Coding Horror: Finishing The Game
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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 )
http://paulbuchheit.blogspot.com/2009/01/question-is-wrong.h...
(and also my own comment from earlier:) http://news.ycombinator.com/item?id=416576
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).
If you had said that a mathematician or a statistician said they have one girl, that would be a different story.
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.
1) X: G, Y: G
2) X: G, Y: B
3) X: B, Y: G
4) X: B, Y: B
Which still seems a little weird to me, that knowing which is a girl, regardless of which one you know about, changes the chances.
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.
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)))))
"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.
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.
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!)
If I am correct about that, then it matches the conditions discussed in the article and the answer would be 2/3 for a boy and a girl.
I don't quite understand it, but apparently anything which can be used to distinguish the children will do.
Possibilities with two children:
Gg, Bg, Gb, Bb
G'g, B'g, G'b, B'b, Gg', Bg', Gb', Bb'
G'g, G'b, Gg', Bg'
Gg, Bg, GbIt makes no difference if they have distinguishing marks or not. It matters if you are told that a particular child is a girl (50%) or if you are only told that at least one child is a girl (66%).
I partially understand your point about G'g vs Gg' now: if having a prime is the only way to distinguish the children, then G and g must be indistinguishable, so G = g.
I now put the coins back into my pocket, shuffle them about, and again take them out inside my hands. This time I look inside both my hands, not letting you see, and tell you (truthfully) that at least one is tails. Given that information, you can deduce three mutually exclusive possibilities each of equal probability - both are tails, only the coin in my right hand is tails or only the coin in my left hand is tails. Hence we have the odds in this situation of 2/3 for a head/tail combo.
It is easy to see that the first situation is akin to knowing that a particular child is female, whilst the second is akin to knowing that at least one of the children is female. Also, in either case it does not matter if the coins are distinguishable - one could be a euro and the other a pound.
Take 1000 two-child families (so as to intuitively ignore fluctuations). Families 1-250 are girl-girl, 251-500 had a girl then a boy, and 501-750 had a boy then a girl. Any of those 750 families could have made the quote in my post, yet there are 500 families with boy-girl, and 250 with girl-girl.
Or another example that I think speaks more directly to your post: you see a woman with a T-shirt reading "Proud Mother of Two" next to a girl who is obviously her daughter. What is the probability of her other child being a boy? Again, since there are twice as many mixed families as pure girl families, the odds are 2/3.
The odds are 50%, do you seriously think different?
Your mistake is that in families 1-250 the girl next to the mother could be either daughter.
In families 1-750 there are 1000 daughters, the daughter standing next to the mother is equally likely to be any of those and half have brothers, half have sisters.
Gambling to the rescue
You would need a special case to determine if you were talking to Johnny Cash's father...
# 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))
# 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))