11T12:37:00.000-07:00Let me think another bit... I have not yet seen th...Let me think another bit... I have not yet seen the other two questions.riemannhttps://www.blogger.com/profile/07014580248413882424noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-20188792385270124392008-04-11T10:58:00.000-07:002008-04-11T10:58:00.000-07:00Hmmm, so far the values 1/3, 2/3, and 1/2 have bee...Hmmm, so far the values 1/3, 2/3, and 1/2 have been ventured as the answer. I think it's worth mentioning that in the Number Guy quiz, items 2 and 3 are very similar (#3 is the one Isabel quotes). If it looks like your model of the problem would fit #2, then I think it's safe to say that your answer will be wrong for #3. (That is, if you accept that those are indeed two distinct questions. The way #3 is worded, I think it's somewhat ambiguous, although I've seen other versions of this problem stated where it is unambiguous what the interpretation should be.)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-61199675127530838262008-04-11T09:12:00.000-07:002008-04-11T09:12:00.000-07:00riemann: And, I wish to add one more (subtle?) poi...riemann: And, I wish to add one more (subtle?) point. We usually construct the sample space <EM>before</EM> making any "observation" (which in this problem is the same as observing that I saw one female child).<BR/><BR/>I think in your solution, you are constructing the sample space <EM>after</EM> making that observation!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-33917410720568132612008-04-11T08:59:00.000-07:002008-04-11T08:59:00.000-07:00Dr Armstrong: Thanks for helping me clarify that p...Dr Armstrong: Thanks for helping me clarify that point! <BR/><BR/>riemann: I read that problem in the following way. The sample space is the set S = {MM, MF, FF, FM}, which takes (in fact, should take) into account if a sibling is younger or older. Now, the statement "given that I saw one female, what is the probability that both are females" is equivalent to computing the probability P({FF}/A), where A = {MF, FF, FM}, <EM>i.e.</EM> we are asking the question, given that event A has occurred, what is the probability of the occurrence of event {FF}? And, from here, it is easy to calculate the required "conditional probability'.<BR/><BR/>-----------<BR/><BR/>I am not saying you are wrong! I think you have constructed the sample space in a different manner, and hence your final answer will be different. In fact, having different sample spaces for the same problem may lead to different answers as is evident in the famous <A HREF="http://en.wikipedia.org/wiki/Bertrand%27s_paradox_(probability)" REL="nofollow">Bertrand's paradox</A>.hydrahttps://www.blogger.com/profile/04730791838590072890noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-59326237486717012732008-04-11T08:26:00.000-07:002008-04-11T08:26:00.000-07:00OK! F = Female, M = Male. Sorry!OK! F = Female, M = Male. Sorry!riemannhttps://www.blogger.com/profile/07014580248413882424noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-77184121850043776162008-04-11T08:21:00.000-07:002008-04-11T08:21:00.000-07:00riemann: Vishal is interpreting "FM" as "older F, ...riemann: Vishal is interpreting "FM" as "older F, younger M" rather than "one F, one M".Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-15073761699867914142008-04-11T08:17:00.000-07:002008-04-11T08:17:00.000-07:00I don't understand your remark, the question is no...I don't understand your remark, the question is not about the age but about the possibility the family has two girls.riemannhttps://www.blogger.com/profile/07014580248413882424noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-50252998842014493412008-04-11T08:10:00.000-07:002008-04-11T08:10:00.000-07:00riemann: That's exactly what I said:"The only sist...riemann: That's exactly what I said:<BR/><BR/>"The <EM>only</EM> sister (which is the same as the <EM>unique</EM> sister) but then now I need to consider the two obvious cases: younger <EM>and</EM> older!"Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-66906306933192024652008-04-11T03:22:00.000-07:002008-04-11T03:22:00.000-07:00No. I told the "unique" sister not the "older" sis...No. I told the "unique" sister not the "older" sister. <BR/>(I'll give a look at your nice blog)<BR/>Byeriemannhttps://www.blogger.com/profile/07014580248413882424noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-60728256816915036232008-04-11T02:26:00.000-07:002008-04-11T02:26:00.000-07:00riemann: Using your argument, option 3 should actu...riemann: Using your argument, option 3 should actually read,<BR/><BR/>3) FM -> I have seen the only sister who is the older one.<BR/><BR/> to which you will need to add a fourth option:<BR/><BR/>4) MF -> I have seen the only sister who is younger. <BR/><BR/>Then, we get 2/4 = 50%Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-42878384433517715602008-04-10T01:23:00.000-07:002008-04-10T01:23:00.000-07:00ERRATA:(IMHO) the answer is 0.66but these are the ...ERRATA:<BR/>(IMHO) the answer is 0.66<BR/>but these are the possibilities:<BR/>1) FF -> I've seen the first of two sisters<BR/>2) FF -> I've seen the second of two sisters<BR/>3) FM -> I've seen the only sister<BR/><BR/>so -> 2/3 -> 0.66riemannhttps://www.blogger.com/profile/07014580248413882424noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-42875878247145455512008-04-09T13:06:00.000-07:002008-04-09T13:06:00.000-07:00Sure is 1/3 = 0.333333...Infact these are the poss...Sure is 1/3 = 0.333333...<BR/><BR/>Infact these are the possibilities:<BR/><BR/>MF - FM - FF -> 1/3Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-65506416746578671572008-04-09T07:09:00.000-07:002008-04-09T07:09:00.000-07:00Anonymous,It certainly is not 50% !Anonymous,<BR/><BR/>It certainly is <EM>not</EM> 50% !Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-86927209656879368432008-04-09T04:06:00.000-07:002008-04-09T04:06:00.000-07:00Forget about names. Just write:"You know that a ce...Forget about names. Just write:<BR/>"You know that a certain family has two children, among them a girl which is blind, but you can't recall . . ."Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-62528050861146270762008-04-08T19:56:00.000-07:002008-04-08T19:56:00.000-07:00The answer I got doesn't seem very surprsing to me...The answer I got doesn't seem very surprsing to me, so I'll look forward to being surprised next week.Anonymousnoreply@blogger.com