The Tuesday Birthday Problem

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Posted by kdawson on Monday June 28, 2010 @09:09PM from the if-it's-tuesday-it-must-be-a-girl dept.

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."

15 of 981 comments (clear)

Ordering and Convergence by eldavojohn · 2010-06-28 21:20 · Score: 5, Informative

First off, I am a huge Martin Gardner fan and if this puzzle intrigues you and you haven't heard of him, get one of his books.

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 two objects 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?

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1. Re:Ordering and Convergence by dakrin9 · 2010-06-28 21:34 · Score: 5, Informative
  
  This is incorrect - the question DOES not disallow the second child being a boy and born on Tuesday.
  
  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."
2. Re:Ordering and Convergence by martin-boundary · 2010-06-28 22:58 · Score: 5, Insightful
  
  "it's more a trick of English converting to statistics than it is a true puzzle". And that is why I don't accept any mathematical answer to the riddle as 100% correct.
  
  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).
3. Re:Ordering and Convergence by bzipitidoo · 2010-06-28 23:13 · 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 first child 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%.
  
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4. Re:Ordering and Convergence by radtea · 2010-06-29 02:09 · 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.
  
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  Blasphemy is a human right. Blasphemophobia kills.
Let's try it without reading TFA by Shin-LaC · 2010-06-28 21:22 · Score: 5, Informative

Let's assume that, if I have two children, it is equally probable that they are born as boy+boy, boy+girl, girl+boy, girl+girl. If I have one boy, girl+girl has probability 0, and the other options are equally likely, so they have probability 1/3. If it is known that I have a boy, there is 1/3 probability that the other is also a boy.

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.
1. Re:Let's try it without reading TFA by jimicus · 2010-06-28 22:54 · 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.
The difference between a man and a woman by dimethylxanthine · 2010-06-28 21:27 · Score: 5, Funny

This reminds me of a famous joke and variations thereof, (at least around eastern europe):

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.
Shorter summary by williamhb · 2010-06-28 21:36 · 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.
Johnny's Mom Has 3 Kids... by mim · 2010-06-28 21:54 · Score: 5, Funny

I used to tend bar and this is not a math puzzle, but fun for messing with the barflies when they've had a beer or 5 and start wanting to tell you their life story. First, as you pose the question, take out 3 coins (this only translates well using USA coins, one being a nickle, the other a penny, the third a quarter, dime or fifty cent piece) and state that "Johnny's Mom has 3 kids, the first one is named 'Penny,' (point to the penny) the second one is name is 'Nicky' (point to the nickle) and then point to the third coin (doesn't matter which you use) and ask What is the third child's name?" Then see how long it takes them to figure it out. And then whether or not they leave you a tip.
I love mathematicians... by itsdapead · 2010-06-28 22:06 · 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 is actually 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!

--
In a survey of 100 programmers, 111111 thought that duck-typing was a good idea.
1. Re:I love mathematicians... by Asmor · 2010-06-28 23:16 · 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...
Man this question pisses me off. by hellop2 · 2010-06-28 23:04 · Score: 5, Insightful

The constraints are not defined.

"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.

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Re:Well? by prionic6 · 2010-06-29 00:50 · 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.
Reminds me of an old joke by RevWaldo · 2010-06-29 03:25 · Score: 5, Funny

It's the 1970's. Two math professors, old friends who both live in London, are on the phone discussing an upcoming conference in Edinburgh they'll both be attending.

- 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'm going 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 two bombs 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)

.

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