Even Odds

The laws of probability, so true in general, so fallacious in particular.
–Edward Gibbon

A puzzle has recently been brought to my attention. It goes like this: “I have two children. One of them is a boy born on a Tuesday. What is the probability that I have two boys?” This puzzle was posed to an audience at the Gathering for Gardner meeting in Atlanta, Georgia, by Gary Foshee. He went on to say, “The first thing you think is ‘What has Tuesday got to do with it?’ Well, it has everything to do with it.”
puzzle
I disagree. The day the boy was born is utterly irrelevant. Perhaps it would be best to look at the “accepted” solution from New Scientist before I tell you why it’s hogwash. Its main point is:

The main bone of contention was how to properly interpret the question. The way Foshee meant it is, of all the families with one boy and exactly one other child, what proportion of those families have two boys?
To answer the question you need to first look at all the equally likely combinations of two children it is possible to have: BG, GB, BB or GG. The question states that one child is a boy. So we can eliminate the GG, leaving us with just three options: BG, GB and BB. One out of these three scenarios is BB, so the probability of the two boys is 1/3.


Wrong already. The probability of two boys is not 1/3, but 1/2. The reason for this is, to me, glaringly obvious. The possible combinations are not BB, GG, BG and GB, because GB and BG are, in the context of this problem, identical. The order in which the children are born is irrelevant. The possibilities are therefore BB, GG and one-of-each. We can, indeed, eliminate GG because we already have one boy, so that just leaves an equal probability of two boys or one-of-each, or a 50-50 chance of two boys.

I’ll simplify further: if I tell you I flipped a coin twice, and one of the results was heads, what is the probability that I flipped two heads? I could obfuscate the problem by telling you that the head on the coin belonged to Tiberius, and that there was a buffalo on the tails side, and that the rim was embossed with a floral pattern, but none of those things would be relevant if they did not alter the balance of the coin. The chances of my having flipped two heads would be 1/2.

Similarly, the day the boy was born on is irrelevant obfuscation, as is the order in which the children were born. Here is why the New Scientist thinks the day is relevant (note that the final solution they arrive at is an approximation of the solution I arrived at without having to twist myself into a mental pretzel):

Now we can repeat this technique for the original question. Let’s list the equally likely possibilities of children, together with the days of the week they are born in. Let’s call a boy born on a Tuesday a BTu. Our possible situations are:
 When the first child is a BTu and the second is a girl born on any day of the week: there are seven different possibilities.
 When the first child is a girl born on any day of the week and the second is a BTu: again, there are seven different possibilities.
 When the first child is a BTu and the second is a boy born on any day of the week: again there are seven different possibilities.
 Finally, there is the situation in which the first child is a boy born on any day of the week and the second child is a BTu – and this is where it gets interesting. There are seven different possibilities here too, but one of them – when both boys are born on a Tuesday – has already been counted when we considered the first to be a BTu and the second on any day of the week. So, since we are counting equally likely possibilities, we can only find an extra six possibilities here.
Summing up the totals, there are 7 + 7 + 7 + 6 = 27 different equally likely combinations of children with specified gender and birth day, and 13 of these combinations are two boys. So the answer is 13/27, which is very different from 1/3.

Yes, 1/3 was wrong in the first place, and 13/27 is a quite close approximation to the correct answer, which is 1/2.

I rest my case.

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Grumpy Old Man by Mark Widdicombe is licensed under a Creative Commons Attribution-Noncommercial-No Derivative Works 2.5 License.

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64 Responses to Even Odds

  1. Con-Tester says:

    Mark, I’m sorry to say when you write that, “The possible combinations are not BB, GG, BG and GB, because GB and BG are, in the context of this problem, identical” and then that, “The possibilities are therefore BB, GG and one-of-each”, and conclude from there that one is dealing with three equiprobable outcomes, you are in fact committing a grave error. Simply, you’re neglecting the fact that in your three-outcomes scenario, the one-of-each combination occurs twice as often as either one of BB or GG, and so 1/3 is indeed the correct answer.

    Another way of solving it is by resort to the conditional probability formula, P(X | Y) = P(X ∩ Y)/P(Y). (In words: The probability that X is true given that Y is true, is equal to the probability that both X and Y are true, divided by the probability the Y is true.) Let X = “Both children are boys” and Y = “At least one child is a boy”. Clearly P(Y) = 3/4 (since three of the four equiprobable possibilities in the sample space {BB, BG, GB, GG} satisfy Y) and P(X ∩ Y) = P(X) = 1/4 (X ∩ Y is automatically true if X is true). Therefore, P(X | Y) = (1/4)/(3/4) = 1/3.

    The New Scientist considerations re “BTu” are curious and counterintuitive, but all the more illuminating for it. To see why, try the problem by specifying that one of the two children is a boy born during daylight hours (6:00 AM and 6:00 PM) without specifying the weekday. If you follow the New Scientist method, your answer will become (2+1)/(2+2+2+1) = 3/7. Similarly, if you specify a birth month, the answer becomes (12+11)/(12+12+12+11) = 23/47. If you now extrapolate this approach and say the boy was born at any time on any weekday, the answer becomes (1+0)/(1+1+1+0) = 1/3, which is not a coincidence. In other words, the answer will depend on how you specify the time-point of the one boy’s birth. What’s going on here is that placing increasing restrictions on the birth time-point of the one boy effectively brings the situation increasingly closer to asking what the probability is that the other child is a boy (P = 1/2) because the sex+weekday sample space for the specified boy diminishes significantly relative to that for the other child. In the limit where you specify the birth time-point with 100% accuracy, you will calculate a probability of exactly 1/2.

    You can test all of this with simple Monte Carlo simulations of several million samples where you randomly generate two genders and weekdays, counting the various combinations and working out the ratios which will approach the theoretical probabilities.

  2. JeffJo says:

    There are two possibilities for this lottery ticket of mine: it could win a million bucks, or it could win nothing. So it seems that I have a 50% chance that it will win that cool million. A good investment for one buck, no?

    This is an extreme example of the misapplication of the Principle of Indifference. It says that if a set of outcomes result from symmetric random occurrences, then they must have equal probability. The misapplication occurs when you ignore the “symmetric” part, and merely count the cases that fit in various categories which have nothing to do with symmetry.

    Winning the lottery requires matching every number, while you lose if any single number is wrong. That’s not symmetric – if there are six digits involved (hey, I can assume a fair lottery if I want to), the first can happen in only 1 way. The second happens in 999,999 different ways that are symmetric with the winning combination.

    The cases where a family of two has two boys, one boy and one girl, and two girls, are not symmetric, either. Two boys and two girls can each occur in one only way, but there are two ways to have a boy and a girl. The probability of each ORDERED combination is 1/4.

    “But order is not a part of the problem, so how can it matter?”, I hear you cry. The answer is that what the problem requires is irrelevant in this case: the Principle of Indifference requires symmetry, and you can get that only through some kind of ordering. If you don’t want to use age, alphabetize the children’s names, or see where they sit clockwise from Mother at the dinner table. The “first” one has a 1/2 chance to be a boy. So does the second. That means the chances of two boys is (1/2)*(1/2). But you get one of each if the first is a boy and the second a girl, or vice versa.

    But the answer of 1/3 is, indeed, wrong. That’s because there are six possible combinations involved here, not four. The “first” child could be a boy or a girl, the “second” could be a boy or a girl, and I COULD TELL YOU ABOUT A BOY OR A GIRL. Assuming I won’t lie, in the 1/4 of all cases where I have two boys, I must tell you about a boy. But in only half of the 1/2 cases where I have one of each will I tell you about the boy. So the chances are equal either way.

    And using this method, the answer won’t change if I tell you the boy was born on Tuesday, or is named Indiana, or is a left-handed, red-haired, weight-lifting boy who was born on a Saturday (which, incidentally, one of my two children is). I had to choose a child before I could formulate a set of facts about him or her, and doing so can’t change the fact that the probability both of my children share the same gender is 1/2.

  3. Con-Tester says:

    Here’s a Delphi function that does a Monte Carlo simulation with 100,000,000 samples of the first part of the problem (i.e. any boy):

    function  MCProbTest01() : Double;
    var       i,
              Child1,
              Child2,
              n1Boy,
              n2Boys  : Integer;
    begin
      // Seed the random number generator
      Randomize();
    
      // Initialise counters
      n1Boy := 0;  // No. of random samples with at least 1 boy
      n2Boys := 0; // No. of random samples with two boys
    
      // Monte Carlo loop
      for i := 1 to 100000000 do
      begin
        Child1 := Random(2); // 0 = girl, 1 = boy
        Child2 := Random(2); // 0 = girl, 1 = boy
        if ((Child1 = 1) or (Child2 = 1)) then
        begin
          // At least one of the two children is a boy
          n1Boy := n1Boy+1;
          if ((Child1 = 1) and (Child2 = 1)) then
            // Both children are boys
            n2Boys := n2Boys+1;
        end;
      end;
    
      // Calculate aggregate Monte Carlo probability
      Result := n2Boys/n1Boy;
    end;

    Here are the results of 10 runs of the above together with the relative errors from the true answer of 1/3:

    0.33338121     +0.0144%
    0.33332612     -0.0022%
    0.33335591     +0.0068%
    0.33338386     +0.0152%
    0.33340081     +0.0202%
    0.33335161     +0.0055%
    0.33340067     +0.0202%
    0.33330235     -0.0093%
    0.33332976     -0.0011%
    0.33333223     -0.0003%

    The average of these 10 runs comes to 0.333356453, an error of less than 70 parts per million.

      
      
      
    

    Here’s a Delphi function that does a Monte Carlo simulation with 100,000,000 samples of the second part of the problem (i.e. at least one boy is born on a Tuesday):

    function  MCProbTest02() : Double;
    var       i,
              Child1,
              Child2,
              nTueBoy,
              nTwoBoys  : Integer;
    begin
      // Seed the random number generator
      Randomize();
    
      // Initialise counters
      nTueBoy := 0;  // No. of rand samples with at least 1 Tues B
      nTwoBoys := 0; // No. of rand samples with 2 B
    
      // Monte Carlo loop
      for i := 1 to 100000000 do
      begin
        Child1 := Random(14); // 0-6 = G, 7-13 = B, 13 = Tues B
        Child2 := Random(14); // 0-6 = G, 7-13 = B, 13 = Tues B
        if ((Child1 = 13) or (Child2 = 13)) then
        begin
          // At least one of the two children is a Tues boy
          nTueBoy := nTueBoy+1;
          if ((Child1 >= 7) and (Child2 >= 7)) then
            // Both children are boys
            nTwoBoys := nTwoBoys+1;
        end;
      end;
    
      // Calculate aggregate Monte Carlo probability
      Result := nTwoBoys/nTueBoy;
    end;

    Here are the results of 10 runs of the above together with the relative errors from the true answer of 13/27:

    0.48179929     +0.0660%
    0.48146023     -0.0044%
    0.48146421     -0.0036%
    0.48119605     -0.0593%
    0.48148890     +0.0015%
    0.48151712     +0.0074%
    0.48152591     +0.0092%
    0.48128200     -0.0414%
    0.48155624     +0.0155%
    0.48150023     +0.0039%

    The average of these 10 runs comes to 0.481479018, an error of less than 6 parts per million.

      
      
      
    

    I have reproduced the functions here so that anyone may verify the results shown. Please feel free to point out exactly where and how these simulations fail.

  4. JeffJo says:

    Con-Tester, you have accurately simulated how the proportion of two-boy families, among two-child families that include a boy, is 1/3. And that the proportion of two-boy families, among two-child families that include a boy born on a Tuesday, is 13/27. But proportion is not the same as probability.

    To see why, let us resort to the definition for conditional probability you cited: P(X|Y) = P(X∩Y)/P(Y). Let B0, B1, and B2 be the events where the family has zero, one, and two boys, respectively. We know that P(B0) = P(B2) = 1/4, and P(B1) = 1/2. Then HB = B1∪B2 is the event where the family has at least one boy, and P(B2|HB) = P(B2∩HB)/P(HB) = P(B2)/P(B1∪B2) = P(B2)/[P(B1)+P(B2)] = (1/4)/[(1/2)+(1/4)] = 1/3.

    But at G4G, Gary Foshee said “I have two children. One of them is a boy (born on a Tuesday).” I’ll ignore the Tuesday part for now, as in the New Scientist article. This is event TB, the event where he tells you he has a boy. You assumed, without justification or even noticing that you did, that it is the same event as HB. That P(TB|HB) = 1. Using the same formula, P(TB|HB) = P(TB∩HB)/P(HB) = P(TB∩( B1∪B2))/P(B1∪B2) = [P(TB∩B1)+P(TB∩B2)]/[P(B1)+P(B2)] = [P(TB|B1)*P(B1)+P(TB|B2)*P(B2)]/ [P(B1)+P(B2)]. (The last part applied the definition in the opposite way.)

    The only way for this to equal one, is if P(TB|B1)=P(TB|B2)=1. Assuming P(T|B2)=1 is justified, but you can’t assume P(TB|B1)=1. You see, P(TB|B1) and P(TG|B1), the probability a man would tell you he has a girl when he has one child of each gender, must sum to one. If P(TB|B1) is already one, then P(TG|B1)=0. And that would make the probability that a man has two girls, given that he tells you he has at least one, 100%.

    So, go back to your first simulation. In the first, change “if ((Child1 = 1) or (Child2 = 1))” to “if ((Child1+Child2 = 2) or ((Child1+Child2 = 1) and (Random(2)=1)))”. Let the man tell us about any gender of a child in his family. You will get the correct answer of 1/2. A similar, but more complex, change to the second simulation will also yield 1/2. If you want to see a Bayesian’s analysis of this problem (well, a predecessor of it), look at this paper: http://economics-files.pomona.edu/GarySmith/Two-Child%20Paradox.pdf .

    Let me expand what you obviously decided not to address: If I have two children, the chances that they have the same gender are 1/2. If I tell you some facts, that include gender but apply to only one of the children, this probability cannot change. Unless I am biased toward telling you about one gender over the other, of course. But you can’t assume that.

    • Con-Tester says:

      Your assertion that “proportion is not the same as probability” is not correct. The probability is exactly the limiting proportion of the number of cases of interest as a proportion of the total number of possible cases in the applicable sample space. I refer you to the Wikipedia article which says:

      When dealing with experiments that are random and well-defined in a purely theoretical setting (like tossing a fair coin), probabilities describe the statistical number of outcomes considered divided by the number of all outcomes (tossing a fair coin twice will yield HH with probability 1/4, because the four outcomes HH, HT, TH and TT are possible).

      (Emphasis added). Elsewhere (STEPS Statistics Glossary), you’ll find that “the probability of the event can be defined as the limiting value of [its] relative frequency”, which is saying exactly the same thing. It is because of this definition that Monte Carlo simulations are such useful experimental validations of theoretical results.

      Where you suggest as follows:—

      In the first, change “if ((Child1 = 1) or (Child2 = 1))” to “if ((Child1+Child2 = 2) or ((Child1+Child2 = 1) and (Random(2)=1)))”. Let the man tell us about any gender of a child in his family. You will get the correct answer of 1/2.

      —you are correct in stating that the limiting probability goes to 1/2. A truth table will show that the construct “((Child1+Child2 = 2) or ((Child1+Child2 = 1) and (Random(2)=1)))” comes up true exactly half the time on average, whereas “((Child1 = 1) or (Child2 = 1))” comes up true three out of four times on average (just as it should: 3 of 4 two-child families have at least one boy), and so you have changed the problem in an obscurantist way. In addition, exactly what the additional “(Random(2)=1)” clause is meant to simulate remains an unexplained mystery, and the overall result of 1/2 would remain unaffected should you change it to “(Random(2)=0)”, which should raise considerable suspicion. However, it has been clearly stated that there are two children, both of whose genders we’ve already picked randomly, but we’re only interested in cases where at least one of them is a boy. The “if ((Child1 = 1) or (Child2 = 1))” construct caters for the latter exactly!

      Since probability is a limiting proportion and there is no purpose to the alteration you suggest other than to produce the desired result, the rest of your post fails to illuminate how the Monte Carlo simulations come so remarkably close to the theoretical answers.

    • Con-Tester says:

      On further reflection, it seems that you’re making an extraordinarily convoluted case for the possibility that the speaker may be lying. Of course, if you assume there’s a 50:50 chance that the bloke is telling the truth when he tells you he has one boy, the GG possibility isn’t entirely eliminated. But this is hardly in the spirit of the problem and changes it completely. There is no indication that we should mistrust the statement that one of them is a boy. But let’s see what happens if we do. Your replacement Boolean construct is rather different to what you suggest. You would have to add “Truth := Random(2);” after the child gender assignments where “Truth” decides 50:50 whether the speaker tells the truth or not. The qualifying sample space admission construct changes to “if (((Truth = 0) and ((Child1 = 0) and (Child2 = 0))) or ((Truth = 1) and ((Child1 = 1) or (Child2 = 1))))” — i.e. our sample space consists of cases where either he is lying (which he can only do in the event of GG), or he is telling the truth. However, with this replacement the simulation returns a probability of ~1/4, and not ~1/2.

  5. JeffJo says:

    Con-tester, you quoted the Wikipedia one of two possible interpretations – not definitions – for unconditional probability. It is more-or-less correct, although you stated it backwards. The limiting proportion is derived from the probability, not the other way around. But look just above where you misquoted it, to the part that says “the word probability does not have a singular direct definition.” Then look at the article for “Probability Axioms.”

    What I meant, for conditional probability, was that probability is not just the proportion of the cases that can exist. You have to consider the probability that each case would produce the observed result. You can learn about it in the article for Bayes Theorem, specifically in the section about the “extended form.” After translating the event names to what I used before, it says that P(B2|TB) = P(TB|B2)/[P(TB|B0)*P(B0)+P(TB|B1)*P(B1)+P(TB|B2)*P(B2)]. You tacitly assumed that the conditional probabilities on the right-hand side of this proven formula could only be zero or 1; possible or impossible. You calculated the proportion of two-boy families from among families with at least one boy, and I repeat that that is not the same thing as the probability of a two-boy family after the father told you one was a boy.

    We are not “only interested in cases where at least one of them is a boy,” we are interested in just those cases where the father tells us one is a boy. The “additional ‘(Random(2)=1)’ clause” is meant to simulate the choice the father has, if he has a boy and a girl, to tell you about the boy and ask the question about boys, or to tell you about the girl and ask the same question about two girls. This is a well-known issue in probability puzzles of this sort, going back to Joseph Bertrand in 1889, in his famous “Box Paradox.” It is identical to this problem, except there are only three cases with proportions exactly as Mark Widdicombe originally thought. The naïve answer there is based on your logic, and the answer Bertrand proved was correct is based on mine. In more modern times, the Three Prisoner’s Problem and the Monty Hall Problem are identical, and the correct solution is also based on my logic.

    And you are the one going through convoluted logic, trying to find a flaw where there is none. I never suggested, or implied in any way, that anyone was lying. You are imagining a reason for me to be wrong, and then accusing me of it. The correct logic is based solely on the fact that sometimes two similar facts are true. The father doesn’t have to tell you about a boy if he also has a girl, so you can’t count all of those cases.

    You are also completely ignoring the other arguments I gave. Try two more: the original problem goes all the way back to Martin Gardner (in 1959, I think). He originally gave your answer to a similar problem, but later that year he withdrew it. The problem wasn’t well formulated and so was ambiguous. He said, and I quote, “If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is 1/3. But there is another procedure that leads to exactly the same statement of the problem. From families with two children, one family is selected at random. If both children are boys, the informant says ‘at least one is a boy.’ If both are girls, he says ‘at least one is a girl.’ And if both sexes are represented, he picks a child at random and says ‘at least one is a …’ naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly 1/2. … That the best of mathematicians can overlook such ambiguities is indicated by the fact that this problem, in unanswerable form, appeared in one of the best of recent college textbooks on modern mathematics.”

    His version of the problem was ambiguous, but this one isn’t. Gary Foshee was not required to have a boy before he could posit this question. If he had a girl, he clearly had the option to tell us about the girl.

    Now, assume that I have two children. Which of these questions do not have an answer of 1/2?

    Q1) What is the probability they have the same gender?
    Q2) If I write the gender of a child of mine on a hidden piece of paper, what is the probability they have the same gender?
    Q3) If you assume I wrote “boy,” what is the probability they have the same gender?
    Q4) If you assume I wrote “girl,” what is the probability they have the same gender?
    Q5) What do you think the probability is that I wrote “boy?”
    Q6) How about “girl?”
    Q7) I show you the paper, and it says “boy.” What is the probability they have the same gender?

    All of what I’ve written before this can be confirmed on Wikipedia (actually, I think the Gardner quote has been deleted, but may be in archives). There’s another one: the law of Total Probability. If A1 thru A7 represent the answers to these question – all of which I claim are 1/2 – you claim either that A3=A4=1/3, that A3~=A4 which is absurd, or that Q3 is not the same as Q7 which is also absurd. But that proven law says that A3*A5+A4*A6=A1. And I don’t care what you think A5 and A6, they must add up to 1 so A3=A1.

    • Con-Tester says:

      In other words, you’ve expanded the problem to include the situation where the speaker has lied (with some entirely unspecified probability), and your Q1 to Q7 are mere sophistry and obfuscation. Very curious. Now you need to explain why you would accept that he’s telling the truth when he says that he has two children, rather than one or six or twenty, but he’s possibly lying about their gender!

      In such contrived and loose circumstances you can always invent a particular combination that meets your desired answer.

      And so you’re obviously and trivially correct.

    • Con-Tester says:

      You wrote:

      You calculated the proportion of two-boy families from among families with at least one boy, and I repeat that that is not the same thing as the probability of a two-boy family after the father told you one was a boy.

      No, you did not make this odd assertion before. But why do you think the two are not the same? Assuming that the father tells the truth about the number and gender of his children, the only possible way this can be so is if your sample space includes all possible two-child families (by including those fathers who say nothing), in which case the answer becomes 1/4 and not 1/2. But the families whose fathers would say nothing or speak about having at least one girl (with the various probabilities of saying certain things still entirely unspecified) are excluded from the sample space by the problem statement, again on the assumption that the father tells the truth about the number and gender of his children!

  6. JeffJo says:

    You don’t get it, you don’t want to get it, and you refuse to try to get it. This will be my last reply because in my experience nobody changes after two like yours.

    Nobody is lying. Everybody is telling the truth. But sometimes there are two possible truths. You have “expanded the problem” by adding the assumption that a father with a boy and a girl cannot tell you about the girl. That he will always choose the one truth you prefer over any other truth.

    I think you believe your answer because it was the only one you were taught. I’ve given you plenty of references to other schools of thought, but you ignore them or call them contrived, but you do all the contriving.

    Good day.

    • Con-Tester says:

      And to you, too.

      So now you have added those families whose fathers who might tell you about their possible girls instead, and so expanded the problem to suit your whim.

      Very good. As said, that makes you obviously and trivially correct.

    • Con-Tester says:

      Sure, the answer of 1/2 is correct if one assumes that a randomly selected BG or GB father will tell you about his boy half the time and his girl the other half. (And why couldn’t you state this simply and concisely without all the overwrought ancillary verbiage!?) In this strict sense there arguably is ambiguity in the problem as stated. But this choice the father has is clearly a convoluted and non-obvious complication in light of how the problem is phrased. Since the focus is unambiguously on boys, the implication is equally clear that all two-child fathers will say that they have a boy if it is indeed the case that they have at least one boy. The sense of the problem is that the questions to a sample two-child father, “Tell me the gender of any of your two children?” and “Do you have two boys?” are not asked in that way; rather, the sample father is asked, “Do you have at least one boy?” and only if he answers in the affirmative is he then asked, “Are both your children boys?”

      However, even in this elaborate view Mark’s treatment remains suspect for not laying it out comprehensively. He does not spell out that the number of one-of-each cases is effectively halved by the father’s discretion in volunteering gender specifics.

      Moreover, the “boy born on Tuesday” situation leads to a result of 8/15 (and not 1/2, please note) because the father can choose to tell you about his other child’s gender+birth weekday (if the other child is not a “BTu”), presumably also with a 50:50 probability. Therefore the sample space admission criterion in “function MCProbTest02()” will become “if (((Child1 = 13) and (Child2 = 13)) or (((Child1 = 13) or (Child1 = 13)) and (Random(2) = 1))) then…” — which again illustrates how jarring and forced this alternative interpretation really is.

  7. JeffJo says:

    I did tell you simply and concisely what I was talking about, right after I confirmed your analysis about the four possible combinations of children: “There are six possible combinations involved here, not four. The ‘first’ child could be a boy or a girl, the ‘second’ could be a boy or a girl, and I COULD TELL YOU ABOUT A BOY OR A GIRL. Assuming I won’t lie, in the 1/4 of all cases where I have two boys, I must tell you about a boy. BUT IN ONLY HALF OF THE 1/2 CASES WHERE I HAVE ONE OF EACH will I tell you about the boy. So the chances are equal either way.” Why you chose to ignore that, I have no idea.

    There is no implication, clear or otherwise, “that all two-child fathers will say that they have a boy if it is indeed the case that they have at least one boy.” Concluding that is, is an example of the logical fallacy called “affirming the consequent.” Look it up on Wikipedia, since you don’t like “ancillary verbiage.” The “sense of the problem” is only as you described if you first assume it is so, and then develop the rationale behind it to support that conclusion.

    Check your math on the “boy born on Tuesday” variant. Of 196 possibilities, 1 has two Tuesday Boys, 12 have two boys but only one born on a Tuesday, and 14 have a Tuesday Boy and a girl. If you assume, as you do without any attempt at justification besides claiming it is obvious, that all fathers of a Tuesday Boy will tell you that fact, the answer is (1+12)/(1+12+14)=13/27. If you assume that a father will choose a similar BUT TRUE fact randomly from the one or two that apply, the answer is (1+12/2)/(1+12/2+14/2) = (1+6)/(1+6+7) = 7/14 = 1/2.

    So please, try assuming I know what I am talking about before you decide I must be wrong. You are still assuming there must be an error, and grabbing onto the first thing you have misunderstood in what I have said as its source.

    • Con-Tester says:

      But I thought you were done with me, owing to my ostensibly obtuse intransigence!? It’s deeply ironic that you seek so simply to dismiss my “sense of the problem” elucidation when your conception is just as debatable. With the knowledge that boys (and/or those born on a Tuesday) are the centrepiece of the problem, all two-child GB or BG fathers will speak in the way Foshee poses the problem because any other question relating to the children’s genders (besides a complete boy-girl interchange) would be absurd. In contrast, your take is that GB or BG fathers are free to volunteer correct info as they wish about one of their children’s genders (and birth weekdays). I disagree that this meets the spirit of the problem.

      Thus, I am not presupposing any errors in anything. My issue is with how the problem statement is interpreted. In fact, I already conceded that the answer of 1/2 is correct on the appropriate assumptions, just as you have acknowledged that 1/3 is correct on a different set of assumptions about how the problem is to be viewed. But this narrow-minded academic doggedness cloaked in garrulous obfuscation is unproductive, and you can be similarly sure that I have a very good idea of what I’m on about.

      Be that as it may, I did make an arithmetic error in my “BTu” probability calculation earlier. On your interpretation, the probability of a two-child “BTu” father (who chooses randomly and with equal likelihood which one of his two children to tell you about) entering the “BTu” sample space is P(Y) = 1*(1/196)+(1/14)*((1/14)*(13/14)+(13/14)*(1/14)) = 1/14. Of the fathers who make it, the probability that any one of them has two boys is P(X ∩ Y) = (7/14)*(1/14) = 1/28, and so P(X | Y) = 1/2 again, as you correctly claim.

      All of which takes us right back to the start again.

    • Jacdac says:

      Dear JeffJo
      I am impressed by you effort to explain this logic, it takes a keen sharp mind to spot exactly where the logic goes astray. I admire the way you have come to math prooth where the fallacy is.
      First of I’m by no means an expert in math I just simply have a good sense of logic but I’m just a layman in math. When i encountered this problem i was sure that 13/27 could not be true and exactly as the author of the above article- btw. is that you?- wrote the point where you devide it up to these groups BG and GB was where i saw the fallacy. It doesnt make sense to add time into it and by that create a larger group.
      And by the sense that tuesday could never influence anything about the gender in any logical sense as it has no impact or influence on the sex.

      I see the point where it goes ad absurdum as you devide it into posibilities in regards to groups and statiscal probabilities of division. Lets say if a family had one boy and no other child and that the mother was pregnant what would the probability be of the unborn child being a boy. well 1/2 why would that change if it had occured already. It has nothing to do with statistical groups of how many families have whatever amount genders in their respective children, simply what is the chance right now, right here for that exact family to get another boy.

      I toss a dice whats the chance of hitting a 6, 1/6 now in a new unit of time i toss a dice what is the chance in that moment of hitting a 6. well for me it seems logical it is 1/6 regardless of hitting a 6 in the last toss. I never got how you can put conditional probability into these situations. It only applies as far as i can see if you have a large array of random numbers then it will add up in the end. My understanding, maybe im wrong but thats what is logical for me.

      When math looses touch of reality, it is mumbo jumbo like it is in this problem and the supposed answer. Math is a tool of logic but if you use the wrong tool it doesnt work and it goes ad absurdum, its not logic anymore. The inventor of a formula knows what that formula is used for, he understands the exact situations these tools apply because he invented the formula to solve the problem he understands what its limitations are. Then other people come around and say hey we can use this for probability and it can become misapplied to situations that it does not apply to.

      Anyways i want to thank you for rehabilitating my confidence in my logical ability. great article and answers.

      And as a curiosum i once saw a movie called 21 that posed a somewhat similar problem. If i remember it correctly the scene is a Math teacher giving his students a problem.
      Your in a show and you have to choose 1 of 3 doors, behind 1 of the doors is a Car you win it if you choose that door. Behind the other 2 doors are a goat respectively. You choose 1 door then the quiz master opens one of the other doors which has a goat behind it. leaving you with with 2 doors the one you choose and another one. Now he gives you the opportunity to change your choice to the other door if u wish to. The question is what is the smart thing to do here?

      In the Movie the students says it doesnt matter except one assumibly the smart kid in the class says it changes everything, he would switch. as now there is 2/3rds of a chance that he was wrong in his first guess and 1 /3 chance that he was right so he chooses to switch and the math teacher applauds him and picks him to be part of his illegal poker team scheme which is all what the movie is about.

      I remember again thinking that is utter bullshit when i saw the movie.To me it seemed completely irrelevant that the host opens the door with a goat behind it. there would always be a door left behind with a goat. he could open, but it wouldnt change the probability. there would be 1/2 chance you win in what ever case you chose using pure logic.
      But i would like to see what you had to say about that.
      I see the falacy there that it should somehow mean a difference that you shift your choice. as you could just as well have chosen that door that could be shifted to from scratch and you would still have a 1/3 probability to win the car. the fact that the host opens a door doesnt change a thing.

      Well what do you say?

      Jacdac

  8. Con-Tester says:

    If you read carefully how the situation is phrased, a simple reductio ad absurdum shows that your view of the problem is highly questionable. In your interpretation, fully half of the one-of-each fathers will ask it as: “I have two children. One of them is a girl (born on X weekday). What is the probability that I have two boys?”

    Short of extensive clowning around and/or irredeemable imbecility, the above won’t happen, and the one-of-each fathers will all mention their boy instead.

  9. Con-Tester says:

    Okay, so let’s recap where things stand.

      
      
      
    

    I. The Short Version

    With a strict literal reading of Foshee’s puzzle (and there is no direct or implied warrant allowing us to do otherwise), the answers of 1/3 and 13/27, respectively, are correct for the set of all earnest and truthful people who ask you the question exactly as Foshee has put it. To avoid ambiguity and controversy, we must proceed on the assumption that Foshee was fully aware that he had to choose his words meticulously. Even if we assume Foshee was speaking hypothetically as if he were such a questioner, the same result is obtained because Foshee’s conduct must remain completely consistent with his hypothetical role. In this view, JeffJo’s analysis completely misses the mark because he has imposed his own view of what he thinks Foshee is asking on the puzzle, instead of treating what Foshee has actually asked, thereby changing the problem.

      
      
      
    

    II. The Long Version

    If it makes it any easier, suppose that instead of any random person, the one who poses Foshee’s question comes from the subpopulation consisting of statisticians and puzzle fiends. Suppose further that they put the question to you exactly as Foshee has worded it. There is no reason to believe that the gender distribution of this subpopulation’s children is any different from the norm. Moreover, all such questioners, unless they are being silly, will mention their boy because to mention any girl they might have leads to an absurd triviality, variously a daft question, and so the requested probabilities remain 1/3 and 13/27.

    Here’s why in yet more detail. Foshee’s two-part question is undeniably framed in the first person: “I have…” and “… probability that I have…”. It is both presumptuous and unwarranted to suppose that Foshee meant anything other than precisely what his words mean, or that he did not quite consciously and intentionally put it exactly the way he had. It should now be clear. Nonetheless I’ll state it plainly anyway. It becomes an altogether different puzzle when worded as follows: “A person has two children. They choose to tell you that one of them is a boy (born on a Tuesday). What is the probability that the person has two boys?” The latter is in line with JeffJo’s proposed interpretation, and the reason it is different is because the person is essentially free to tell you about either one of their children, whereas Foshee, as he asks it and if he is being reasonable, does not enjoy this latitude.

    The gist of it is that one can neither compartmentalise Foshee’s two-part question into its two constituents because it is an error to assume that they are independent of one another, and nor can one blithely divorce it from the person who’s doing the actual asking because a straightforwardly logical inference about the questioner on the assumption of his/her truthfulness constrains the question that they can ask, and thereby also the answer.

    If, as JeffJo asserts, “Nobody is lying. Everybody is telling the truth.” (February 9, 2013 at 20:39, 2nd paragraph) then his earlier assertion that “Gary Foshee was not required to have a boy before he could posit this question” (February 9, 2013 at 20:01, 6th paragraph) is either an obvious self-contradiction, especially in view of his simultaneous assertion that Foshee’s wording is not ambiguous, or a way of speaking hypothetically as a questioner would. In the first case, if Foshee does not in fact have exactly two children of whom at least one is a boy, his question contains at least one lie. In the second case, Foshee must remain true to his hypothetical role as a questioner, and therefore is still constrained to mentioning his hypothetical boy for the same reason the real questioner would be.

    Over the past few decades, we have been so inundated with the postmodernists’ subterfuge where we can play fast and loose with the meaning of words as needed that we hardly notice it anymore — or object to it when we do. In this particular instance, Foshee’s use of “I” really must be taken to mean “me, the one who’s asking”, “boy” really must be taken to mean “child of male gender” and “one of them is a boy” really must be taken to mean that it is a hard fact (rather than that Foshee is telling you about it) because we have no indication from Foshee that he meant anything more or different from exactly what he said, and any kind of more loose reading cannot but lead to exactly the type of controversy we’ve seen here. Furthermore, it’s decidedly ham-fisted to imply that Foshee was unaware of the difficulties with Gardner’s original puzzle and that he didn’t knowingly compensate for them in his own version.

    So, you cannot have it both ways. Either you follow Foshee’s wording (and all alike questioners’) literally, unwaveringly and precisely in all respects (with everything that doing so entails), or you can change the problem by arbitrarily deciding how and where you think he’s speaking an odd and suspiciously English-like language in which “I” can mean “some random person”, “one of them is X” can mean “the person chooses to tell you one of them is X” (note here again the subtle trick where an unadorned statement of fact morphs into “a choice to be told” about it), and one in which “two boys” can mean “either gender”. And once you have realised that a strictly literal reading of Foshee’s words as plain English completely eliminates ambiguity, you should also then realise that for the set of all askers of Foshee’s question (i.e., all those foolishness-avoiding, truth-speaking I’s), the answers 1/3 and 13/27, respectively, are correct.

  10. PalmerEldritch says:

    This problem was conceived by mathematician Michael Starbird and presented by Gary Foshee at the 2010 G4G. I have no idea whether Mr Foshee has any children let alone a boy born on Tuesday.

    Here is what Gary Foshee himself said about the controversy surrounding the answer to this problem:
    “There is definitely an argument to be made based on choice. My solution was based on set theory. Look at the entire set of all families with two children. Then look at a subset: those with two boys. Then look at another subset: those with a boy born on Tuesday. If you look at it that way, then 13/27 is the correct answer.

    “If you start putting in factors about how the children were chosen, from which set, then yes there is an argument the answer could be different. It’s a very tricky and controversial subject.”

    Another mathematician present at the same G4G conference had this to say about the puzzle:
    “The Tuesday boy was, to me, an idea taken a little too far. It takes real imagination to concoct a situation where the required information is received in a manner that makes the suggested calculation correct”
    Dr. Peter Winkler – Research Mathematician at Dartmouth College

    There is plenty of support from other published mathematicians for the argument put forward by JeffJo in his analysis of this problem, which I agree with.

  11. JeffJo says:

    Thank you. It’s good to know some people will think about problems, instead of blindly agreeming with what somebody else said.

    The problem with Foshee’s solution is that set theory is a tool that has to be used correctly. Every solution uses it, but Foshee did it incorrectly. Specifically, you have to calculate the probability that you would receive the evidence you have for each element of your set. Foshee used the probability of existence for each element that *could* produce the evidence, not the probability it *would* produce that evidence. And Con-Tester refuses to recognize the difference.

  12. Gormful says:

    JeffJo is on a mission about this problem. He accuses his critics of ignoring how the evidence presented could arise and of refusing to see things his way. Con-tester showed how in reality JeffJo is ignoring and refusing to see the restriction on who can qualify to ask this puzzle. It’s a subset of all possible (role playing) fathers and you can immediately deduce things about the asker’s kids when asked the question.

  13. JeffJo says:

    Sure am – a mission to get people to open their minds about the problem. People like Con-tester, who showed absolutely nothing of value. He claims that a man, who has a boy born on a Tuesday and a girl born on a Thursday, is lying if he tells you “I have a girl born on a Thursday.” Stop and read that again – Con-tester is literally saying that a man who states a true fact is lying. Where is the lie?

    This is an ancient problem, and a well-understood one except by people who keep their minds closed. It is related to the Bertrand Box Paradox, the Monty Hall Problem, the Principle of Restricted Choice (a real-life use often applied in the game of Bridge, not a puzzle), the Three Prisoner’s Problem, and many others.

    In all of these problems, there is a difference between the set of cases where the facts, as presented, are true; and the set of cases where those facts would be what gets presented. And in every one, except – for some inexplicable reason – with variations of the Two Child Problem, the universally-accepted solution is based on the latter set.

    That solution has been recognized by mathematicians and puzzle solvers the world over – including Martin Gardner, whose memory Gary Foshee defiled by ignoring what he said about the problem.

    • Gormful says:

      Bwahahahahaha! JeffJo, your mission now makes it looks like you’re so full of your own BS you only see how clever you are. You’re either a slack reader, stupid or a liar. Con-tester didn’t say that man is lying. He said a bloke is an ass if he asks you “I have two children, one is a girl born on a Thursday. What is the probability that I have two boys?” First person makes all the difference, read Foshee’s version again. So it’s you who is ignoring and refusing to see.

  14. JeffJo says:

    No, friend,I was referring to things like February 9, 2013 at 19:14: “it seems that you’re making an extraordinarily convoluted case for the possibility that the speaker may be lying.” My quite simple case was that a father of a boy and a girl is equally likely to tell you about the girl, as the boy. If he does, he is not lying as Co-tester clearly claims.

    Or February 9, 2013 at 20:20: “In other words, you’ve expanded the problem to include the situation where the speaker has lied.” I *allowed*, not *expanded* into, a situation where a man who has a boy and a girl could tell you about the girl. This is not lying. It is being realistic.

    Or February 9, 2013 at 20:42: “the families whose fathers would say nothing or speak about having at least one girl … are excluded from the sample space by the problem statement, again on the assumption that the father tells the truth about the number and gender of his children!” A father of a boy and a girl, who says he has a girl, is excluded from the *condition*, but not the sample space (GG are also in the sample space, and excluded from the condition). He is excluded because he does not satisfy the condition, not because such a statement is not the truth.

    Anyway, I understood the invalid point Con-tester was making there, that a man who says “I have a Thursday girl” won’t ask for the probability of two boys. The obvious response is that he will ask for the probability if two girls. (And the point is completely misplaced if he says “I have a left-handed, red-haired, weight-lifting Saturday Boy.” Which I do, among my two children.)

    Con-tester’s error was concluding that since this particular man mentioned boys, or mentioned Tuesdays, or asked about boys, implies that he was forced to mentioned boys, or mention Tuesdays, or ask about boys. That’s the absurdity here. Con-tester realized half of this error,but his closed mind prevented him from accepting the whole thing. And you seem intent on not being illuminated, either.

    • Gormful says:

      Jeez, you’re like a religious nut with a bible verse, preaching and proselytizing to a stupid world. I understand Con-tester’s view and why he wrote the stuff about possible lying, given all the unintelligible BS you churned before. I also understand where you’re coming from but your error is right there in your second last paragraph of your latest post. Not that you’ll ever admit that or even see it, so it’s pointless going on against a mind as closed as yours. Just promise to have fun talking to yourself now, okay? Bye. 😆

      • PalmerEldritch says:

        Gormful, it’s you who appears to have the closed mind, refusing to consider any arguments that don’t agree with your pre-concieved notion of what the answer should be.
        Just promise to have fun wearing those blinkers of yours 🙂

      • Gormful says:

        Yup, appearances… :mrgreen:

  15. Con-Tester says:

    Gormful, as we say in my country, “Laat hulle maar hul kak praat.

    The relevant Wikipedia article reads as follows:

    “Commenting on Gardner’s version of the problem, Bar-Hillel and Falk note that ‘Mr Smith, unlike the reader, is presumably aware of the sex of both of his children when making this statement’, i.e. that ‘I have two children and at least one of them is a boy.’ If it is further assumed that Mr Smith would report this fact if it were true then the correct answer is 1/3 as Gardner intended.”

    This is what I’ve said all along, namely (1) that the man uses first person (“I have two children…”) just as Foshee did, and (2) that the man isn’t lying (“Mr Smith would report this fact if it were true”). JeffJo’s singular and continued failure to present any convincing case whatsoever for why the asker of the question doesn’t by direct implication reduce the subset of fathers who can legitimately ask the question will no doubt continue to hang around him like a mantle of dead fish. His dodge? That a GG father is still included just as BB/BG/GB fathers are because he would ask about the probability of two girls as if the question was actually the same. As you can see, that totally clinches it… 🙄

    But Bar-Hillel and Falk are very obviously of a much lower intellectual calibre than JeffJo and his eldritch apostle, so let’s sit back and enjoy the show as these two OCD morons jerk one another off about how stupid, blind, unenlightened, closed-minded and misguided we dissenters are with a new brace of piss-poor, recycled and contrived excuses that completely miss the point. I virtually guarantee it’ll be a huge hoot.

    • PalmerEldritch says:

      “…as these two OCD morons jerk one another off….”
      Very classy. Are you always this abusive whenever someone points out you’re wrong?

      In an earlier post you said:
      “Sure,the answer of 1/2 is correct if one assumes that a randomly selected BG or GB father will tell you about his boy half the time and his girl the other half. In this strict sense there arguably is ambiguity in the problem as stated.”

      And again:
      ” My issue is with how the problem statement is interpreted. In fact, I already conceded that the answer of 1/2 is correct on the appropriate assumptions”
      So even you are prepared to admit there is another interpretation to the problem than just yours.

      The answer therefore comes down to what are the “appropriate assumptions” to make. You and Gormfull disagree with JeffJo and me. I think the assumptions you make are unwarranted and aren’t in any way implied by the problem statement (whatever Foshee’s intention may or may not have been). I think it’s the lack of any assumptions that lead to the 1/2 answer.

      The main difference though – is that I won’t call YOU a moron for disagreeing with me.

      • Gormful says:

        You’re morons because you and your preacher aren’t able to get what Con-tester been saying, not because you disagree.

        Give us more of the same. You’re also freaking hilarious. 😆

      • Con-Tester says:

        Apostle-of-TRUTH™ PalmerEldritch compliments:

        “Very classy.”

        True — compared to a lot of the other blatant circumlocution we’ve seen here.

        Apostle-of-TRUTH™ PalmerEldritch whinges:

        “Are you always this abusive whenever someone points out you’re wrong?”

        No, only when repeatedly confronted by obdurate and unrepentant dingbats whose pretence to wisdom doesn’t withstand scrutiny and who continually attempt to insult my intelligence with a bouquet of evasions, irrelevancies, quote-mines, obfuscations and other verbose varieties of deceptive BS, and who crown their folly by crowing how open their minds are, how theirs is the correct view, how they’ve shown me to be wrong and how insightful they are when exactly the opposite is clearly the case. All of these aforesaid aspects merely betray an obstinate unwillingness, variously an inability, to engage with what I have laid out at length in considerable detail.

        The assumption that a guy who gets up in front of an audience (even an audience of one) to pose Foshee’s riddle, speaking in the first person singular and all the while knowing that the focus is male children — barring a complete gender reversal of the problem which by symmetry doesn’t change the answer — that this guy will arbitrarily decide whether to tell you about his boy or his girl, is simply and plain-as-a-smack-in-the-mouth absurd, as I repeatedly pointed out to no apparent avail. The guy knows he’s posing a puzzle with a particular answer in mind, and the very nature of that puzzle constrains what he can say, and so to yammer on and on and on about how he can choose what to tell you reveals no more than a deep ignorance of, or a reckless disregard for that constraint. All of this I have pointed out before but still you wankers won’t hear it, let alone address it.

        When you jokers finally manage to pull your smug thumbs from your conceited ringpieces and furnish a compelling argument for why this is wrong, rather than your usual blunt unmotivated denials and more inconsequent babble about how the riddler has a choice — rephrased, to be sure — you’ll easily find me a whole lot more polite and receptive.

        Good luck with that. Until then, my labels stands, as does my amusement with you myopic yokels’ ham-handed antics at attrition.

      • JeffJo says:

        Con-tester wrote:

        “… all the while knowing that the focus is male children …”

        And this is where you go (fill in whatever descriptions you used) wrong. You can’t deduce that the focus is “male children.”

        If the focus is “my children,” then a truthful poser with two boys will ask about boys, one with two girls will ask about girls, and half of those one with one of each will ask about either one. He will use phrases starting with “I have…”

        This isn’t complicated, unless you need to justify an incorrect answer. There are two hypotheses that lead to the evidence we see. One is biased towards boys, and one is unbiased. We can’t assume a bias, so we must use the unbiased one.

        “The main difference though – is that I won’t call YOU a moron for disagreeing with me.”

        Yet you did. Your credibility is very low because you keep lying, and rudely attacking.

      • Con-Tester says:

        Another torrent of irrelevant BS from the pontiff of probability that yet again completely misses the point about it being … wait for it … yes, it’s a puzzle!!!

        Keep going, that credibilityof yours is at an all time high. 😮

      • PalmerEldritch says:

        Con-tester. that saying which you mentioned in an earlier post, someone translated it as:
        “I might be pig-ignorant but I’m never wrong”
        Is that accurate? Somewhat apt if it is.

      • Gormful says:

        Jeffjo said “You can’t deduce that the focus is “male children”.”

        Bwahahahahahaha! Yup, Foshee puzzle question “What is the probability that I have two boys?” includes girls as boys… 😳

        Like you said, not complicated unless. But give us more, pope JeffJo. You’re still priceless. 💡

        Bishop PalmSandwich, looked in a mirror lately to see how brown an unoriginal your nose is? I guess no. Almost as brown and unoriginal as your posts.

        But give us more, bishop PalmSandwich. You almost as priceless as your pope. 💡

      • PalmerEldritch says:

        6 comments Gormful, and zero intelligent content. Priceless ( or should that be gormless).

        Why do you bother? Is Con-tester paying you? LOL

        Riddle me this Einstein
        I’ll flip 2 fair coins:,
        1) What’s the probability both coins land the same side up?
        2) If I tell truthfully you 1 landed heads , what’s the probability both coins land the same side up?
        3) If I truthfully tell you 1 landed tails , what’s the probability both coins land the same side up?

        If your answer to 2) and 3) is different to your answer to 1) then I’ve got a betting proposition that you might be interested in. Perhaps you ought to ask Con-tester before you decide.

      • Gormful says:

        That’s why you a moron bishop PalmSandwich. You don’t/can’t think. Your 3 questions all 1/2 probability but unrelated to Foshee problem cos you didn’t decide ahead of time to focus only on H (or T) like Foshee did with boys.

        But give us more, bishop PalmSandwich. Your now as priceless as your pope. 😆

      • PalmerEldritch says:

        And where in the problem statement does it say that it’s been “decided ahead of time” to only mention boys? Oh ….that ‘s right it doesn’t, and Gary Foshee never said that either,. But Con-tester says it’s “absurd” to think otherwise, so it must be true.
        ROTFLMFAO

      • Con-Tester says:

        PalmerEldritch, or should I say Bishop PalmSandwich 😆 (Brilliant!), you manage to show your own exceptional class really well, boet. 😈

          
          
          
        

        Unlike Gormful, you obviously are incapable of getting it. The overall context of Foshee’s puzzle setting and its wording make it abundantly clear that his focus is on the occurrence of boys, just as the overall context of your three-question comment (November 8, 2013 at 16:17) makes it clear that you want to impose your no-specific-focus view on Foshee for no convincing reason other than to feel superior about how it supposedly proves your case, which it definitely doesn’t. And equally obviously, I keep repeating the same point using different words while you still fail utterly either to grasp it or to provide any persuasively reasoned refutation whatsoever thereof. Until you and/or Pope JeffJo show some insight into my argument, rather than the customary brain-dead flat denial, you can rest assured that you’re wasting your time and mine.

  16. JeffJo says:

    Maybe you need to read it better: “*If* it is further assumed that Mr Smith would report this fact if it were true…” It’s a conditional statement. Not “Mr Smith would report this fact if it were true.” It was necessary to make it conditional, because it isn’t implied by the text.

    Or maybe note the fact that Bar-Hillel and Falk only said the “presumably aware” part. The conditional was from whoever wrote that part of the article. And quoted them out of context. B-H&F were comparing the problem to one where you meet Mr. Smith in the street. walking with a son: “Gardner … is not, however, in disagreement with our analysis. He is merely addressing a different problem! Having Mr. Smith tell us that he has a son, denoted Bs, is altogether different than discovering that fact on the basis of observing only one of his children. For Mr. Smith, unlike the reader, presumably is aware of the sex of both his children when making this statement.”

    And B-H&Falk themselves misquoted Gardner. His problem didn’t include “I have…”, it was “You know…”. Gardner is in no way suggesting that the problem should be interpreted as though you can only know about a boy. In fact, he said exactly the opposite. His said his problem was ambiguous about how you learned the fact, and so either 1/2 or 1/3 could be correct. Foshee’s isn’t ambiguous this way: he chose a gender to tell you about.

    This is the kind of information you can discover when you look with an open mind.

  17. Gormful says:

    Yup, you called it Con-tester. Clown JeffJo just can’t get his totally closed mind to understand that all BG/GB/BB fathers asking Foshee’s puzzle to an audience will mention their boy knowing the question is about boys, and all other fathers are excluded.

    Now it’s pulling conditionals and out-of-context and misquoting and new meaning from his ass.

    Give us more, JeffJo. You’re freaking hilarious. 😆

    • PalmerEldritch says:

      5 comments by “Gormful the Brown Noser” and zero content.

      “all BG/GB/BB fathers asking Foshee’s puzzle to an audience will mention their boy knowing the question is about boys, and all other fathers are excluded”,where do you get that ridiculous idea from? Oh that’s right, from Con-tester, and whatever he says is OK by you. Don’t let an original thought interrupt your blinkered viewpoint.

      Give us less Gormful, you’re boring as batshit

      • Gormful says:

        Hey bishop PalmSandwich, maybe I understand Con-testers argument and you don’t. Nope, that’s too accurate for someone “pig-ignorant but never wrong” like you, hey? 😯

        But give us more, bishop PalmSandwich. You almost as priceless as the pope. 😀

      • PalmerEldritch says:

        You’re not very good at this malarkey are you Gormful? You have nothing constructive to say, and repeat the same lame insults in every post.

        Your only contribution to this discussion is using a variety of emoticons in your comments. But top marks for that. LOL

  18. JeffJo says:

    Con-tester wrote:

    “The overall context of Foshee’s puzzle setting and its wording make it abundantly clear that his focus is on the occurrence of boys.”

    Riddle me this: suppose you go into a situation where you want to pose such a question, but your “focus” is “my children,” not “boys” or “girls.” If you have two boys, you will ask about boys (“I have…”). If you have two girls, you will ask about girls. If you have one of each, you will flip a coin and ask about heads=boys, tails=girls.

    Now turn it around. You hear someone ask such a question about boys. How do you deduce what the “focus” is about, “boys” or “my children,” when both an lead to the same statement?

    • Con-Tester says:

      Like your brown-nosing toady, one Bishop PalmSandwich :lol:, you’re too dim to contemplate the eminent possibility of being wrong, so you once again resort to your stock response of blunt, unreasoned denial cloaked in a new guise of pseudo-intellectual questions that merely rehash your long-broken BS.

      Let’s try a slightly different angle. The entire premise from which you proceed is flawed from start to finish. Your premise is a speculative, imaginary world of four quadrillion two-child Foshees divided into four equal groups of BB, BG, GB and GG Foshees, i.e. a quadrillion of each. Of the two quadrillion BG plus GB Foshee groups, you further speculate that half of each will preferentially mention their B while the other half will choose to mention their G.

      Okay so far?

      Good.

      Now here’s the crux: None of the above is in any way in evidence. It’s your imagination and your speculation that you seek to impose on the reality of the situation.

      And that’s where you pair of buffoons fall off the bus without even realising it. You can only legitimately go on the evidence that is actually to hand, not your speculations about what might or should or could be the case.

      The only evidence that is actually to hand is one single concrete Foshee claiming he’s a two-parent father with at least one B. In other words, he’s not a GG Foshee – real, imagined, speculative, hypothetical or otherwise. That’s why Foshee’s first person singular mode of talking (“I have…”) makes all the difference, as I have repeatedly pointed out.

      Okay so far?

      Good.

      Now what this means is that your flights of imagination and your speculations about four quadrillion Foshees and what they might have said are totally irrelevant; indeed, they are a hindrance and a fiction and an unnecessary complication that serve only to obfuscate.

      Moving on, if we believe that the one specific Foshee before us is speaking the truth (and we have no reason to doubt his word), then he can only be a BB, BG or GB Foshee (since we have no further info about his children), each those possibilities being equiprobable.

      Okay so far?

      Good.

      Therefore, the answer to this one specific Foshee’s question of what the probability is that he is a BB Foshee must be 1/3.

      Now I’m sure you’ll confabulate a whole new brace of verbiage concealing the same hackneyed BS about how all those other hypothetical Foshees might have spoken differently. The fact is that he didn’t speak differently. He spoke in a precise, unambiguous way, and that’s all you have to go on. Anything more is pie-in-the-sky fantasy. You do not get to reinterpret meaning and reality on guesses and might-have-beens, and that’s exactly what you’ve unwaveringly attempted here.

  19. JeffJo says:

    “Now I’m sure you’ll confabulate a whole new brace of verbiage concealing the same hackneyed BS …”

    No, I’ll leave that to you. You seem so much better at it than I.

    G’day.

    • Con-Tester says:

      Yes, appearances can be awfully deceiving, especially when your mind’s so closed that all you see is what suits you, preventing you from entertaining even the slightest possibility of weakness in your unshakeable fundamentalist belief about how right you are. But tell me, are you giving up because you realise that you have nothing cogent to offer, only a naïve, ill-conceived and misdirected mantra about choices, or have you grown tired of evangelising to the world about this problem? It couldn’t be because you realise I may have a point, now could it? No, that would take some coherent thought about, some actual insight into the process through which you came by Foshee’s facts.

      Cheers, boet.

      • Jacdac says:

        Seriosly Mister! You need to work on your sense of behavior and conduct. I have now read through all of the above comments pertaining to the discussion between you and JeffJo. You have a good sense of math and a well spoken english, but that behavior!…
        What you are accusing JeffjJo of e.g. insulting your intelligence is in fact what you are doing. You are insulting his intelligence. He comes with a well founded proof of his result. While you pretend to ask you are really just putting words in his mouth without understanding his arguement. All in the meantime you are dishing out a fine tirade of intelectual euphemisms – Clearly insulting his intelligence. Well why do you need to do this?
        Is your arguement so weak that you need to lesten his intelligence for it to stand clear? this is what Authoritism is and is the way you degrade science.

        I am not adequately well versed in either English nor Math to take up the discussion wether who is right. But i see the truth in what JeffJo says.

        The problem is a perfect example of Math put to ad absurdum. It is simply misapplied. Tuesdays cannot and will never influence anything about a gender, it defies any logic that you could get another answer than 1/2 to this question. It goes wrong at the time you add time into the equation e.g. that there are 4 groups BG GG BB GB as the author of the article wrote. There are not 4 groups in this problem, it is math misapplied.Mat h is a branch of logic, if it goes astray from logic it is wrongly applied and goes ad absurdum. Math is a set of tools to apply in logic but if you use the wrong tools it goes to heck. And thats why the whole choice of saying boy or girl comes in to elucidate the misapplication of the math. And corrects the answer to 1/2
        Your first comment was a good one but from the point you start insulting JeffJo you start an eggfight.

        Please consider next time you see someone having a well founded arguement. It’s not wether or not YOU are right or wrong, Its Wether you can change your position or are able to see if something is right or wrong.
        Consider:
        Is there something to this?

        Remember a true genious is never closed on his position, if something shows up that seems to alter therories he is interested, he is not starting an eggfight about it.

    • Jacdac says:

      After rereading some of your answers i came up with an even easier way of explaining the fallacy than the probability of telling its a boy or a girl.

      When you devide it up into these 4 categories, BG GG BB GB one of the groups can not be counted if we are talking in amount of probabilities like we are in this method of solving it.

      WHY? because we know one is a boy but we have no information wether its the eldest or the youngest who is a boy. So BG is the group where the boy is the eldest and GB is the group where the girl is the eldest. 1 of them cant be true. and cannot be used as a probability cause its impossible for both to be true at the same time in probability that means the probability of one of these groups BG GB is 0, CANT HAPPEN. and we are talking about one family here one situation. both cant be true in 1 case which one we dont know but we have to discount 1 of the groups as the amount is the same it doesnt matter which one we choose. and then we get the right answer 1/2. namely we have a group that is 1 of each and a group of BB the 1 of each cannot both be BG or GB as one of them is a lie and then we get an equal size of each group leaving it to be 1/2

      this one really cant be denied theres no way around it. It is a fallacy. that the result is 13/27.

      Con-tester should be blushed now for being so impolite on enforcing a fallacy while trying to sound very educated throwing eggs at you 🙂

      • Con-Tester says:

        No, you and your friends JeffJo and PalmerEldritch should go stand in a corner and blush. What none of you seems able to comprehend or even entertain is the simple fact that the group of fathers who get to ask this puzzle question (or those who play the role of such a father) is not a random sampling from the entire population of fathers. The answer P = ½ hinges critically on that flawed assumption.

  20. PalmerEldritch says:

    Con-Tester said:”the group of fathers who get to ask this puzzle question…is not a random sampling from the entire population of fathers.”

    That’s your interpretation. Why would you assume the father posing the question is NOT randomly drawn from the population?

    • Con-Tester says:

      It’s not an assumption. It’s an obvious fact — if only you actually bothered to think about it. I’m simply not going to waste my time trying yet again to explain what I have previously wasted my time on explaining in many different ways. Go read my posts above. Or continue to keep your eyes shut. Your choice.

      • PalmerEldritch says:

        It IS an assumption, and a critically flawed one at that. I’m not going to waste my time explaining again why that is. Go read the posts above or continue to keep your mind closed. 🙂

      • Con-Tester says:

        I understand your dim-witted argument perfectly well: Of all one-of-each fathers, on average half will ask about a boy and the other half about a girl. On that critically flawed assumption, P = ½ is the correct answer. I have admitted that point several times and have never denied it — on that critically flawed assumption. In contrast, none of you have displayed either the slightest understanding or even the faintest desire to understand what I’ve said in so many different ways. And until you actually engage with what I’ve said, your “arguments” will continue to fall flat.

        But you must do whatever strokes your ego for you, boet.

  21. JeffJo says:

    Con-Tester, you have proven that you do not understand this long-established (Joseph Bertrand, 1889) argument. What a father ASKS ABOUT is driven by what he TELLS YOU. One-of-each fathers have to decide whether to TELL YOU about their boy, or their girl, BEFORE they can resolve what to ASK YOU.

    BB fathers will TELL YOU about a boy, GG fathers will TELL YOU about a girl, and {BG,GB} fathers will split evenly about which to TELL YOU about. Each will then ASK YOU a question appropriate to what they TOLD YOU.

    Your answer to the puzzle begs the question (“begs the question” means “Assumes in the premise what is hoped to be established in the conclusion”) by making what the father asks the premise. But you are so entrenched in this conclusion-based premise that I hold no hopes that you will even consider any other point of view.

    • Con-Tester says:

      Read your post and evaluate yourself against the charges you level at me, boet.

      As said many, many times before: Until you engage with the content of my argument, you’re just blowing your usual pretentious and irrelevant smoke.

  22. JeffJo says:

    Read your post and evaluate yourself against the charges you level at me.

    • Con-Tester says:

      Yup, I guess the fundie-cult proselytising of The One True Probability Gospel According To JeffJo™ compels you to think you’re the only show in town. That, plus a hearty dose of the academic’s smug conceit.

      The telling difference is that I’ve conceded several times that P = ½ is correct if one accepts your assumptions, whereas you’re not even prepared to consider my argument. Instead, you just drone on endlessly with your irrelevant mantra, thereby revealing either an inability or an unwillingness to understand the basis of my argument.

      Guess whose mind is really closed here, boet. But win you must at all costs. You’re a total hoot, you are.

  23. JeffJo says:

    And I’ve always said that P=1/3 or 13/27 is correct if one accepts your assumptions. The point I am making – and that you will not allow yourself to consider – is that those assumptions lead to internal contradictions, and so cannot be accepted.

    • Con-Tester says:

      More contrived self-congratulatory bullshit. Your pompous pontifications reveal that you haven’t the first clue how I come by my “assumptions” as you label them.

      You’re getting funnier by the day.

      • Con-Tester says:

        —which is to say, yet again, that you can “question-beg” me and “internal contradiction” me and “Joseph Bertrand” me or whatever else takes your bloviatory fancy, but until you show a rudimentary comprehension of the basis on which I maintain what I do about the puzzle-asking fathers, you’re just compounding your hilariously obstinate folly.

  24. Pick 6 Leak Scam

    Even Odds | Grumpy Old Man

  25. Ron Osmond says:

    Probably far too late now to impact the debate, but I would just like to say that Jeff-Jo knows what he is talking about. I had a similar debate with him on this topic in 2011, and learnt a lot. What irritated me about the Foshee problem is the implication that the day of the week that the boy is born impacts the probability of a two boy family. And of course it does not.

    As this thread has shown the answer will always be 1/2 if you do not force a 50% chance to be 100%. Thus if you “force” every Boy/Girl family to be a Boy then the answer to the simple question will be 1/3. If you don’t it will be 1/2.

    In the Tuesday Boy scenario, if you force the Boy over the Girl AND force Tuesday over the other day of the week when you have a BB family with two different days, then you end up with 13/27.

    If you force the Boy over the Girl and do NOT force the Tuesday, then the answer remains at 1/3.

    If you force neither the boy nor the Tuesday then the answer is always 1/2, and in truth this is the most sane answer.

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