## Friday, August 27, 2010 ... //

### One son on Tuesday: a probabilistic puzzle

John Baez, a savior of the Earth, discusses an interesting puzzle that was sent to him by Greg Egan:

A few months ago I read about a very simple but fun probability puzzle. Someone tells you:

“I have two children. At least one is a boy born on a Tuesday. [And if it were not the case, I would have told you.] What is the probability I have two boys?”
Try to solve it yourself. John Baez mentions that you would think or he would think that the information about Tuesday is irrelevant because the days of the week are independent of the sex and we only care about the latter.

So you would think that there are 4 equally represented groups of 2-kid families, namely boy-boy, boy-girl, girl-boy, and girl-girl families where the two hyphenated words refer to the younger and older kid, respectively. Only the girl-girl families are eliminated, and 1 of the remaining 3 groups is a two-boy family, so the conditional probability is 1/3.

However, that's a wrong result. The information about the Tuesday actually does matter. Here's why:

Correct solution

In all families with exactly 2 children, one may label the children as the "younger" and "older" one, even if the difference is just in seconds.

Each kid may be born on any day and have any sex, so there are 14 equally likely possibilities for each child. The two children are independent (forget that the phenomenon of twins tends to increase the same-day pairs), so there are 14 x 14 possibilities for two kids. Each of these 14 x 14 possibilities is equally likely. So 1/196 of the world's families with exactly 2 kids fits each condition.

Among the 196 types of the families, how many of them contain at least one Tuesday son? Well, in 14 of them, the younger kid is a Tuesday son (the older one may be anything chosen from the 14 possibilities). In 14 other of them (the younger can be anything), the older one is a Tuesday son. However, I have counted the families with two Tuesday sons twice. So there are 14+14-1 = 27 possibilities among the 196 for which the condition "at least one kid is a Tuesday son" is satisfied.

This is the assumption which is a part of the calculation of the conditional probability. We need the other part, too. Among these 27/196 of the families, 13/196 of all families have two boys, by pure counting, so the result is
P = 13/27
as the fraction of the families that satisfied the condition. Note that it is just slightly less than 1/2 = 13.5/27 i.e. much more than 1/3. I had to highlight the result because almost no one reads the full article and almost no one notices that the right results is neither 1/3 nor 1/2.

Indeed, the large difference of the right result from 1/3 appears because one de facto identifies one of the sons by mentioning that it is the kid from Tuesday. If you assumed there were infinitely many days in a week and you would take any family with at least one Tuesday kid, the "Tuesday" information would identify this kid completely (two Tuesday kids would be infinitesimally unlikely), and the question what is the probability of 2 sons would be reduced to the question what is the probability that the other, equally specific kid - the non-Tuesday kid - is male - which is of course 1/2.

I will discuss this "identification" and reasons why the result is close to 1/2 at the very end.

Indistinguishable kids' bound states

With kids that would satisfy the Bose or Fermi statistics, the counting would be different but equally straightforward. Instead of 14 x 14 = 196 possibilities, one has 14 x 15 / 2 = 105 for bosons (the symmetric triangle) and 14 x 13 / 2 = 91 (the antisymmetric triangle) for fermions. Among the 105 or 91 options, how many of them contain at least one Tuesday son? Well, in these two cases, we can't say which of them is older and younger: they're identical.

So if there is at least 1 Tuesday son, the number of states with at least 1 Tuesday son is 14 for the bosons - we can just create the other particle into the 1-particle state - or 13 for the fermions - we can also add the second creation operator, but with another Tuesday son, the state will vanish because of Pauli's exclusion principle.

Among these 14 or 13 states respectively, for bosons and fermions, 7 or 6 are two-son states, respectively. So the odds are 7/14 = 1/2 for the bosons and 6/13 for the fermions. Note that the bosons literally saturate the 1/2 bound while the fermions are just slightly below it.

Why not one third?

Finally, I want to comment on "why the information about Tuesday matters". If we sum up the probabilities for the problems where the son is born on Sunday, Monday... and up to Saturday, shouldn't we get the same result? And by symmetry, the result must be equal for all 7 days, so doesn't each term have to be 1/3?

The answer is that we can't add the probabilities in this way because the "at least one Monday son" etc. are assumptions, not propositions conditioned by these assumptions, and they're not disjoint. At any rate, the calculation is nonlinear because the conditional probabilities have the probability of the assumption in the denominator rather than the numerator, so you can't simply add the possibilities in any way.

The word term in the previous paragraph is therefore incorrect.

How and why 1/3 gets enhanced to nearly 1/2

If you were only told that "one of the kids is a boy", the mixed families would be overrepresented over the two-boy families by the 2-to-1 ratio because boy-girl and girl-boy families are as likely as boy-boy families; again, the kids notation is younger-older.

However, if you're told that "one of the kids is a Tuesday boy", this overrepresentation almost disappears. Why? Because 1/7 of the boy-girl and girl-boy families have a Tuesday boy. But (approximately) 2/7 of the boy-boy families have at least one Tuesday boy because each of these two boys has a chance to be born on Tuesday.

In this way, the boy-boy families (nearly) compensate the factor of two by which they were underrepresented relatively to the mixed families.

Bonus: this puzzle and crackpot Sean Carroll's misunderstanding of logic

This logical puzzle is actually a very precise pedagogical example showing what's wrong with the thinking of various people about the arrow of time. Some people - those who say that the information about Tuesday doesn't matter and who typically end up with the result 1/3 - think that
Prob(cond,any_day) = Prob(cond,Monday) + ... + Prob(cond,Sunday)
where "cond" is an extra condition. So if we make a statement about a specific object and if this statement doesn't prefer any day of the week, then adding the information about "its" day of the week doesn't matter. It only reduces the probability by a factor of 7 if the probability is day-blind.

That's right for "conclusions" or "outcomes". However, the error that these people are making is that they think that this "additive" counting of the probabilities also holds for the probabilities of assumptions, i.e. probabilities of conditions in the conditional probability. But no such a linearity exists over there. Conditions (and initial states) don't follow the same maths as the outcomes (and final states)!

There is no condition-outcome or past-future symmetry in mathematical logic! That's why it matters for the probabilities whether the information about Tuesday is specified even though there is nothing special about Tuesday.

#### snail feedback (43) :

reader Unknown said...

Dear Sir,

I fail to see the problem. Leaving out such a nasty things and genetics, boy/girl birth rate differences, etc. - isn't this just coin tossing?
You toss twice. Flip 1: "Heads". What is the probablity that flip 2 gives "Heads" too? 50 %.

Cheers,
a usually lurking reader

reader Unknown said...

You should link to http://gregegan.customer.netspace.net.au/ESSAYS/TUESDAY/Tuesday.html , who "good job of making it more intuitive".

reader Luboš Motl said...

Dear Gregor,

just read the solution to see that the right answer is not quite 50 percent.

You failed to appreciate that the kid that was revealed to be a Tuesday son is not necessarily "the first one" according to your arbitrary ordering but it can also be the second one.

Frank,

I didn't know the colorful page. Well, you have linked it yourself. To see how the apparent answer 1/3 gets enhanced almost to 1/2, I personally prefer the last 3 short paragraphs of my text over the convoluted page you linked.

Cheers
LM

reader Phelps said...

It's exactly 50-50. This is just a variation on the Monty hall paradox. The other child is either a boy or a girl; everything else is known.

(unless you want to argue that transgendred kids don't count as one or the other)

reader Luboš Motl said...

Phelps, again, no, the right answer is 13/27. ;-)

reader Unknown said...

Firstly we must assure what exactly the speaker says to us.

Either, he says 'I have selected one child and this child is a Tuesday boy'.
Then he already ordered the children for us (the described one and the other).
Here he says nothing about the other one and therefore the other is 50% boy.

Or, he says 'I have randomly chosen one child and this child is a Tuesday boy.
Then he did not order the children, but we must see that in this scenario
family with two Tuesday boys has higher chance about being referred to
as 'family with Tuesday boy'. Here, we must order the children ourselves.
Either he randomly (50%) chose the older child, then the other is 50% boy,
or he randomly (50%) chose the younger child, then the other is 50% boy.
To sum it up (0.5 x 0.5) + (0.5 x 0.5) gives us again 50% boy.

If he says only that he as a Tuesday boy and we don't know how he had
chosen the child about which he speaks, we cannot compute anything.

reader Alejandro Rivero said...

Readers should notice (perhaps you should underline the relevant parragraph) that both limits, with infinite days in the week and with only one day in the week, are remarkable, because they interpolate between 1/3 and 1/2.

reader Unknown said...
This comment has been removed by the author.

reader bbear said...

I believe your result and I suppose Bayes formula applies here but, for the sake of argument, look: You say, "If you were only told that 'one of the kids is a boy', the mixed families would be overrepresented over the two-boy families by the 2-to-1 ratio…"

But, in effect, that's all we are told. The choice of Tuesday is arbitrary, right? We could as well have specified Wednesday. Or Monday. Or Thursday. But of course we know the boy child we're told about was born on some day of the week, and since we agree that the name of that day doesn't matter, no additional information is actually adduced and the two cases merge into one…

reader Unknown said...
This comment has been removed by the author.

reader chuck said...

Two sexes, seven days, 14 possibilities. Make a square grid 14 on a side. The combined lenght of the top and left edge is 27, the combined length for the same edges restricted to the upper left quater is 13. The answer is 13/27, which is easily seen to be almost 1/2 because it is only the square in the corner that isn't counted twice. Note that the days start with Tuesdays and the sexes with boys, so the row and columns are labled btues,bweds,bthur,...gtues,...gmon.

reader Luboš Motl said...

Chuck, excellent!

Bbear, I did the calculation in a purely frequentist fashion. You may say that I was estimating the number of actual families in this real world that satisfy certain conditions (neglecting the boy-girl imbalances etc.). There are no arbitrary Bayesian priors anywhere.

Indeed, it doesn't matter whether it's Tuesday or Wednesday, but it *does* matter whether the day of the week is specified or not. Just do the correct calculation - or check mine. There is no linearity when we consider probabilities of *assumptions*.

This reminds me of the debates with the crackpots of the Sean Carroll type who don't understand this point when they freely revert the arrow of time in logic, failing to realize that assumptions and their possible consequences don't play a symmetric role in logic which is why the knowledge about the past and the knowledge about the future don't play a symmetric role in thermodynamics, either.

Alejandro, isn't the word "remarkable" for interpolating between two extremes a little bit overhyped? Are the extrema +1 and -1 of the sine function "remarkable"? There is just the question whether the information given to us effectively distringuishes which kid we're talking about. If it specifies nothing or everything, we get 1/3 or 1/2, and partial specifications unsurprisingly give answers in between. What's so "remarkable" about it? It's correct but I wouldn't say "remarkable".

reader Unknown said...

Lubos,

You say, "Indeed, it doesn't matter whether it's Tuesday or Wednesday, but it *does* matter whether the day of the week is specified or not."

I disagree. The puzzle could also say "One is a boy and I just rolled a six on this die." In the context of normal conversation, we know that the person is stating two separate facts, not talking about logical intersection of those facts.

Of course the puzzle could be re-written to use terms from set theory, and then your counting would be correct, but then it wouldn't be a puzzle!

reader Luboš Motl said...

Dear Tom Shaw, OK, your answer was wrong, it continues to be wrong, and you have nothing else to say, do you? It does matter whether *a* specification about the day is made or not.

This is not a matter of "agreements" and "disagreements" - it is a mathematical fact where only one answer is right and it is not yours so it is not wise for you to pretend that this is a discussion of two equally valuable opinions.

So please shut your mouth, carefully read the correct solution, learn it, and meanwhile, I will be rejecting your equally vacuous comments such as your last one that only add noise here, OK?

reader Unknown said...

Lubos,

I agree that my first comments didn't add anything so I have deleted them.

However, my disagreement is not with your mathematics (in which there is only one answer) but with your interpretation of the puzzle. Surely you would agree that conversational language can have multiple interpretations, and those of us who are not Sheldon Cooper or Spock would find my interpretation more natural.

You are like the pedant, who, when asked "Shall we turn left or right?" responds "Yes".

reader Unknown said...

Greg Egan agrees with me that there are multiple interpretations, depending on whether or not you assume the birth day is spontaneously disclosed (in my words, the person is "stating two separate facts"):
http://johncarlosbaez.wordpress.com/2010/08/24/probability-puzzles-from-egan/#comment-715

I look forward to your update.

Best regards
Tom

reader Luboš Motl said...

Dear Tom, that comment by Egan makes no sense.

He's completely changing the problem by adding a completely arbitrary and previously unspecified procedure in which the parent is "tossing a coin" to decide whether or not he will reveal some information, and what information. Moreover, even when he does so, the correct result would still be 13/27.

There is nothing ambiguous about the original problem whatsoever. If a truth-speaking person says "I have [at least one] son born on Tuesday", it does mean the same thing as "the answer to the [pre-existing] question whether I have [at least one] son is yes". Claiming that this equivalence doesn't hold is a denial of mathematical logic. The propositions "A" and "A is true" are always equivalent.

If a person *without* a Tuesday son would say the sentence "I have at least one Tuesday son", he would be lying. If a person *with* at least one Tuesday son decided to not to say the sentence "I have at least one Tuesday son", he wouldn't satisfy the condition in the conditional probability we're asked to calculate - because the condition does include the revelation of the fact that he has a Tuesday son.

Because the condition wouldn't be satisfied, such a person would be outside the larger set from which the conditional probability of "having two sons" is computed, so the conditional probability wouldn't be affected at all.

So this Egan's comment is just about spreading fog. In particular, the sentence "I have a son born on Tuesday" is always exactly equivalent to "I have at least one son born on Tuesday". For the people with exactly two sons, the probability of this statement is always almost 2/7 (probably 13/49?) and never 1/7. There is absolutely no linguistic ambiguity here.

What leads you and Egan to the foggy comment is not a linguistic ambiguity but a logical mistake in the evaluation of one of the equivalent ways to phrase the sentence. His error is the incorrect assumption that the statements "I have a son born on Tuesday" and "I have a son born on Friday" are mutually exclusive. But for the people with two sons, it is obvious that these two statements are not mutually exclusive.

When someone has two sons, the probability that he has a son born on Tuesday is simply bigger (almost by a factor of two) than the probability that a one-sonned person has a Tuesday son. You can't change this fact by confusing comments about additional tossing of a coin. It's been said in the formulation of the puzzle that the person *did* say the sentence - so all the conditions that were necessary for him to say the sentence (he has to be able to speak, he got a permission from God or government to speak, if he decided to toss a coin to see whether he should speak about this information, the coin also told him Yes, and so on) were clearly satisfied. So it's completely inconsequential for the calculation of the conditional probability what happened before he said "I have at least one Tuesday son" - and the result is therefore always 13/27 regardless of the previous history.

reader Luboš Motl said...

Dear Tom, that comment by Egan makes no sense.

He's completely changing the problem by adding a completely arbitrary and previously unspecified procedure in which the parent is "tossing a coin" to decide whether or not he will reveal some information, and what information. Moreover, even when he does so, the correct result would still be 13/27.

There is nothing ambiguous about the original problem whatsoever. If a truth-speaking person says "I have [at least one] son born on Tuesday", it does mean the same thing as "the answer to the [pre-existing] question whether I have [at least one] son is yes". Claiming that this equivalence doesn't hold is a denial of mathematical logic. The propositions "A" and "A is true" are always equivalent.

If a person *without* a Tuesday son would say the sentence "I have at least one Tuesday son", he would be lying. If a person *with* at least one Tuesday son decided to not to say the sentence "I have at least one Tuesday son", he wouldn't satisfy the condition in the conditional probability we're asked to calculate - because the condition does include the revelation of the fact that he has a Tuesday son.

Because the condition wouldn't be satisfied, such a person would be outside the larger set from which the conditional probability of "having two sons" is computed, so the conditional probability wouldn't be affected at all.

So this Egan's comment is just about spreading fog. In particular, the sentence "I have a son born on Tuesday" is always exactly equivalent to "I have at least one son born on Tuesday". For the people with exactly two sons, the probability of this statement is always almost 2/7 (probably 13/49?) and never 1/7. There is absolutely no linguistic ambiguity here.

What leads you and Egan to the foggy comment is not a linguistic ambiguity but a logical mistake in the evaluation of one of the equivalent ways to phrase the sentence. His error is the incorrect assumption that the statements "I have a son born on Tuesday" and "I have a son born on Friday" are mutually exclusive. But for the people with two sons, it is obvious that these two statements are not mutually exclusive.

When someone has two sons, the probability that he has a son born on Tuesday is simply bigger (almost by a factor of two) than the probability that a one-sonned person has a Tuesday son. You can't change this fact by confusing comments about additional tossing of a coin. It's been said in the formulation of the puzzle that the person *did* say the sentence - so all the conditions that were necessary for him to say the sentence (he has to be able to speak, he got a permission from God or government to speak, if he decided to toss a coin to see whether he should speak about this information, the coin also told him Yes, and so on) were clearly satisfied. So it's completely inconsequential for the calculation of the conditional probability what happened before he said "I have at least one Tuesday son" - and the result is therefore always 13/27 regardless of the previous history.

reader Luboš Motl said...

Dear Tom, that comment by Egan makes no sense.

He's completely changing the problem by adding a completely arbitrary and previously unspecified procedure in which the parent is "tossing a coin" to decide whether or not he will reveal some information, and what information. Moreover, even when he does so, the correct result would still be 13/27.

There is nothing ambiguous about the original problem whatsoever. If a truth-speaking person says "I have [at least one] son born on Tuesday", it does mean the same thing as "the answer to the [pre-existing] question whether I have [at least one] son is yes". Claiming that this equivalence doesn't hold is a denial of mathematical logic. The propositions "A" and "A is true" are always equivalent.

If a person *without* a Tuesday son would say the sentence "I have at least one Tuesday son", he would be lying. If a person *with* at least one Tuesday son decided to not to say the sentence "I have at least one Tuesday son", he wouldn't satisfy the condition in the conditional probability we're asked to calculate - because the condition does include the revelation of the fact that he has a Tuesday son.

Because the condition wouldn't be satisfied, such a person would be outside the larger set from which the conditional probability of "having two sons" is computed, so the conditional probability wouldn't be affected at all.

[to be continued]

reader Luboš Motl said...

[continuation]

So this Egan's comment is just about spreading fog. In particular, the sentence "I have a son born on Tuesday" is always exactly equivalent to "I have at least one son born on Tuesday". For the people with exactly two sons, the probability of this statement is always almost 2/7 (probably 13/49?) and never 1/7. There is absolutely no linguistic ambiguity here.

What leads you and Egan to the foggy comment is not a linguistic ambiguity but a logical mistake in the evaluation of one of the equivalent ways to phrase the sentence. His error is the incorrect assumption that the statements "I have a son born on Tuesday" and "I have a son born on Friday" are mutually exclusive. But for the people with two sons, it is obvious that these two statements are not mutually exclusive.

When someone has two sons, the probability that he has a son born on Tuesday is simply bigger (almost by a factor of two) than the probability that a one-sonned person has a Tuesday son. You can't change this fact by confusing comments about additional tossing of a coin. It's been said in the formulation of the puzzle that the person *did* say the sentence - so all the conditions that were necessary for him to say the sentence (he has to be able to speak, he got a permission from God or government to speak, if he decided to toss a coin to see whether he should speak about this information, the coin also told him Yes, and so on) were clearly satisfied. So it's completely inconsequential for the calculation of the conditional probability what happened before he said "I have at least one Tuesday son" - and the result is therefore always 13/27 regardless of the previous history.

reader Unknown said...

Hi Lubos

That wooshing sound was the point going right over your head.

Greg Egan explains it more clearly here. Note especially his distinction between what he calls Puzzle 1 and Puzzle 2.

http://gregegan.customer.netspace.net.au/ESSAYS/TUESDAY/Tuesday.html

I was hoping that a few sentences would be enough to catch your intuition, but maybe his pictures will help too.

reader Luboš Motl said...

Nope. Puzzle 1 and Puzzle 2 are exactly equivalent.

Exactly the same information is revealed as a condition of the problem, and we're asked to calculate the conditional probability of exactly the same outcome. So the result is inevitably the same in both cases and it is 13/27.

He says that as a part of the condition, the parent "simply reveals the day of the week when one of his children was born" in Puzzle 2. But that's exactly what he or she is doing in Puzzle 1, too.

The main difference between the two formulations of the same puzzle - representations called "Puzzle 1" and "Puzzle 2" - is that Egan was able to correctly solve the first one but not the second one.

If one is giving all possible wrong answers to a problem besides the correct answer, it may be wrong to say that he actually knows how to solve the problem.

reader Unknown said...

Lubos,

What happens if, as I suggested, the person says "One is a boy and I just rolled a six on this die"?

Does the probability of having two boys change? Does it matter if the die is six-sided or twenty-sided?

What if the person says "One is a boy. Let me roll the die: I just got a six"?

Tom

reader Unknown said...

I think this has to do with the 'measure' of the probability space. The two following probability spaces:

(1) No day is specified.->4 discrete outcomes.

(2) The birthday of the boy is specified.->196 discrete outcomes.

have different measure. (1) can be mapped into (2). But (2) cannot be mapped into (1) if the day is specified. (2) can only be mapped into (1) if you 'integrate out'the day (i.e.: sum over all possible day).

This paradox reminds me of another probability paradox.

http://en.wikipedia.org/wiki/Bertrand_paradox_(probability)

reader Luboš Motl said...

Dear Tom,

you begin to be immensely obnoxious. Couldn't you please dedicate more time to thinking and less time to flooding this blog with rubbish? You got an F from solving this problem, can we please close it in this way?

If he said that he just rolled a die and got six, it wouldn't influence the counting and the result would be 1/3. However, if he says that he has a Tuesday-born son, it does matter and the probability jumps to 13/27.

The difference between the two additional pieces of information is that boy-boy families have a higher chance (13/49) of having a Tuesday-born son than boy-girl families (1/7), almost by a factor of two. On the other hand, the boy-boy and boy-girl families have no difference in the probability that their father rolls a six (1/6 in both cases).

Your latest comment showed that the problem behind your wrong statements is 100% made out of your stupidity and 0% of any hypothetical linguistic confusion. There is no confusion in the formulation of the problem. The only thing that there exists here is your flawed thinking.

Cheers
LM

reader Luboš Motl said...

Dear Haryo,

the 196 choices are made out of 49 boy-boy, 49+49 boy-girl and girl-boy options, and 49 girl-girl options.

One can divide the 4 states into 4 x 49 finer states, but of course, one cannot divide 196 boxes to 4 boxes because 196 is greater than 4. Is it really so shocking? Is it so necessary to call it a "paradox"?

Once some information about the days is specified, we obviously need to work with the boxes that do care about the days, i.e. with the 196 boxes. It doesn't matter which day was pronounced but it does matter that a day was pronounced.

The assumption that we can simply forget about the extra information and pretend that it wasn't said has been proved wrong by an explicit calculation. So any argument leading to the conclusion that the Tuesday information may be simply ignored is wrong, too. All questions in this kind of mathematics have sharp and unique yes/no answers and there's no need to talk about "paradoxes".

Cheers
LM

reader aptiko said...

Lumo,

I think that Egran is right and that you are wrong. Let me explain in other words: I tell you "I have two children and at least one of them is a boy". At this stage, your probability is 1/3. You ask me: "What day was he born?" I reply a name of a day. It can be Monday, or Tuesday, or Wednesday, or whatever. After I reply, would that change the probability of 1/3 for you?

The only case in which that would be true is if you had pre-decided the day before I gave you my answer; if you had pre-decided that the day that interests you would be, for example, Tuesday. In that case, if my reply were "Tuesday", your probability would be 13/27, and if my reply were anything else (i.e. "not Tuesday"), your probability would likely go a tiny bit higher than 1/3 (10/27 perhaps, I haven't calculated). In that case, where you have pre-decided the day, Puzzle 2 is reduced to Puzzle 1.

reader Unknown said...

On the contrary, we're close to finding the point of distinction.

There is a difference between:
A) "One is a boy who was born on a Tuesday"

and

B) "One is a boy. By the way, he was born on a Tuesday"

In case (A), you are correct. In case (B), it is analogous to the thrown die.

Whether or not you agree with my position that (B) is more natural in the context, you must agree that the two produce different results.

reader Luboš Motl said...

Tom, please stop it already!! I can't believe that you are *that* stupid.

The statements A and B are manifestly 100% equivalent. They say that the person has a son and that the son was born on Tuesday - i.e. that the person has a son born on Tuesday. It's the same thing.

Among other stupid people, you may lead them to different emotions if you decide to make a rhetorical demagogic exercise and if you say the same thing in two different ways. For example, when you say "by the way", you may want them to ignore the information that follows (which some of them will do while others will, on the contrary, focus on this proposition haha).

But it is *logically* the same statement! You still said that the son was born on Tuesday so you said the same thing. There exists at least one kid among your children that is male and that was born on Tuesday.

Don't tell me that you don't understand it. You may be honest but the stupidity beyond a certain threshold is insufferable, too.

reader Luboš Motl said...

Aptiko, this is clearly an attack of the cloned idiots. Please take it personally if necessary. ;-)

I have formulated a very clear problem where the solver was first given the information that the person had a Tuesday-born son, and *then* he was asked what is the probability of two sons.

I can't believe that you can't distinguish this problem from the problem when the solver is asked about the probability *before* he learns that the/a son was born on Tuesday.

And indeed, the only way how you can get back to 1/3 is when you cut the Tuesday statement out of the game entirely.

Whenever the information about the Tuesday is a part of the problem, before the person is asked about the answer (about the probability), and it was surely the ordering in my formulation of the problem, the answer is 13/27 simply because the probability that a father of 2 boys can say that "I have a son and the son was born in Tuesday", in any order, is always 13/49 and never 1/7.

reader chuck said...

Let's make the problem more interesting. Suppose the information given was "at least one is a boy born on a Monday, Wednesday, or Friday". Using the same sort of grid as before, nine cells are counted twice, so the probability is (6*7 - 9)/(6x14 - 9) = 33/75. In general,
(2*n*7 - n^2)/(2*n*14 - n^2) if n days are possibilities. In particular, if all days are possible then n=7 and we get 1/3.

reader clazy said...

The confusion here seems to come down to a point made by Briggs: "Probability, and its brother randomness, are measures of information." For the full discussion: http://wmbriggs.com/blog/?p=2227

reader aptiko said...

Luboš,

sorry for not signing my previous post properly - it was my first post on this site and I didn't know how it works. I then tried to create a profile, but I didn't like the terms it required me to accept. I am Antonis Christofides, and I am grateful to you because nine months ago you advertised a site of mine.

Back to the issue. I don't mind if we disagree, but at least I want to understand what we disagree about.

If I go to the state registry and pick up all families which have exactly two children one of which is a boy born on a Tuesday, then 13/27 of those families will have two boys, and the other 14/27 will have a boy and a girl. I think we agree in that. Of course, if we repeat the experiment using Monday, we'll get the same results.

What I think we disagree in is this: I think that your initial phrasing of the problem means something different, and that it leads to a different result. You think either that it means exactly the same thing, or, maybe, that it is a different thing but leads to the same result.

My sister has two children, and one of them is a boy born on a Thursday. My wife's aunt has two children, and one of them is a boy born on a Friday. My uncle has two children, and one of them is a boy born on a Wednesday. In that case, I did not sample in the same way as before; I sampled people that I know and who have two children of which at least one is a boy (of which I know the birthday), and I just told you the weekday they were born. I don't think this will give you 13/27; I think it will give you 1/3. And I also think that your initial phrasing of the problem can be interpreted this way; and that it is the most natural way of interpreting it.

reader Luboš Motl said...

Dear aptiko,
I can't believe it.

You told me the days of the week when children of some people you know were born. None of them was Tuesday so your comment has clearly absolutely no relevance for the problem considered in this article in which one was asked what is the probability of a two-boy family given at least one Tuesday boy.

This probability is always 13/27.

If you look at all 2-kid families you know and count how many of them can tell you that they have a Tuesday kid that is also a boy, or vice versa, or in any order - it absolutely doesn't matter in which order the sex and day of a kid is said because it's an identical package of information - I assure you that with a high enough number of families you know (and with neglecting biological biases), the fraction of two-boy families will be 13/27. Everything else is wrong.

The information about the existence of a Tuesday boy in the family simply does matter. For boy-girl and girl-boy families, it reduces the number of families that satisfy the condition by a factor of 1/7. For boy-boy families, however, it reduces them by a factor of 13/49. So the original 2-to-1 ratio in favor of the mixed families (relatively to 2-boy families) is reduced to 14-to-13, i.e. to 13/27 of two-boy families. If we replaced "Tuesday" by any other day, we would get the same result. But if we omitted the remark about the day completely, we would get a different result.

All of you are just doing your best to suffocate your brains in mist. I have erased 2 new comments with a similar garbage and I will keep on erasing comments that flood this thread with nonsense more systematically.

Best wishes
Lubos

reader andim said...

So, there are many similar examples out there where the inclusion of seemingly irrelevant information makes the calculation of probabilities even for smart people counterintuitive or impossible.

Do you have examples from "real life"? Clinical trials come to mind. Should I be worried?

reader Paul Rimmer said...

I also don't understand Lubos's explanation (maybe I'm not smart enough for this problem).

The way I understand it is by adding some extra information.

“I have two children. [At least] One is a boy born on a Tuesday, and his name is Harold. If I had another boy, his name would be Kumar. What is the probability I have two boys?”

Harold (H) is a given. Kumar (K) and a Girl (G) are the possibilities.

So (ignoring the Tuesday statement), we have four possibilities: H then K, K then H, H then G, G then H. So 2/4, or 1/2.

Paying attention to the Tuesday statement, I get 14/(14+14) = (13+1)/(27+1) = 1/2.

It is quite likely I'm doing something wrong, but I don't see what it is.

reader Paul Rimmer said...

I thought about the problem a bit longer, and have determined that your solution is absolutely correct. You can't double-count the boys. (14-1)/(28-1) or 13/27 is right.

Please disregard my previous statement.

reader BobM said...

Would this not be true only if there is one boy born on a Tuesday?
If you allow for the possibility that two sons were born on Tuesday wouldn't that raise the chance to 1/2?

reader Luboš Motl said...

Dear Bob, nope, the two Tuesday-son case was carefully included, but it's important that you cannot double count it. This is exactly the point that Paul right above your comment was concerned with and figured out after a minute, too.

If you modified the problem so that it would also be said that it's not true that both of them are Tuesday sons, the probability would drop from 13/27 to 12/26=6/13.

reader Unknown said...

Yes I think I've cracked it!

The two meanings can be completely distinguished by using the numeral "1" when you mean the number, and saying "a particular" when you mean the adjective "a certain".

If you went through your blog post and changed the word "one" accordingly, we could stop having these silly arguments about interpretation and instead focus on the logical argument!

reader ForNow said...

"A certain" is better than "a particular" in the sense of an unknown definite (unknown constant), though I guess either one will do in a pinch. "A certain" is a lot like that which Quine called a dummy letter as opposed to a variable. Still, readers only lightly acquainted with probability problems may not immediately pick up on the significance of the distinction.

The main reason for the confusion about Egan's statement of the puzzle is that it seems to mention the two children in an ordering - firstly, an assured Tuesday boy, and secondly, an indirectly mentioned boy who makes them be two boys (people are reading it in a bit of a connect-the-dots way, taking the seeming allusiveness perhaps as misdirection, like in a coin trick). Actually Egan singled out properties or classes, not individuals, and didn't clearly say which child is which. But it is common in ordinary, casual language for the ordering of mentions to represent, in a kind of casual verbal diagram, an ordering of things. Formal language often formalizes that habit. Anyway, that’s how some have been reading Egan's ordinary-language statement of the puzzle. From a bit of slippery stuff grows a monstrosity like in an old Sinbad the Sailor movie.

There's a second and subtler reason, occasioned by the first and reinforcing it, for some of the confusion. The seeming exclusion even from tacit consideration of non-two-child families as if by fiat (when in fact they merely would make no difference) leaves some in the frame of mind that it's settled that all the families have two children the same as with the known outcome of a coin toss, and they feel a mis-cognitive consonance (as opposed to a cognitive dissonance) with thinking of the Tuesday boy in the same way. It gets pretty messy. How did we get here? The monstrosity grows from a start that feels innocent and natural start in the habitual reading of ordering of mentions as representing diagrammatically an ordering of objects.

Actually I've been analyzing some of my own initial thoughts and generalizing them to other folks.

As for questions "changing" the probability, the objective relationships of the givens remain the same no matter what questions are actually asked. As two children such that one or the other is a boy, the children have a 1/3 probability of both being boys. As two children such that one or the other is a Tuesday boy, the children have a 13/27 probability of both being boys. Both those statements remain true about the same two children before and after their father provides spotty info to the cheerful young temp from the census bureau. Same two children, but as representatives of various sets. As I (mis-)understand it, in a strict sense probabilities are for the future, just as in a strict sense facts and data are from the past. But sometimes I find it less confusing to think of objective frequencies than of objective probabilities. The frequency of a Tuesday boy for a family given that it has a Tuesday boy is of course 100%. But the frequency of such an outcome among two-child families in general remains 27/196.

Aptiko's version with the roll of the die isn't similar enough to Egan's version of the puzzle unless you get two throws of the die (seeking, say, at least one three) for the case of two boys, and only one throw of the die (seeking the same result, a three in the example) for each case of two children different in gender. That is, the throw of a die is to be distributed to the pairings just as the Tuesday-boy property is. Let's say, for even greater similarity to Egan's version, that it's a fair 7-sided die - a heptagon in Flatland, and you're looking to roll a three. The solution will be familiar: 13/27.

Boy, Boy, one or two threes in two throws
1/2 x 1/2 x 13/49
Boy, Girl, a single-throw three
1/2 x 1/2 x 7/49
Girl, Boy, a single-throw three
1/2 x 1/2 x 7/49

13/(13+7+7) = 13/27.

reader Kmeson said...

In my thinking there is an odd importance to whether you ask "Do you have at least one son born on Tuesday?" or you are told "I have at least one son born on Tuesday." If you ask the question then I agree with the 13/27 answer.

If i'm in a room with 14*14*2 people with two children (with freakishly even distribution of sexes and days of the week) and an announcer says "go tell that guy a true statement of the form 'I have at least one $sex born on a$day'" then I expect 28 of them to tell me (son, Tues.) and a full half of them will be of the two son variety. Note that the full crowd of people with a Tuesday son don't approach me since most of them had a choice of statements and didn't pick the son on Tuesday option.

If the announcer instead says "go tell that guy a true statement of the form 'I have at least one son born on a \$day'" then I expect fewer visitors (3/4) but now the one boy set has no choice of what to tell me so I get 42 people telling me (son, Tues.) and 14 of them will be of the two son variety.

Finally, if the announcer says "if you can truthfully, tell that guy 'I have a son born on Tuesday'" Now I get a full compliment of 54 people giving this statement, since the two boy crowd no longer has a choice of statements, and 26 of them are in the two boy cohort.

I think the part I don't understand is how to pick which of these cases is similar to the unsolicited statement.

reader Moggsy Melboune said...

If this whole puzzle was not a logical fallacy you would get the same result by swapping Tuesday for January - a twelve count instead of a seven count, etc etc etc
The intriguing matter is to determine the exact identity of the fallacy - I nominate the term 'empty clothes-pegs' fallacy, if new, because at each stage redundant material is subtly added to a straightforward proposition - I have two children, a and b, what probability that b is a boy -50% - finito
Moggsy Melbourne

(function(i,s,o,g,r,a,m){i['GoogleAnalyticsObject']=r;i[r]=i[r]||function(){ (i[r].q=i[r].q||[]).push(arguments)},i[r].l=1*new Date();a=s.createElement(o), m=s.getElementsByTagName(o)[0];a.async=1;a.src=g;m.parentNode.insertBefore(a,m) })(window,document,'script','//www.google-analytics.com/analytics.js','ga'); ga('create', 'UA-1828728-1', 'auto'); ga('send', 'pageview');