## The Tuesday Birthday Problem 981 981

An anonymous reader sends in a mathematical puzzle introduced at the recent Gathering 4 Gardner, a convention of mathematicians, magicians, and puzzle enthusiasts held biannually in Atlanta. The Tuesday Birthday Problem is simply stated, but tends to mislead both intuitive and mathematically informed guesses.

*"I have two children, one of whom is a boy born on a Tuesday. What's the probability that my other child is a boy?"*The submitter adds, "Believe it or not, the Tuesday thing is relevant. Well, sort of. It's ambiguous."
## Ordering and Convergence (Score:5, Informative)

a hugeMartin Gardner fan and if this puzzle intrigues you and you haven't heard of him, get one of his books [amazon.com].This problem hinges very greatly on how it is phrased and I think it's more a trick of English converting to statistics than it is a true puzzle. If you were to rephrase this problem as "My first child was a boy born on Tuesday now what are the odds that my next child is a boy?" But they don't. They phrase it as after the fact of both births we are now studying a set of

twoobjects that have been determined prior to me asking the question. This ordering causes the set to be enumerable. Which brings in an interesting piece of information theory to this game. Whenever you say something about one child that is exclusive to that child and that trait is enumerable than you have just affected the outcome of that second child. In the original two childs problem this is just gender. But in the above problem it pairs gender with day of week the child was born on. Now since ordering matters you recognize that whether or not the older or younger child is the one in question creates a different scenario and you can't have twins because only one was born on Tuesday. The article does a good job of explaining this.The interesting thing is that the answer to this comes down to 13/27. And the larger the enumerable set is of possibilities, the closer this converges to 1/2. If you did this with a specific day of the year, your answer would be 729/1459 which is even closer to 1/2. The general rule is that for a set with N possibilities you would have (N*2 - 1)/(N*4 - 1). Now, what's interesting is if N is unbounded or unenumerable? What if I said "I have two children, one of whom is a boy that likes the number 1835736583. What's the probability that my other child is a boy?" Wouldn't it converge to 1/2?

## Re: (Score:2)

What if the second

childwas born on Tuesday but was not a twin of the Tuesday boy?Say, the Tuesday 75 weeks later. It's still a Tuesday...

## Re: (Score:2)

What if the second

childwas born on Tuesday but was not a twin of the Tuesday boy?Say, the Tuesday 75 weeks later. It's still a Tuesday...

According to the quote that would have to be a girl born on Tuesday because the quote in question states that one and only one of the children is a boy born on Tuesday. It cannot be a boy born on Tuesday. The only information you get form that statement is that one of the two ordered children cannot be a boy born on Tuesday. We have two genders. We have seven days. That's fourteen permutations. We have two children that can possess those fourteen permutations but we know that one of those permutations

## Re:Ordering and Convergence (Score:5, Informative)

Here's a reply to the article: (I haven't verified for mathematical correctness)

"The "(and only one)" qualification suggested by Ralph Dratman is _not_ required. Indeed, in the first case of the analysis, "older child is a boy born on Tuesday", the possibility that the younger child is also a boy born on Tuesday is explicitly included and counted. The hypothesis for the second case does exclude the possibility of both being boys born on Tuesdays. The two cases are mutually exclusive and exhaustive.

Note that if the puzzle had included the "(and only one)" qualification, then the possibility count would have been 13 (6 for boy and 7 for girl) in both cases, and the probability drops to 12/26."

## Re: (Score:3, Funny)

I have never seen "DOES not disallow" in my entire life.

You haven't been watching the officiating during the World Cup.

Example: "The referee does not disallow goals that are scored fairly unless they are scored by the United States."

## Re: (Score:3, Informative)

You are wrong.

>Label the child we know to be a boy as A.

Lets do that.

>The child without a known gender is B.

>Boy-A Boy-B

>Boy-A Girl-B

>Boy-B Boy-A

>Girl-B Boy-A

But, why did you list the Boy - Boy combination twice? If you have two children, there are 4 possible orders for them to be born, all equally likely:

1 Boy-Boy

2 Boy-Girl

3 Girl-Boy

4 Girl-Girl

Just because in case (1) you can label any of the two boys as boy A, is not a reason to list that combination twice in your list of possible birth

## Re: (Score:2)

## Re:Ordering and Convergence (Score:5, Insightful)

Normal English strongly implies it does. If you say to someone I have several pieces of fruit and one of them is a banana, when in fact two of them are bananas, most people would call that lying. One could argue that strictly speaking the statement was true: you did have one banana, you just also had an additional banana, but that level of honesty is only tolerated in politicians. If you had said "at least one of which is a banana" that would be fine, otherwise the statement is deliberately misleading.

## Re: (Score:2)

## Re:Ordering and Convergence (Score:5, Informative)

The problem is stated thus:

I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?

One of whom is not "exactly one of whom", so 'one' might be opposed to 'the other' or 'two'. All we know is that one of the problem poser's children is a boy born on a Tuesday. It states nothing about the relationship between the two children in time or space, so the probabilities are independent.

Further, the problem doesn't ask about any probability related to the second boy's birthday. The problem doesn't ask, e.g.,

What is the probability that my other child is a boy not born on Tuesday?. That makes the birth weekday completely irrelevant.## Re: (Score:3, Insightful)

This problem (and related problems) are almost always poorly phrased. However, your assertion that failure to ask about the day-of-week of the 2nd child's birth makes the day-of-week of the 1st child's birth irrelevant is incorrect, if the problem is understood as it was intended. And actually my own phrasing there is a bit poor there, because the crux of the problem is that there is no ordering - no "1st child" and "2nd child".

A better (or in any event less misleading) phrasing would be: "There are two c

## Re:Ordering and Convergence (Score:5, Insightful)

These kind of problems require competence in both English and Maths which is why so few people get them right.

Mostly they require competency in psychology, so you can figure out how the twit posing the problem is deliberately trying to mislead you by using ambiguous English and claiming on the basis of their poor communication skills to be clever.

## Re: (Score:3, Informative)

Yeah, say what you will about the Monty Hall problem, but at least it is stated clearly and is actually trying to trip up your intuitive understanding of probability, not your intuitive understanding of English. :P

Though of course, the Monty Hall problem as given could be considered

wrongbecause Monty later said he would usually only offer the contestant the chance to switch their choice if their first guess was correct. But hey, that's just funny.## Re: (Score:2)

the quote in question states that one and only one of the children is a boy born on Tuesday.

The quote is, and I quote:

"I have two children, one of whom is a boy born on a Tuesday. What's the probability that my other child is a boy?"

Please point, click or otherwise indicate the word "only" in that quote.

## Re:Ordering and Convergence (Score:5, Insightful)

## Re: (Score:3, Informative)

Sounds vaguely similar to the same failure-of-intuition behind the Monty Hall problem [wikipedia.org]: you fix several quantities, then you reveal one of them, and then ask for a guess for the probabilities of the remaining, unrevealed quantities. Since the unrevealed quantities are completely independent of the revealed ones, it seems like the revealed information shouldn't matter, and your guess should still be that they're uniformly randomly distributed. But, it isn't.

## Re: (Score:3, Informative)

Since the unrevealed quantities are completely independent of the revealed ones,But they are related. When you eliminate one wrong answer, the probability of the remaining answer being right is 50%. When you eliminate one answer at random, the chance that the remaining answer is the right one is 33%. The chance of the one you chose being right is always 33%. When you reveal an answer known to be wrong, you change the probability of the ones that was chosen from. Since the door opened can't be the one

## Re:Ordering and Convergence (Score:5, Insightful)

Yes, it is like the Monty Hall problem. What is the probability that 2 children are both boys? 25%. Knowing that one of the children is a boy does not change that probability as much as might be thought. The answer then is 33.3%, not 50%. This is because the additional information has cleverly NOT specified which child is the boy. If a particular child is picked out, eg. the

firstchild is a boy, then it is 50% the other is a boy, because it always was 50% likely that a child is a boy. The bit about "born on Tuesday" does matter, because it comes close to specifying a particular child. The more improbable it is that both children fit some criteria, the closer the probability gets to 50%. If the info had been "one of whom is a boy born on Feb 29", the answer would be nearly 50%.## Re: (Score:3, Informative)

No it does't, because the problem as stated in the summary DOES NOT EXCLUDE the case that the other child is also a boy born on Tuesday. You are ASSUMING that because one child was "A boy born on Tuesday" the other child is not also a "boy born on Tuesday", a condition that was never set forth in the problem.Yes it does matter, and no he's not assuming that.

Even thoughit is possible for both children to have the qualification "born on Tuesday" or "born on Feb. 29th", this informationdoes"come close" t## Shorter summary (Score:5, Funny)

As the article notes, it depends what you mean by "one of", (specific one vs "at least one"), and quibbling mathematicians don't always pick the most common interpretation.

In other news, an aeroplane carrying a hundred mathematicians crashed with no survivors; their university made a press release stating that one of its mathematicians died in the crash.

## Re: (Score:3, Insightful)

This problem hinges very greatly on how it is phrased and I think it's more a trick of English converting to statistics than it is a true puzzle.

100% correct. All "paradoxes" of this form are very carefully, almost ritually, stated because they are almost entirely linguistic tricks designed to mislead and confuse rather than educate and enlighten.

The very first thing anyone analyzing a problem of this type should do is restate it in several different ways, making as much implicit information explicit as possible. Such as:

I have two children.

I'm going to tell you some things about one of my children.

I have not chosen that child randomly. For examp

## Re: (Score:2)

Saying one of the children is a boy who's favourite number between 1 and 7 (inclusive) is 3 would give the same probability of 13/27.

You're just narrowing down which boy you're talking about - the more specific you get the closer the probability tends towards one.

"The general rule is that for a set with N possibilities you would have (N*2 - 1)/(N*4 - 1)."

The wisdom of EldavoJohn [slashdot.org]

## Re: (Score:2)

Gah!

## Re:Ordering and Convergence (Score:5, Insightful)

Not necessarily. Any reading of an English sentence is an exercise in interpretation. We don't just read the words alone, we actually interpret them and use a mental model to help dismiss obviously incorrect ambiguities.

Let's say the sentence has several interpretations. For each interpretation, we could solve for an interpreted probability answer. Then we could look at each answer and ask if it makes sense. If that answer doesn't make sense, we could dismiss that particular interpretation of the sentence. If a single answer remains after that, it would be THE answer (Sherlock Holmes style).

In the example, here are some possible disambiguations:

1) "I have two children, (exactly) one of whom is a boy born on a Tuesday."

2) "I have two children, (at least) one of whom is a boy born on a Tuesday."

3) "I have two children, one of whom is a boy born on a (particular) Tuesday."

4) "I have two children, one of whom is a boy born on a (generic) Tuesday."

There is also an ambiguity in the second sentence, which is only obvious to statisticians and probabilists:

a) "What's the (Bayesian subjective) probability that my other child is a boy?"

b) "What's the (objective) probability that my other child is a boy?"

In case a), the problem is underspecified as it requires the full set of personal beliefs of the reader to be used for an answer (Bayesian subjectivists propose that a probability is merely a degree of personal belief, such that two people will not agree on the probability for the same event, because, being different people, they have different prior beliefs.)

In case b), the problem can (should) be solved solely from the problem and general common knowledge of the world (which is still required to interpret the question).

## Let's try it without reading TFA (Score:5, Informative)

X=one boy is born on a tuesday

P(X|boyboy) = 1/7 + (6/7*1/7) = 13/49

P(X|boygirl) = 1/7

P(X|girlboy) = 1/7

P(boyboy) = P(boygirl) = P(girlboy) = 1/3

P(X) = (1/7 + 1/7 + 13/49)/3 = 9/49

Using Bayes's theorem:

P(boyboy|X) = P(X|boyboy)*P(boyboy)/P(X) = 13/49 * 1/3 * 49/9 = 13/27

Which is different from 1/3. So yes, the weekday of birth is significant.

## Re: (Score:2)

You miss one thing: if you have two children, there is a X% chance that they are twins. IF they are twins, they have a close to 100% probability of sharing a birthday, and a >50% probability of sharing gender (identical twins share gender, non-identical share gender in 50% of cases).

Some googling says: .4%

P(identical twins) ~

P(non-ident twins) ~ 2.6%

The maths involved are left as an exercise to the reader :-)

## Re: (Score:2)

## Probability (Score:5, Informative)

This is kind of like saying "I flip a coin. What is the chance it lands heads facing up?"

And you say "50%."

And I say, "Incorrect. There is a very small chance it will land balanced perfectly on it's side, so both the chance of heads and the chance of tails is under 50%."

## Re: (Score:3, Interesting)

There is a very small chance it will land balanced perfectly on it's side

Has anyone ever seen that happen?

Yes, I've had that happen once with a 10Fr coin (very similar in shape to the current 1 euro coin). The ground was irregular which probably helped a lot.

## Good job. (Score:2)

Nice work! I read the article and it agrees with your math. 13/27 is exactly what the author concludes.

## Re: (Score:2, Insightful)

## Re:Let's try it without reading TFA (Score:5, Interesting)

It's playing games with words and attaching significance to two sets that in any practical case I can think of would be considered one.

The argument is that if you were to consider it as a set, there are four possible ways for your children to be distributed:

1. (Boy, Boy)

2. (Boy, Girl)

3. (Girl, Boy)

4. (Girl, Girl)

We already know that your children can't possibly fall into the fourth set, and so looking at the sets it appears that the probability should be 1/3. But this misses one minor point - you've added an extra set which only makes sense if you wish to attach significance to the order in which the children were born (Sets 2 and 3). But as soon as you do attach that significance, the information you are given in order to establish the probability of any particular outcome (eg. the boy is older) allows you to eliminate two sets rather than just one.

## Re: (Score:2)

Shouldn't ordering matter in the first part? You're told that the

firstchild was a boy, so of the four ordered possibilities (boy, boy), (boy, girl), (girl, boy), (girl, girl), you can eliminateboth(girl, boy) and (girl, girl).## Re: (Score:3, Informative)

## Re: (Score:2)

Hmm, somehow I misread that several times in the problem statement. I could've sworn it said, "he has two children, the first of which was a boy born on a Tuesday". But then clearly the problem would be much less paradoxical, so makes more sense now.

## Re: (Score:2)

So yes, the weekday of birth is significant.

To clarify: I mean, of course, that

the fact that the weekday of birth is specifiedis significant. It doesn't matter whether that day is Tuesday, Friday, or even Monday. (Assuming uniform probability of childbirth across weekdays, etc.)## Re: (Score:2, Insightful)

You are good:) I have a question.

I met my girlfriend on Thursday. She is receptionist. What is the probability of my second girlfriend being supermodel?

## Re: (Score:2)

If it is known that I have a boy, there is 1/3 probability that the other is also a boy.I'm confused. Say we took 900 families with two children, one of whom is a boy. Are you saying that we would expect 300 of them to be (boy,boy) and 600 would be (boy, girl)?

## The difference between a man and a woman (Score:5, Funny)

A man is asked on the street: What is the probability you will come across a dinosaur on the street today?

The man replies: less than 0,000000001%

When a woman is asked the same question, she replies:

50% - I either will or I won't.

So, really, it depends on who you ask.

## When the OP Is a spoiler (Score:2)

> The submitter adds, "Believe it or not, the Tuesday thing is relevant. Well, sort of. It's ambiguous."

Why even say that? Anyone who does these puzzles likes to figure it out by themselves.

## What's counterintuitive about it? (Score:2)

## Re: (Score:3, Informative)

Compare these two questions:

"I have two children, one of whom is a boy born on a Tuesday. What's the probability that my other child is a boy?"

"I have two children, one of whom is a boy. What's the probability that my other child is a boy?"

Most people would think that the being born on a Tuesday bit was irrelevant and would make no difference to the answer. In fact it makes a big difference to the answer.

## Re: (Score:2)

## Re:What's counterintuitive about it? (Score:4, Insightful)

This is a question written in purposely misleading English.

This, in other words, is a shit question.

## Re: (Score:2)

Don't try to think too hard about it, statisticians think about these kinds things in order to convince themselves that they're more intelligent than people who actually do something worthwhile with their lives. From the original problem:

"Suppose that Mr. Smith has two children, at least one of whom is a son. What is the probability both children are boys?"

They then go on to list options of:

Boy, girl

Boy, boy

Girl, boy

when birth order has nothing to do with the original question. Options one and three are the

## another way of looking at it? (Score:2)

## Other problems (Score:3, Insightful)

"I have two children, one of whom is a boy born in the first day of the year. What's the probability that my other child is a boy?"

"I have two children, one of whom is a boy born in January. What's the probability that my other child is a boy?"

"I have two children, one of whom is a boy born in Winter. What's the probability that my other child is a boy?"

Do they give different probabilities?

## Re: (Score:2)

It's P=(4x-1)/(2x-1) where x is the number of options for the extra information.

So first day of year, x=365, (ignoring leap years),

born in January x=12,

winter x=4.

## Principles of Restricted Choice (Score:4, Insightful)

This is related to the Principle of Restricted Choice [wikipedia.org] often seen in Contract Bridge.

If the parent has two boys born on a Tuesday, he could equally have declared the other boy as being born on a Tuesday. In a parallel universe, the other boy would have been declared as being born on a Tuesday, whereas if only one of the child was a boy born on Tuesday nothing would have changed in any of the other parallel universes. Therefore the effect is the probability of 2 boys borne on Tuesday has been halved, resulting in 13/27 probability of the second child being a boy.

## Johnny's Mom Has 3 Kids... (Score:5, Funny)

## Re: (Score:3, Informative)

i've obviously not had enough to drink - what's the answer?

The riddle starts out as "Johnny's Mom has 3 kids" therefore one of the kids must be named Johnny.

## OK I'm stupid (Score:2)

Once you're told that there's at least one boy, doesn't that mean that one result is fixed and has no outcome on the second result?

The calculations for a 1/3 chance would make sense in the following situation: You flip two coins, if you get two tails, you flip again. What is the chance of you ending up with two heads?

To me the question doesn't match the answer, the questio

## Re:OK I'm stupid (Score:4, Informative)

Close. The situation with coins is this:

I flip a coin twice and record the answer. I repeat this many times. I discard all the pairs where both coins came up tails. I then select a pair where at least one toss came up heads. What is the probability that the other is also heads?

The selection criteria screw up the probabilities. If instead I flip a coin, see that it comes up heads, and ask what the probability of it coming up heads again is, the answer is 50%.

## I love mathematicians... (Score:5, Funny)

Take an abstract mathematical problem, invent a pseudo-real-world context, rephrase the problem very sloppily and ambiguously in plain English then laugh smugly when people get the wrong answer.

The correct answer to the question, by the way, is "I don't know - you have not given me enough information, and I'd have to go check that the gender of successive offspring from the same couple

isactually independent, but its probably gonna be somewhere between 1/3 and 2/3 - and since you'd have to somehow re-formulate it as a viable experiment and run it 100 times to confirm that result, only the Bayesians give a flying fuck what the precise value is".In other news: you can't actually build a hotel with an infinite number of rooms - you'd run out of bricks - so don't try and engage my interest in all the weird thing that would happen if you did something impossible. And stop hiding goats behind my door!

## Re:I love mathematicians... (Score:5, Funny)

In other news: you can't actually build a hotel with an infinite number of rooms - you'd run out of bricks

Don't let the venture capitalist who's funding me know that. I duped him into paying me an infinite amount of money! He paid me $1000 for the first room, $500 for the second, $250 for the third, etc. I'll be rich! Rich, I tells ya! I've been eying this nice $2000 watch, and I should have enough for it any day now...

## Re: (Score:3, Insightful)

So the correct answer is "That depends. Is your other child a boy?"

## The last name is even MORE important (Score:5, Funny)

If the dad is Schrödinger the other kid is both born and unborn at the same time.

## Man this question pisses me off. (Score:5, Insightful)

"I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?"

I have three children. But I also have 2 children. See the problem? I have a son born on a Tuesday, and I have another son born on a Tuesday. See the problem?

It doesn't say I have *only* two children. It doesn't say the other child can't be a son born on a Tuesday. It assumes the birth rate is 50/50, but most statistics agree it's not even. FTA, it assumes there's no such thing as twins. It assumes you have only one wife. But none of this shit is specified.

Pisses me off. Use coins and cards. Not assumed biblical customs.

## As always, ambiguous language (Score:3, Insightful)

"I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?"

As always, the challenge is the assumptions intentionally hidden in the problem statement.

"I" - was your family chosen at random, and if so, from what set?

"two children" - exactly or at least?

"one of whom" - exactly or at least?

"son" - was the sex to say chosen at random, or did you pick a child and announce his/her sex?

"Tuesday" - was the day chosen at random, or did you pick a child and announce his.her birthday?

"What is the probability..." - Some parent you are! Don't you know the sex of your own children?

Simply and honestly reveal the assumptions and the math is straightforward.

"Given a family, chosen at random from the set of all families that have exactly two children and have at least one son born on a Tuesday, what is the probability that both children are boys?"

To make the math easier, let's start with 196 families with two children, with the expected mix of boys and girls. 49 (25%) have two boys and 98 (50%) have a boy and a girl. Of the 98 boy-girl families, 84 do not have a Tuesday-Boy, leaving 14 that do. Of the 49 boy-boy families, 36 do not have a Tuesday-Boy, leaving 13 that do. That leaves a total of 27 families, of which 13 have at least one son born on a Tuesday.

So the probability is 13/27.

Reveal different assumptions, and the answer changes.

## Reminds me of an old joke (Score:5, Funny)

- Hey, we could fly over together if you'd like.

- Thanks, but I'll be driving.

- All that way? It'd take you most of the day! Whatever for?

- Well, I recently made a study of the statistics of bombs being smuggled on board passenger planes. And while the odds of it occurring on any particular flight are high, the possibility still makes me uncomfortable with flying.

- Well, suit yourself.

I'mgoing to take the plane.A short time later, the one professor is boarding her flight out to the conference, and who should be sitting in the adjacent seat but her old friend! They're both pleasantly surprised, and the first professor settles into her seat. She leans in and quietly asks her friend -

- So what about that whole probability issue? Was your math off, or did you just work up the nerve?

- Wrong on both counts! I did have a breakthrough, however.

- Really? How do you mean?

- Well, I went over the statistics again, and worked out the odds of

twobombs being separately smuggled on board the same flight.- High?

- Astronomical! You've a better chance of being struck by lightning!

- So how does knowing that make you more comfortable with flying?

- (singsongs) Guess what I've got in the briefcase...(pats the case on her lap)

.

## Two Slashdot Readers (Score:4, Funny)

I have two slashdot readers and one of them did not read the article before they posted a reply on a Tuesday. What is the probability that the other didn't read the article as well?

100%

## Dear /. (Score:3, Funny)

Can you please create a flag for everyone that posted a wrong conclusion and then filter them out of all my views in the futures.

I wish to do this because it will eliminate people who don't read the articles and people who can't do math.

Thank you.

## Re: (Score:3, Informative)

13/27

## Re: (Score:2, Funny)

Only if you believe in randomness. If the other child in fact is a boy the probablility for it is 1.

## Re:Well? (Score:4, Insightful)

My older brother and I were both born on Tuesdays.

## Re: (Score:3, Funny)

My older brother and I were both born on Tuesdays.

Nothing to see here, just a systemic anomaly. Move along, now.

## Re:Well? (Score:4, Funny)

...what that actually represents is both having one rock in your presence butt also...

I'm sure glad I don't have a rock present in my butt...

## Re: (Score:3, Insightful)

Here is the list of outcomes if it is possible for both boys to have been born on a Tuesday:

Girl on Monday, Boy on Tuesday

Girl on Tuesday, Boy on Tuesday

Girl on Wednesday Boy on Tuesday

Girl on Thursday, Boy on Tuesday

Girl on Friday, Boy on Tuesday

Girl on Saturday, Boy on Tuesday

Girl on Sunday, Boy on Tuesday

Boy on Monday, Boy on Tuesday

Boy on Tuesday, Boy on Tuesday

Boy on Wednesday Boy on Tuesday

Boy on Thursday, Boy on Tuesday

Boy on Friday, Boy on Tuesday

Boy on Saturday, Boy on Tuesday

Boy on Sunday, Boy on

## Re:Well? (Score:5, Informative)

Assuming 50/50 chance of boy or girl and a 1/7 chance of each weekday, let's look at the general population of families with two childs:

B1B1 B1B2 B1B3 B1B4 B1B5 B1B6 B1B7

B2B1 B2B2 B2B3 B2B4 B2B5 B2B6 B2B7

B3B1 B3B2 B3B3 B3B4 B3B5 B3B6 B3B7

B4B1 B4B2 B4B3 B4B4 B4B5 B4B6 B4B7

B5B1 B5B2 B5B3 B5B4 B5B5 B5B6 B5B7

B6B1 B6B2 B6B3 B6B4 B6B5 B6B6 B6B7

B7B1 B7B2 B7B3 B7B4 B7B5 B7B6 B7B7

B1G1 B1G2 B1G3 B1G4 B1G5 B1G6 B1G7

B2G1 B2G2 B2G3 B2G4 B2G5 B2G6 B2G7

B3G1 B3G2 B3G3 B3G4 B3G5 B3G6 B3G7

B4G1 B4G2 B4G3 B4G4 B4G5 B4G6 B4G7

B5G1 B5G2 B5G3 B5G4 B5G5 B5G6 B5G7

B6G1 B6G2 B6G3 B6G4 B6G5 B6G6 B6G7

B7G1 B7G2 B7G3 B7G4 B7G5 B7G6 B7G7

G1B1 G1B2 G1B3 G1B4 G1B5 G1B6 G1B7

G2B1 G2B2 G2B3 G2B4 G2B5 G2B6 G2B7

G3B1 G3B2 G3B3 G3B4 G3B5 G3B6 G3B7

G4B1 G4B2 G4B3 G4B4 G4B5 G4B6 G4B7

G5B1 G5B2 G5B3 G5B4 G5B5 G5B6 G5B7

G6B1 G6B2 G6B3 G6B4 G6B5 G6B6 G6B7

G7B1 G7B2 G7B3 G7B4 G7B5 G7B6 G7B7

G1G1 G1G2 G1G3 G1G4 G1G5 G1G6 G1G7

G2G1 G2G2 G2G3 G2G4 G2G5 G2G6 G2G7

G3G1 G3G2 G3G3 G3G4 G3G5 G3G6 G3G7

G4G1 G4G2 G4G3 G4G4 G4G5 G4G6 G4G7

G5G1 G5G2 G5G3 G5G4 G5G5 G5G6 G5G7

G6G1 G6G2 G6G3 G6G4 G6G5 G6G6 G6G7

G7G1 G7G2 G7G3 G7G4 G7G5 G7G6 G7G7

Each outcome has P = 1/(7*7*4) = 1/196

Let's only look at the families with (at least one) tuesday boy:

B1B2

B2B1 B2B2 B2B3 B2B4 B2B5 B2B6 B2B7

B3B2

B4B2

B5B2

B6B2

B7B2

B2G1 B2G2 B2G3 B2G4 B2G5 B2G6 B2G7

G1B2

G2B2

G3B2

G4B2

G5B2

G6B2

G7B2

Of these 27 families, 13 have another boy. So P = 13/27.

## Re: (Score:3, Informative)

Assuming 50/50 chance of boy or girl and a 1/7 chance of each weekday, let's look at the general population of families with two childs:B1B1 B1B2 B1B3 B1B4 B1B5 B1B6 B1B7B2B1 B2B2 B2B3 B2B4 B2B5 B2B6 B2B7

B3B1 B3B2 B3B3 B3B4 B3B5 B3B6 B3B7

B4B1 B4B2 B4B3 B4B4 B4B5 B4B6 B4B7

B5B1 B5B2 B5B3 B5B4 B5B5 B5B6 B5B7

B6B1 B6B2 B6B3 B6B4 B6B5 B6B6 B6B7

B7B1 B7B2 B7B3 B7B4 B7B5 B7B6 B7B7

B1G1 B1G2 B1G3 B1G4 B1G5 B1G6 B1G7B2G1 B2G2 B2G3 B2G4 B2G5 B2G6 B2G7

B3G1 B3G2 B3G3 B3G4 B3G5 B3G6 B3G7

B4G1 B4G2 B4G3 B4G4 B4G5 B4G6 B4G7

B5G1 B5G2 B5G3 B5G4 B5G5 B5G6 B5G7

B6G1 B6G2 B6G3 B6G4 B6G5 B6G6 B6G7

B7G1 B7G2 B7G3 B7G4 B7G5 B7G6 B7G7

G1B1 G1B2 G1B3 G1B4 G1B5 G1B6 G1B7G2B1 G2B2 G2B3 G2B4 G2B5 G2B6 G2B7

G3B1 G3B2 G3B3 G3B4 G3B5 G3B6 G3B7

G4B1 G4B2 G4B3 G4B4 G4B5 G4B6 G4B7

G5B1 G5B2 G5B3 G5B4 G5B5 G5B6 G5B7

G6B1 G6B2 G6B3 G6B4 G6B5 G6B6 G6B7

G7B1 G7B2 G7B3 G7B4 G7B5 G7B6 G7B7

G1G1 G1G2 G1G3 G1G4 G1G5 G1G6 G1G7

G2G1 G2G2 G2G3 G2G4 G2G5 G2G6 G2G7

G3G1 G3G2 G3G3 G3G4 G3G5 G3G6 G3G7

G4G1 G4G2 G4G3 G4G4 G4G5 G4G6 G4G7

G5G1 G5G2 G5G3 G5G4 G5G5 G5G6 G5G7

G6G1 G6G2 G6G3 G6G4 G6G5 G6G6 G6G7

G7G1 G7G2 G7G3 G7G4 G7G5 G7G6 G7G7

Why are you assuming that order matters, but boys are interchangable? If we label the child that was mentioned in article as A and the other one as B then B2B2 splits into two cases when A has an older brother and when he has a yonger brother, therefore we get:

B2B1 B2B2 B2B3 B2B4 B2B5 B2B6 B2B7

B1B2

B2B2

B3B2

B4B2

B5B2

## Re:MOD PARENT UP (Score:5, Insightful)

In my experience, many non-intutive probabilty results are easier to understand if you spell out the full population. For example, I coudn't understand http://en.wikipedia.org/wiki/Berkson's_paradox [wikipedia.org] until I drawed it up graphically.

## Order is irrellevent but uniqueness is not. (Score:4, Informative)

You are neglecting to treat the two children as independent non-exchangeable objects. It is easier to think about if you consider that you have two pets a dog and a cat, each of which can be male or female. Then the enumeration of possibilities:

Male Dog, Male Cat

Male Dog, Female Cat

Female Dog, Male Cat

Female Dog, Female Cat

If you select a family where (at least) one of them is a Male that leaves the following options:

Male Dog, Male Cat

Male Dog, Female Cat

Female Dog, Male Cat

So the chances of both being male are 1/3.

Now moving onto the date question. You have selected a family from families known to have one male pet born on Tuesday (and one dog and cat each). The options are:

Male Cat born Tuesday, Male Dog born Monday

Male Cat born Tuesday, Male Dog born Tuesday *

Male Cat born Tuesday, Male Dog born Wednesday

Male Cat born Tuesday, Male Dog born Thursday

Male Cat born Tuesday, Male Dog born Friday

Male Cat born Tuesday, Male Dog born Saturday

Male Cat born Tuesday, Male Dog born Sunday

Male Cat born Tuesday, Female Dog born Monday

Male Cat born Tuesday, Female Dog born Tuesday

Male Cat born Tuesday, Female Dog born Wednesday

Male Cat born Tuesday, Female Dog born Thursday

Male Cat born Tuesday, Female Dog born Friday

Male Cat born Tuesday, Female Dog born Saturday

Male Cat born Tuesday, Female Dog born Sunday

Male Dog born Tuesday, Male Cat born Monday

Male Dog born Tuesday, Male Cat born Tuesday *

Male Dog born Tuesday, Male Cat born Wednesday

Male Dog born Tuesday, Male Cat born Thursday

Male Dog born Tuesday, Male Cat born Friday

Male Dog born Tuesday, Male Cat born Saturday

Male Dog born Tuesday, Male Cat born Sunday

Male Dog born Tuesday, Female Cat born Monday

Male Dog born Tuesday, Female Cat born Tuesday

Male Dog born Tuesday, Female Cat born Wednesday

Male Dog born Tuesday, Female Cat born Thursday

Male Dog born Tuesday, Female Cat born Friday

Male Dog born Tuesday, Female Cat born Saturday

Male Dog born Tuesday, Female Cat born Sunday

* Note that I enumerated the case where both are males born on Tuesday twice. These are redundant and one must be discarded else I will double count that situation. After doing so there are 13/27 cases where both are males.

Notice that if you ignored the fact that one was a dog and the other was a cat you would have merged the two lists, ending up with your original list, and double counting the case where both are boys born on Tuesday.

In other words your mistake is that you assumed you had been given the sex and birth date of child A, and enumerated the sex and birth date of child B. However, you don't know the sex and birth date of child A or B, just that one of child A or B have that sex and birthdate. That is a subtly different problem.

So the order that they are born in is irrelevant, but keeping track of the fact that they are the two unique items while enumerating the cases is vital (and older and younger is a simple label to use while doing so).

## Re: (Score:3, Informative)

It's not that the order in time is especially relevant. It's that the identity of the child matters. You could call them child A and child B, and list the permutations, and the answer would still be 13/27. Child A being born on a Tueday are NOT the same set of permutations as child B being born on a tuesday. They have to be listed separately so that duplicate permutations can be removed from the list. That's what makes the difference.

## Re:Well? (Score:4, Informative)

What you (or the website you copied and pasted the ratio from) fail to take into account (and why it's a Gambler's fallacy problem) is that when involving chance, anything that happened in the past is completely irrelevant to future probables. I could roll a die 99 times, and get 6, the probability of getting 100 6's when I've already got 99 6's is still 1 out of 6, not 6^100.

The reason the chi square doesn't come into play here is because it doesn't MATTER the order. Has she said "What is the probability my SECOND-BORN was a boy?" it would be perfectly logical to write the square because the boy who was born on Tuesday could be either the first born or the second born, she never stipulated.

We can say that the boy, who was born on a tuesday, was also a Gemini. Does this change the ratio? No, the probability of having two boys is still 50-50%, because the unknown only has two possible outcomes: boy or girl.

## Re:Well? (Score:5, Informative)

Failed to read the article, hey?

You're absolutely correct IF you assume that the man is just describing one of his children. That is the logical interpretation of the problem. However, if someone went out and specifically selected a family with at least one boy, the probability of the other child being a girl is not 50% - the population in question has been artificially depleted of girls because the criteria excluded all two-girl families.

## Re:Well? (Score:5, Informative)

Read the article. Whether the answer is 50% or not depends on the context, and that context is not specified in the problem.

If you meet the person with his son, and he tells you that that son is born on a Tuesday, then the chance of the other kid being a boy is 50%. If you see him as one of the thousands of parents who have two children, at least of which is a boy who's born on a Tuesday, the chance of the other kid being a boy is 13/27.

## Re:Well? (Score:4, Funny)

Did I win the lottery last week? The unknown only has two possible outcomes: I won or I lost.

Therefore, based on your math, my odds are 50-50%.

University of Phoenix online wants their diploma back. :-)

## Re: (Score:3, Informative)

Or to put it another way: If it was 25% chance like you say then so is FM and MF. There are no other possibilities so it should add up to 100%, but you only get 75%...

## Re:Well? (Score:5, Informative)

You're right! Pretty amazing, considering the article gives the correct answer and explains it pretty thoroughly. Even when you think: "But wait! Can't you look at it some other way?", the article does just that.

Really, this time it pays to RTFA.

## Re: (Score:3, Informative)

I'm still able to reach the site. Here's the entire text:

When intuition and math probably look wrongA twist on the Two Children Problem shows how information can steer what looks probable.

By Julie Rehmeyer

I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?

Gary Foshee, a puzzle designer from Issaquah, Wash., posed this puzzle during his talk this past March at Gathering 4 Gardner, a convention of mathematicians, magicians and puzzle enthusiasts held bian

## Re: (Score:2)

Maths fail..

.. this is the same as sayingNo it isn't. Go and read the article again.

## Re: (Score:3, Interesting)

## Re:Rubbish (Score:5, Informative)

"I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?" .. this is the same as saying "I have just tossed a 10 pence coin and it has come up heads, what is the probability that another coin toss will come up heads?

No, it isn't. It's the same as saying, "I have just tossed a quarter twice, and given that it came up heads at least once, what's the chance it that it came up head both times?"

We can do an analogue, but not in the way you're saying. I.e., "I have just tossed a coin and it came up heads, what's the chance it will come up heads if I toss it again" == "I have just had a kid and it's a boy, if I have another kid, what's the chance it will be a boy" == 50%

If you don't understand the problem, that's fine. It's counterintuitive, these things usually take a second look. If you don't understand the problem and you want to claim the solution is incorrect because you don't get it, well, that's something else entirely.

## Re: (Score:2)

Actually, I made a mistake up there...on the first line, I meant to take the Tuesday thing out. For reasons that are explained in the article.

## Re: (Score:3, Informative)

Nope. Wrong. Age does matter. Not because it correlates with gender or anything, but because it makes the two children distinguishable. (E.g., one of them must be the older child, one must be the younger.)

You don't have to distinguish them with ages. You could distinguish them with height, e.g., "the taller child is a boy" or anything else.

Probability in this case is about how much information you have. The more information you have, the more you can refine your results. If I know that a parent has two chil

## Re: (Score:3, Informative)

. this is the same as saying "I have just tossed a 10 pence coin and it has come up heads, what is the probability that another coin toss will come up heads?"

Nope, it is equivalent to "I have just tossed a 10 pence coin

twice, and I tell you that it has come up headsat least once, what is the probability that it has come up heads twice".The 2/3 vs 1/3 probability hinges on the fact that the ordering of the kids is not defined.

If the kid's father told you "my

oldestchild is a boy", then you would be right.Unfortunately, ...), which m

anydefined order can play that role ("the first of his kids that I met in person", "the first of his kids that he mentioned",## Re:Rubbish (Score:5, Informative)

Nope.

"I have just tossed a 10 pence coin and it has come up heads, what is the probability that another coin toss will come up heads?"

Is *not* the same as:

"I just tossed two coins and one of them was heads, what is the probability that the other one was heads as well?"

## Re: (Score:3, Informative)

""I have just tossed a 10 pence coin and it has come up heads, what is the probability that another coin toss will come up heads?""

Umm no, if you look at the original problem, "one of my 2 kinds is a boy, what's the probability both are boys" is like saying "Ive've tossed a coin twice, at least one of which was heads, what;s the probability that both were heads?" - so here the possibilities are H H / H T/ T H of which only one of the 3 is both heads so it's 1/3rd.

It's not saying, "the first guy was a boy -

## Re: (Score:2, Insightful)

## Re: (Score:3, Insightful)

I'm at a company picnic and I have 5 employees sitting around. I need 2 more players for the women's volleyball team, so I take 2 of the women away. What is the sex distribution of the rest? The issue is I'm cherrypicking based on a condition. Here the condition is clear: I'm picking based on sex.

The confusion in the Tuesday problem comes in because the condition doesn't appear relevant. Who cares about Tuesday?? The assumption is he picked one of his kids to talk about.

## Re: (Score:2)

Ah right.. I get it.. while previous statement is 100% correct and true, the Tuesday statement does put down restrictions on the second child and thus reducing the 50% chance..

## Re: (Score:2)

Exactly. The second child is not a boy born on a Tuesday.

## Re: (Score:2)

I don't know about the chance of it landing heads up, but if you threw it

up, it has a rather high chance of being damaged by your stomach acid...## Re:Summary misstates the problem (Score:5, Insightful)

"I have two children, one of whom is a boy. What's the probability that my other child is a boy?" ... it is given that the FIRST child is a boy.

I must admit that English is not my native tongue but I fail to see how this gives that the FIRST child is a boy. Doesn't "one of whom" implies that it can be either the first or the second?

## Wrong (Score:3, Informative)

"Select the gender of all second children where the first child was born on a Tuesday and the first child was male." Yes, it will be 50/50 male and female. "Select the gender of all first children where the second child was born on a Tuesday and the second child was male." Again it will be 50/50.

But the gender split in the union of those two groups will *not* be 50/50. You have counted the families with two boys born on a Tuesday twice. 1 in 14 of the first group will also be in the second group. Takin

## Re:Science and Intuition defeating Fun Math (Score:4, Informative)

No joke, you break out the SQL.

I did so.

I generated a bunch of SQL records, into a table with two relevant fields, both binary. (Male was indicated by false, female was indicated by true).

Then I counted how many SQL records were returned WHERE gender1 = male OR gender2 = male

Then I counted how many SQL records were returned WHERE gender1 = male AND gender2 = male.

The results? Out of a population of 100,000 records: 24940 male-male, 74893 at least one male. Yielding 1/3 to the third decimal place.

The mistake you made (and it is a mistake, because you claim your experiment would show a result that it empirically does not) is that you failed to count properly. There will always be approximately twice as many records with one boy as there are two boy records.

It's not the BIRTH ORDER that turns this problem into something interesting.

As a side note, in the future, actually run your experiments if it is actually feasible to do so.

## Re: (Score:3, Insightful)

#!/usr/bin/perlmy @set;for my $gen (1 .. 100000) {my $sex1 = rand > 0.5 ? "m" : "f";

my $sex2 = rand > 0.5 ? "m" : "f";

push @set, [$sex1, $sex2];}

my $count = 0;my $total = 0;

foreach my $pair (@set) {next if ($$pair[0] ne "m" and $$pair[1] ne "m");

## Re: (Score:3, Informative)

The first step is to see that if one of the children is a boy, then the probability that they both are is 1/3 not 1/2.

## Re:The other problem posed in TFA (Score:4, Insightful)

In 1000 pairs of children you'll have 250 girl/girl, 250 boy/boy, 500 girl/boy.

Of those, the ones that have at least one boy are the 250 boy/boy and 500 girl/boy pairs. So there's a 33% chance it's boy/boy if you know one is a boy.

The whole point is you could be talking about either of the boys in the 250 boy/boy pairs - it doesn't increase the probability that it's boy/boy instead of girl/boy (you're still twice as likely to have a girl/boy pair relative to a boy/boy pair). If you specify more about the boy you're talking about - for example (ironically) saying his name is Peter - then the boys are no longer interchangable and the probability tends towards 1/2.

It is tricky

## Re:The other problem posed in TFA (Score:5, Informative)

Your mistake is in believing that, by virtue of naming one of the boys Peter, the probabilities are magically equalized. They're not. The correct probabilties for your table are:

Peter, Boy = 1/6

Boy, Peter = 1/6

Peter, Girl = 1/3

Girl, Peter = 1/3