tag:blogger.com,1999:blog-15005476.post7567287901191667280..comments2023-05-19T09:11:39.993-04:00Comments on Rationally Speaking: Julia's PicksUnknownnoreply@blogger.comBlogger25125tag:blogger.com,1999:blog-15005476.post-17568240217982506022010-07-08T02:38:00.260-04:002010-07-08T02:38:00.260-04:00Hector,
If all I say is that I flipped two coins,...Hector,<br /><br />If all I say is that I flipped two coins, then there are 4 possibilities (HH,HT,TH,TT), and the probability of HH is 1/4.<br /><br />But I provide additional information, which reduces the possibilites. <br /><br />If I say that both coins came up heads, there is only one possibility (HH), and its probability is 1. Agree?<br /><br />Or if I say that the first coin came up heads, there are 2 possibilities (HH,HT), and the probability of HH is 1/2.<br /><br />Or if I say that at least one of the coins came up heads, there are 3 possibilities (HH,HT,TH), and the probability of HH is 1/3.Maxhttps://www.blogger.com/profile/12483245818327188536noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-7059213737984697182010-07-05T22:10:38.099-04:002010-07-05T22:10:38.099-04:00Max, according to your description of the problem,...Max, according to your description of the problem, it is 1/4, not 1/3. You have FOUR possibilities (HH,HT,TH,TT) and only one is HH.Hector M.https://www.blogger.com/profile/10008738285159771679noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-44845926633764088592010-07-05T19:27:19.329-04:002010-07-05T19:27:19.329-04:00For simplicity, just flip two coins.
I flipped two...For simplicity, just flip two coins.<br />I flipped two coins.<br />(4 possible outcomes: HH, HT, TH, TT)<br />At least one of the coins came up heads.<br />(Rules out TT, leaving 3 possible outcomes: HH, HT, TH)<br />What's the probability that both coins came up heads (HH)?<br />HH is one of 3 possible outcomes, so the probability is 1/3.Maxhttps://www.blogger.com/profile/12483245818327188536noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-64327185443977392102010-07-05T13:01:38.603-04:002010-07-05T13:01:38.603-04:00Hector,
You are absolutely right when you say tha...Hector,<br /><br />You are absolutely right when you say that consecutive events are independent. The result of throwing a die does not depend on the results of the previous throws; same for boys and girls (ignoring possible biological factors).<br /><br />But this is not what I am talking about in the experiment I describe in my previous comment. In this case you throw two dice, check if <i>any one of them</i> is a five and keep only these throws. In the case of boys and girls you assume two children, check if <i>any one of them</i> is a boy and you consider only these pairs. You can simulate the experiment on a computer if you wish and you will get values close to the theoretical probabilities of 1/11 for dice and 1/3 for boys & girls (if you do it precisely as I described). I expect at least a few thousands throws are needed for the dice case before values stabilize enough.JPhttps://www.blogger.com/profile/12609837930361362269noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-91200517149597538132010-07-05T09:55:32.862-04:002010-07-05T09:55:32.862-04:00JP, your example is IMHO not good. For compound pr...JP, your example is IMHO not good. For compound probabilities to apply, all events should not have occurred (they must be all only possibilities). If you have already a child (say, a boy), that is a fact occurred in the past, and has no probability at all (or its probability equals one). The next child, instead, is affected by probability (1/2). The same with die throwing: if you already got a 5, the next throw has only 1/6 chances of getting a 5 (or any other number).<br />Compound probabilities apply only to still uncertain outcomes. E.g., the probability of getting two girls in two different pregnancies is about 0.5 x 0.5 = 0.25. But if you already have one baby girl, the prob of the next being a girl is 0.5. The circumstances of the previous births or die throws are irrelevant.Hector M.https://www.blogger.com/profile/10008738285159771679noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-63205150742568727722010-07-05T07:38:09.375-04:002010-07-05T07:38:09.375-04:00Jekrmj,
There are 36 possible outcomes of throwing...Jekrmj,<br />There are 36 possible outcomes of throwing two dice, all having the same probability (1/36). Among these, 55 occurs once. We agree on this, as you said the same in a previous post.<br /><br />I interpret the problem in terms of the following experiment (maybe you have something else in mind): someone throws two dice repeatedly. In the long term, he will get each of the 36 pairs about 1/36 of the time (including the now unique 55). Among these 36, keep only the ones for which either die shows a 5 (these are the ones satisfying the condition of the problem). There are 11 such cases (because 55 is only one of the 36 possibilities), each having the same probability. Among these 11, only one shows two fives, yielding the 1/11 probability.JPhttps://www.blogger.com/profile/12609837930361362269noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-69794729876803155892010-07-05T06:41:50.594-04:002010-07-05T06:41:50.594-04:00JP:
Not 11, but 12 possibilities: 15, 25, 35, 45,...JP:<br /><br />Not 11, but 12 possibilities: 15, 25, 35, 45, 55, 56, 51, 52, 53, 54, 55, 65. <br />Apparently you discard one 55 because you think that this would amount to counting double. No, it is not. The series consists of two subseries, one where the first die was a 5 and one where the second was a 5. (The pivotal word in your problem is <i>either</i>.) <br /><br />In both cases there is a chance of 1:6 that the other die too, is a 5. <br />2:12= 1/6. <br /><br />If 55 is double, you might as well say that 15 and 51 is double. OK, then we also remove 52, 53, 54, and 56, (because of 25, 35 ,45, and 65). Leaves us 15 25 35 45 55 56. <br /><br />Probability: 1/6.<br /><br />Q.E.D.<br /><br />(This is how you arrived at 13/27 for the original example. The 'Tuesday' thing must be a similar ploy. I call swindle!)Jekrmjhttps://www.blogger.com/profile/18276854898593567192noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-56448594087485206042010-07-04T20:45:51.516-04:002010-07-04T20:45:51.516-04:00Jekrmj, Max,
The equivalent problem with dice is t...Jekrmj, Max,<br />The equivalent problem with dice is the following: <i>throw two dice; if either one is 5 what is the probability that the other one is also a 5</i>? There are 11 possible results, all having the same probability (1/36): 15, 25, 35, 45, 51, 52, 53, 54, 55, 56, 65. Among these, only 1 has two fives. Thus, the probability is 1/11.JPhttps://www.blogger.com/profile/12609837930361362269noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-23884075071373984422010-07-03T16:06:54.435-04:002010-07-03T16:06:54.435-04:00Max,
it doesn't matter. Each pregnancy is a wh...Max,<br />it doesn't matter. Each pregnancy is a whole new game, with about 1/2 chances of bearing a boy. Other children (before or after) may have been male or female, born on Tuesday or Friday or any other day: any pregnancy has 1/2 probability of bearing a boy. All the rest is irrelevant.<br />Even if the mother has already given birth to a boy, this is an accomplished fact, not a probability. Thus the probability of a boy in the next pregnancy is NOT the compound probability of one boy + another boy: the first one has already been born, and thus the other is a completely different affair, with just 1/2 chances of being male.<br />Of course, the actual chance is not 0.50 but about 0.503 because of the slightly higher probability of male births.Hector M.https://www.blogger.com/profile/10008738285159771679noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-91739014967150292112010-07-03T12:40:16.046-04:002010-07-03T12:40:16.046-04:00Jekrmj,
Your examples are analogous to saying, &q...Jekrmj,<br /><br />Your examples are analogous to saying, "My first (or older) child is a boy. What's the probability that my second child is also a boy?" The answer is 1/2. But Julia didn't say which child was first, second, older, or younger.Maxhttps://www.blogger.com/profile/12483245818327188536noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-52002379134166161792010-07-02T17:59:43.463-04:002010-07-02T17:59:43.463-04:00Max,
(1a) I just threw a die and it was a 5. What...Max,<br /><br />(1a) I just threw a die and it was a 5. What is the chance that at the next throw I will again throw a 5?<br />(1b) I just threw a die and it was a 5. What is the chance that the previous throw was also a 5? (Having nothing else on my mind, I throw dice all day long, day after day.)<br />The answer is 0.5 and 0.5, respectively. The average of these two values is 0.5.Jekrmjhttps://www.blogger.com/profile/18276854898593567192noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-27321786833998963112010-07-01T22:59:39.197-04:002010-07-01T22:59:39.197-04:00Discussed this at work and at dinner - I stand cor...Discussed this at work and at dinner - I stand corrected. The single answer was agreed by most to be <br /><br /> "either 1/2 or 13/27"<br /><br />, depending on why you were given this info pertaining to the child's sex and birth-day-of-week.DaveShttps://www.blogger.com/profile/15840516954793215700noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-31271997843807892282010-07-01T17:43:23.203-04:002010-07-01T17:43:23.203-04:00Jekrmj,
Your first example gives the order of the...Jekrmj,<br /><br />Your first example gives the order of the events, but Julia didn't say that the "other" child is the younger one.Maxhttps://www.blogger.com/profile/12483245818327188536noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-22350327239206285162010-07-01T16:40:33.512-04:002010-07-01T16:40:33.512-04:00Max,
Compare:
(1) I just threw a die and it was a...Max, <br />Compare:<br />(1) I just threw a die and it was a 5. What is the chance that at the next throw I will again throw a 5?<br /><br />to:<br /><br />(2) What is the chance that I will throw two fives in two throws?<br />Answer to (1): 1/6; answer to (2): 1/36. <br />These calculations are extrapolations from what we know to what we donâ€™t. Therefore, by removing knowledge, we change the outcome.<br /><br />Obviously, I might see it in a totally different way if I were not born on a Thursday. Let me guess, are you born on a Saturday?Jekrmjhttps://www.blogger.com/profile/18276854898593567192noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-42342295982426805662010-07-01T07:47:38.200-04:002010-07-01T07:47:38.200-04:00The Venn diagram isn't clever enough to warran...The Venn diagram isn't clever enough to warrant the insult to Christians. I wouldn't be so sensitive about this, except that I would like them to read this blog and the diagram is not helping. However, if it were an extremely clever Venn diagram, I may feel differently.Austin Landhttps://www.blogger.com/profile/07734213784566148226noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-63958383168660114612010-07-01T04:22:09.246-04:002010-07-01T04:22:09.246-04:00jekrmj,
Given that one of my two children is a bo...jekrmj,<br /><br />Given that one of my two children is a boy, if the other one is a boy, then I have two boys. It's saying the same thing.Maxhttps://www.blogger.com/profile/12483245818327188536noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-81508083975213267912010-06-30T19:11:38.989-04:002010-06-30T19:11:38.989-04:00As with all such probability questions it is neces...As with all such probability questions it is necessary to formulate the problem more precisely before finding the answer. As stated, the problem has many interpretations. Maybe the most natural formulation (others are discussed in the link) is this: <i>Among all 2-child families with at least a boy born on a Tuesday, how many have two boys?</i> The proportion would be the probability. To find out is is simple enough: simply enumerate all the possibilities and count. In this case, there 14 cases for both the first and second child giving a total of 14*14 = 196 possibilities, each having the same probability (assuming a 50-50 distribution of boys versus girls, equal probability for each day of the week and no twins). If one cares to count, one will find 27 cases with at least a boy born on a Tuesday, among which 13 with two boys, giving the probability of 13/27 as stated in the link. There are easier ways to get to the number 27 however: count 14 cases with the first child satisfying the condition and 14 for the second, giving 28. One case is counted twice (both children satisfy), giving 27. In fact it is the presence of this double case that explains the difference between the calculated probability and the "intuitive" one of 1/2.<br /><br />For the mathematically inclined here is a small generalization. If there are <i>n</i> possibilities for each boy or girl (n=7 in our case), the resulting probability will be (2n-1)/(4n-1). So, if the question is <i>I have two children, one of whom is a boy born on June 30th, what is the probability that I have two boys?</i>, the answer (assuming 365 days a year) is 729/1459 ((2*365-1)/4*365-1), which is more than 49.9%. The more possibilities we have, the closer we get to one half. Returning to the original question (<i>If I have two children, one of whom is a boy, what is the probability that I have two boys?</i>), n is 1, giving a probability of 1/3.JPhttps://www.blogger.com/profile/12609837930361362269noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-11651579168781474492010-06-30T14:26:31.895-04:002010-06-30T14:26:31.895-04:00I know it's hard to believe, but the specifica...I know it's hard to believe, but the specification that the one of the babies was a boy born <i>on a Tuesday</i> actually is relevant to the probability the other one is a boy.Julia Galefhttps://www.blogger.com/profile/05020069129381463375noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-68462137754391525602010-06-30T14:04:48.611-04:002010-06-30T14:04:48.611-04:00what jekrmj saidwhat jekrmj saidDaveShttps://www.blogger.com/profile/15840516954793215700noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-11009944745024499462010-06-30T11:42:12.895-04:002010-06-30T11:42:12.895-04:00"The reason women are more selective than men..."The reason women are more selective than men in the world of dating may be partly because of the convention that men are expected to ask women out, and not vice versa."<br /><br />But the convention has probably emerged a RESULT of the choosiness of women, which in turn is a well-known consequence of our genetic make up. Mammal females make the bigger parental investment, whilst males make a much smaller one, so that natural selection would have favored the kind of male that tries to "do it" with every female, and rapidly forget about further responsibilities, while at the same time favoring choosy females looking for the more caring males, who are more likely to care for children (besides looking for signs of reproductive capacity and good health). This would have produced the kind of males and females we actually see around.Hector M.https://www.blogger.com/profile/10008738285159771679noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-63329278189823097132010-06-30T11:35:02.399-04:002010-06-30T11:35:02.399-04:00" which is worse, a large harm for a few, or ..." which is worse, a large harm for a few, or a tiny harm for many?"<br /><br />Worse for whom? Possibilities:<br />From the point of view of the few, their situation is worse. To the many, the other situation is worse. If you do not know in which situation you are, there are two answers:<br />1. According to the economic theory of rational choice, it would depend on the expected harm, which in turn is a function of the actual magnitudes of harm and the probability of either outcome. The expected harm (or disutility) is p(H) x amount of harm + [1 - p(H) x zero harm, where p(H) is the probability of being harmed in either situation. Which is preferred, in this approach, depends on the actual numbers involved. For instance, a 0.01 probability of suffering a loss of $1000 (and zero otherwise) has an expected disutility of $10. A 0.80 probability of suffering a loss of $10 has an expected outcome of $8, and therefore the second is better. But if the losses were different (say $800 and $10) the preference would be reversed.<br />On the other hand, experimental results (Kahnemann and Tversky, and others) show that people have a stronger aversion to big losses, even if they have very small probability of occurring. This loss aversion factor may tilt the preferences towards the small loss for many, even if the eventual (average) outcome may be worse in that case. These experimental results suggest there would be a bias towards the small-loss game, more than is granted by the probabilities and rewards.<br /><br />Another look at the problem is from the SOCIAL point of view. Which is better FOR SOCIETY AS A WHOLE? But establishing a "rational" social ordering of preferences that fulfills certain very simple conditions is IMPOSSIBLE (Arrow's impossibility theorem, 1950). Thus the latter approach (social preference) may be nonsensical since the question is not answerable.Hector M.https://www.blogger.com/profile/10008738285159771679noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-71785655194642172672010-06-30T11:19:40.871-04:002010-06-30T11:19:40.871-04:00I agree with jekrmj: the sex of every newborn is d...I agree with jekrmj: the sex of every newborn is determined at conception, quite independently of previous (or successive) siblings, just like a new throw of a coin is independent of previous throws.<br />The actual probability of a boy is somewhat higher than 0.5, i.e. approx. 0.503, because slightly more males than females are born. The reason for the difference is apparently that XX sperm are slightly heavier (and therefore slightly slower) than their XY counterparts, and therefore the latter have a slightly higher probability of reaching the egg and fertilize it.Hector M.https://www.blogger.com/profile/10008738285159771679noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-55643701390143549292010-06-30T06:40:08.873-04:002010-06-30T06:40:08.873-04:00If you say "I have two children, one of whom ...If you say "I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?" the answer is 1/3; the way you phrase it, the answer is 1/2.Jekrmjhttps://www.blogger.com/profile/18276854898593567192noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-57404823485636755682010-06-29T23:51:04.562-04:002010-06-29T23:51:04.562-04:00The idea that a game can be both boring and addict...The idea that a game can be both boring and addictive has been overlooked. It is simply not true that people always play games because they find them exciting. I've had numerous experiences in my life in which I kept doing something even though I did not have to and did not find it even remotely exciting.Ritchie the Bearhttps://www.blogger.com/profile/10249784344018510589noreply@blogger.comtag:blogger.com,1999:blog-15005476.post-54195409305809035272010-06-29T23:03:05.376-04:002010-06-29T23:03:05.376-04:00"Everything depends, he points out, on why I ...<i>"Everything depends, he points out, on why I decided to tell you about the Tuesday-birthday-boy."</i><br /><br />Neat puzzle. Allow me to volley the ball back to Philosophy... This is a case of <a href="http://en.wikipedia.org/wiki/De_dicto_and_de_re" rel="nofollow">de dicto</a> versus <a href="http://en.wikipedia.org/wiki/De_dicto_and_de_re" rel="nofollow">de re</a>, right? <br /><br />De dicto, it interprets as "a-boy-born-on-a-Tuesday"; De re, as "a boy (born on a Tuesday)." Did I get that right?I.Strangehttps://www.blogger.com/profile/15689899019458033716noreply@blogger.com