Coding Horror: Finishing The Game

http://www.codinghorror.com/blog/archives/001204.html

by Anon84 • 18 years ago

52 39 18 years ago

codinghorror.com

chime2• 18 years ago

URGH! I just lost :(

chime2• 18 years ago

Aw come on... I was just following Rule#3: http://en.wikipedia.org/wiki/The_Game_(mind_game)

pchristensen2• 18 years ago

Sorry Jeff, I understand the problem, but the language of the setup was wrong. Since a PERSON told you they have one girl, the GB and BG are equivalent and collapse to one case instead of two in the way that normal people talk.

If you had said that a mathematician or a statistician said they have one girl, that would be a different story.

dinosaur5• 18 years ago

I'm not sure what you mean be equivalent, but the relevant point is that a family is twice as likely to have one boy and one girl than have two girls.

barrkel• 18 years ago

I don't think you do understand the problem. The thing you're taking issue with - GB vs BG - is the vitally important to the statistics, and is the thing that "normal" people get wrong when they see this kind of problem for the first time.

The only real objection to the problem as it's posed is that it's pretty unlikely that anyone would say that one of their children was a girl when both of their children are girls. For a normal conversation, the ambiguity wouldn't be present; "two children" combined with "one is a girl" means the chances of one boy and one girl are 100%.

It's a bit like the old joke about "which month of the year has 28 days?" - answer being, they all have 28 days, just some have more.

However, in other less contrived situations, this is important. People risk discounting permutations, and only considering combinations, when the permutations are necessary to get a correct view of the odds.

seano7• 18 years ago

His point is that a person telling you they have one girl would be accepted as referring to a particular child. Thus, BG and GB merge and the odds are 50% of a boy and a girl.

joe_the_user• 18 years ago

The only collapse I see is that it's possible that "I have one girl" would imply ONLY one girl.

BG and GB as possibilities that often collapse in the minds of those thinking about this problem but that is exactly why the problem is paradoxical, why a certain group of people not only get it wrong but also can't let go of their wrong answer.

danprager• 18 years ago

Another possible collapse would be to say "I have two children, and the oldest [or youngest] is a girl". That would disambiguate GB & BG, by effectively stating I have G? [or ?G]. Then the odds would be 50%.

Niten2• 18 years ago

Wrong, wrong, WRONG. You DO NOT understand the problem, if you think that this means the probability of the other child being a boy is 1/2. GB and BG are equivalent linguistically, but not mathematically; they do not represent identical possibilities in the space of all conceptual representations of a family of two.

Here, I'll prove it for you:

    ;; Have a child, with an equal probability of the child being a boy or a
    ;; girl.
    (defun make-child ()
      (if (eql (random 2) 0)
          'boy
          'girl))

    ;; Make a family of n members, with each child having an equal probability
    ;; of being either a boy or a girl.
    (defun make-family (n)
      (do ((family nil)
           (i 0 (+ i 1)))
          ((>= i n) family)
        (push (make-child) family)))

    ;; As in the story, meet someone at a party who informs us that (1) she or
    ;; he has two children, and (2) that at least one of these children is a
    ;; girl.  E.g.,
    ;;
    ;; Me: "Hi there, what's your name?"
    ;; Cute lady: "Jennifer"
    ;; Me: "Nice to meet you, Jennifer, I'm Mark.  What brings you here?"
    ;; Jennifer: "Well my husband is out of town on a business trip, so I
    ;;            wanted to come here and catch up with some old friends of
    ;;            mine.  Fortunately I was able to get a baby sitter for my
    ;;            two kids on such short notice.  One of the kids, Meg, has
    ;;            to be up early in the morning for dance practice and..."
    ;; Me: "Husband?  Damn, all the good ones are taken."
    ;; Jennifer: "What?"
    ;; Me: "Nothing.  Hey, hang on while I work out the probability that your
    ;;      other child is a boy, based on the information that you've already
    ;;      given me."
    ;; Jennifer: "You're weird.  Have a good evening."
    ;;
    ;; This function works by calling (make-family 2) repeatedly until we get a
    ;; family that meets both of these criteria, then returns a representation
    ;; of said family.  This is a precise analogy for the story: just as the
    ;; fictional person at this party, here we take the space of all possible
    ;; conceptual representations of a family of two, then discard any of these
    ;; that *does not* include at least one female child.  THIS IS THE ONLY
    ;; REASONABLE WAY TO HANDLE THE INFORMATION THAT JENNIFER RELATES IN THE
    ;; ABOVE EXCHANGE.
    (defun meet-at-party ()
      (do ((family (make-family 2) (make-family 2)))
          ((find 'girl family) family)))

    ;; Perform the (meet-at-party) scenario num-total times, and then return
    ;; the fraction of those times in which the family contained a boy.
    (defun run-test (num-total)
      (let ((num-with-boys 0))
        (dotimes (i num-total (float (/ num-with-boys num-total)))
          (if (find 'boy (meet-at-party)) (incf num-with-boys)))))

Just call (run-test 1000000) or something. The result is approximately 2/3.

pchristensen2• 18 years ago

seano nailed the way I interpreted the (poorly-phrased) version of it:

"Both of my kids are driving me crazy! Just yesterday I had to pick one of them up from the police station--I grounded her for a month!" - given just the information in your quote, the odds are 50% of a boy and a girl.

http://news.ycombinator.com/item?id=416555

snowbird122• 18 years ago

This kind of problem comes up all the time in a poker tournament. Each person has a different number of chips in front of them, and when they reach the final table, since so much money is at stake, they want to make a deal. How do you determine a fair deal when each person has a different number of chips and prizes are radically different for first, second, and third place. Fun problem. Answer: The independent chip model

Gambling to the rescue

raganwald• 18 years ago

It is very clearly mirrored in backgammon theory, where touneament players memorize "match equity" tables and use them to govern their use of the doubling cube.

oscardelben• 18 years ago

Interesting theory, I'd never thought of every possibility, especially in this context, where we are talking about a person, I was thinking about human psychology or behavior when I first read that article.

spolsky• 18 years ago

Jeff didn't phrase the question carefully enough. In English if you say, "I have two children, one is a girl" that CANNOT mean both are girls. If both were girls you would never say that. Saying you have 1 girl implies that you have 1 boy. Or maybe 1 girl and one hermaphrodite.

100% was the right answer.

It's easy to get people to argue when you give them an almost-ambiguous word problem; they're not arguing about the math, they're arguing about the mapping from ambiguous words to math.

dinosaur5• 18 years ago

I don't agree; Jeff was not giving a quote. Instead it is just the relevant information abstracted from whatever the person said. By choosing the quote you did, you have added more information to the problem (at least when reading it with conversational English).

I think this would be a better quote of what the person might have said:"Both of my kids are driving me crazy! Just yesterday I had to pick one of them up from the police station--I grounded her for a month!" Pulling out the information corresponding to gender and family size would give only the information given in Jeff's post.

When applying math to the real world, you have to pull out the important information and deal with just that information. But here you are doing the opposite--trying to find a real world situation that applies to the math problem. In my opinion, your example does not quite apply.

(I don't know how the probabilities change when you account for hermaphrodites, but if it changes significantly enough so that approximately 66% is a bad answer, I would find that very interesting!)

seano7• 18 years ago

lyesit• 18 years ago

Saying "A person has two children, at least one of which is a girl. What is the probability that both of the children are girls?" would more clearly explain the problem.

brfox3• 18 years ago

I think that people are debating between 3 choices: (a) 100% chance that there is a boy and girl, (b) 2/3 chance that it is a boy and a girl, or (c) 50% chance that it is a boy and a girl.

It depends very much on the exact wording of the problem.

Like many people in the comments of Jeff's post said that if the person says that "one of my kids is a girl," then common sense makes it sound like ONLY one kid is a girl and the other must be a boy (case a).

If the question is worded precisely like question #2 in this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl#Second_question (where you have a population of families and you randomly pick a family which has a girl), then you use the Bayesian approach to conclude that there is a 66% chance that the other kid is a boy (case b).

Finally, I think that for case c, if you are just walking around and you're not specifically looking for a family with a girl, and you strike up a conversation and you hear that the person randomly decides to talk about his daughter, then I think the odds are 50% that this person's other kid is a boy. This is because you did not require that you choose a family with a specific gender (like in the wikipedia #2 scenario). First you picked a person to talk to, and then this person just happened to mention he had a daughter. Basically, we did not require that the person has a girl (case b), we only learned that he has a girl (case c).

brfox3• 18 years ago

http://paulbuchheit.blogspot.com/2009/01/question-is-wrong.h...

pfedor2• 18 years ago

This can be used as a warm-up question for another fun problem (which a friend of mine was asked during a phone screen for a position at DE Shaw): What if instead of telling you that they have two children at least one of them a girl, someone told you that they have two children at least one of them named, say, Linda? The assumption is that there are no boys named Linda, that the likelihood for a girl to be named Linda is not affected by having an older brother or sister, etc. It's not a trick question, nor a sociological problem requiring any knowledge about names, it's a purely mathematical puzzle. And, as you may have guessed, the solution is not the same as in the previous case (with two children at least one of them girl).

brk• 18 years ago

someone told you that they have two children at least one of them named, say, Linda

You would need a special case to determine if you were talking to Johnny Cash's father...

vlad• 18 years ago

I'm not sure what you mean. Is the question still the same? Then why wouldn't the result be 2/3rds in either case?

pfedor2• 18 years ago

It is not the same question, they only seem that way. The information given is different. The answer in the second case depends on the popularity of the name Linda.

fhars• 18 years ago

The question is ill posed, because the anwser is highly dependent on how the couple told you that they have one girl.

If they tell you "we have two kids", you ask them if at least one of them is a girl and they answer "yes", then indeed the probability of the other kid beeing a boy is 2/3.

If they tell you "we have two kids, and one of them is a girl", then by all rules of rational discourse the probability for the other kid being a boy is 1.

Things start to get interesting if the talking member of the couple is a mathematician (so the rules of ordinary discourse do not apply) and he says "We have two kids, and at least one of them is a girl". Now you might say that this is equivalent two the first case. But you could also go bayesian and look at the possible conversations with couples containing at least one mathematician that have two kids. Say you have 100 couples. On average, 25 of these will have two boys and tell you "We have two kids, and at least one of them is a boy", another 25 will have two girls and tell you "We have two kids, and at least one of them is a girl". The remaining 50 couples will have one boy and one girl. Assuming no gender bias on behalf of the speaking mathematician (ha!), 25 of these will tell you "We have two kids, and at least one of them is a boy", while the remaining 25 will say "We have two kids, and at least one of them is a girl." So out of the 50 couples that tell you "We have two kids, and at least one of them is a girl," 25 will have two girls, and 25 will have a boy and a girl. So in this case, the correct answer is 1/2.

(This is essentially the argument from the intro on this page that was linked from the discussion on coding horror: http://www.overcomingbias.com/2008/10/my-bayesian-enl.html )

brfox3• 18 years ago

Yes! Here is more agreement:

http://paulbuchheit.blogspot.com/2009/01/question-is-wrong.h...

(and also my own comment from earlier:) http://news.ycombinator.com/item?id=416576

ynd2• 18 years ago

# The hard part: THIS IS AN 'A POSTERIORI' PROBLEM.

# The fact as ALREADY happened.

# BG/GB being DIFFERENT is NOT THE TWIST.

# Whether they are or not it will amount to 50% total.

# TWIST: BB is not possible.

  def boy_girl_problem():
    from random import choice

    families = {}
    for i in range(1000):
      families[i] = []
      families[i].append(choice(['BOY', 'GIRL']))
      families[i].append(choice(['BOY', 'GIRL']))
  
    had_both_sex = filter(lambda f: set(f) == set(['BOY', 'GIRL']), families.values())
    had_two_boys = filter(lambda f: 'GIRL' not in f, families.values())

    return len(had_both_sex) / float(len(families) - len(had_two_boys))

# average(boy_girl_problem) == 2/3

ynd2• 18 years ago

# Here's a less crude rewrite

# It reflects conditional probability better

  def boy_girl_problem():
    from random import choice

    families = {}
    for i in range(1000):
      families[i] = []
      families[i].append(choice(['BOY', 'GIRL']))
      families[i].append(choice(['BOY', 'GIRL']))
  
    had_both_sex = filter(lambda f: set(f) == set(['BOY', 'GIRL']), families.values())
    had_one_girl = filter(lambda f: 'GIRL' in f, families.values())

    return len(had_both_sex) / float(len(had_one_girl))

Coding Horror: Finishing The Game

Related Stories