## 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.”

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.

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

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### 17 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.

• Con-Tester says:

Correction: P(Y) = 1*(1/196)+(1/2)*((1/14)*(13/14)+(13/14)*(1/14)) = 1/14.

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.

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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.

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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.