If it’s a word game, then the answer is 0% because she said [only] one kid is a girl. But if it’s a math game, the answer is 50%.
We start with 2 booleans, and then one of them is already solved. The simplified phrasing of this question is: I have 2 kids. Guess the gender of ONE in particular.
Everybody who tries to solve a more complex problem is overengineering it.
But if it’s a math game, the answer is 50%.
Then you lost the game. (Sorry if anyone else lost it just now…)
We start with 2 booleans, and then one of them is already solved.
Which one is solved? You don’t know. All you know is that they can’t both be boys.
Hm. Back to my original statement:
Given that we can throw away one of the likelihoods (i.e. B+B)…
P(b|g) = P(b AND g)/P(g) = 0.667
P(b|g) = 0.667/1 = 0.667
So… 66.67%
Is this really still be debated??
…met someone who told you they had two children, and one of them is a girl…
Read carefully…
The fundamental issue I think lies in the fact that, having one un-age-defined girl means that a state with two girls is at half the probability of a BG mixture.
I looked back at the question, thought about it for another minute, and I see what you 66% people are saying now. There is a difference between what are the chances that she has a boy AND girl and what are the chances she has a boy as her next baby.
I’m changing my answer from 50% to 66%.
Before the guy told us that he had a girl, the probability of him having a boy and a girl are 2/4 (4 permutations B-B, B-G, G-B, G-G out of which two have one boy and one girl). But now he gave us additional information that one of them is a girl so it has to be B-G/G-B/G-G which is three permutations and out of which two have one boy and one girl. So the answer is 2/3 = 66.67%
Before the guy told us that he had a girl, the probability of him having a boy and a girl are 2/4 (4 permutations B-B, B-G, G-B, G-G out of which two have one boy and one girl). But now he gave us additional information that one of them is a girl so it has to be B-G/G-B/G-G which is three permutations and out of which two have one boy and one girl. So the answer is 2/3 = 66.67%
@Therac-25
Again, if we did Trial A 1000 times, and Trial B 1000 times, are you saying that the number of times you’re right would be statistically different?
… or do you still think that probability exists anywhere outside of a mind? Yes, the number of times you’re right between Trials A and B will exist outside of your mind, when we measure how often you are correct.
This is exactly why I love developers. On the surface this is a trivial question, assuming my assumptions are correct. But immediately the first several posts shot holes in my assumptions. Leave it to developers to overcomplicate a question just to make sure they are correct.
@PunchAndPie and everyone else in the 50% camp
Consider it this way: we have 1000 mothers with two children.
250 mothers will have two boys.
500 mothers will have one girl and one boy.
250 mothers will have two girls.
I think we can all agree on that, yes?
Now, if we know that one of a mother’s children is a girl, we eliminate the 250 mothers who have two boys. That leaves us with 750 mothers, 500 of which have one girl and one boy.
500/750 = 66%
It depends why the parent revealed that one child is a girl.
If its because you asked is (at least) one of the children a girl? and you’re told yes then its 66%.
But if you stumble upon the information (phone rings, parent has conversation that’s obviously about a girl, you then ask and its confirmed that at least one of the children is a girl) then its 50%.
The difference is that, whilst there are twice as many BG households as GG, you’re twice as likely to stumble upon one child being a girl when in a GG household.
Jeff Atwood,
Let’s say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?
This question might have several answers depending upon what the purpose of asking the question was.
If the question was to show that Bayesian Reasoning is very difficult and many people get it wrong {as the link would imply}. Then the answer is 2/3.
If the question was to show that normal people don’t think like CS majors or Mathematicians. Then the answer is near 100%, depending upon the number of percentage of people in a random sample that would happen to be Mathematicians. I wonder what percentage of the populace are Mathematicians.
If the question is posed to make us ask, Are we really given enough information. I should get more detailed information before drawing a conclusion. Then the answer is I don’t know, need more info.
If the question was posed just to hit 1000+ comments, well done.
Please Jeff, stop the madness. What was the point?
Here’s a little joke for you all…
An astrophysicist, a physicist, and a mathematician were on a train riding through Ireland. They looked out the window and saw a black sheep grazing on the pasture.
The astrophysicist says, All of the sheep in Ireland are black.
The physicist chimes in, Your wrong, you only know that there exists in Ireland at least one black sheep.
Your both wrong, the mathematician says, You only know what there exists in Ireland at least one sheep that black on one side.
I’m all for reductionism, but not to the point of re-writing the problem statement.
Unless I’m missing something, the answer 2/3.
People are right that order doesn’t matter and that the gender of each child is completely independent, but as someone mentioned, this is a matter of filtered information, and the 50% crowd is mis-reading the problem.
This isn’t a simple question of the likelihood of a particulay child being a boy (with some extra rhetorical icing); this is a different question altogether.
In a two-child family, there are – presumably – 4 combinations of gender outcomes, each statistically even:
Boy then Girl (25%)
Boy then Boy (25%)
Girl then Boy (25%)
Girl then Girl (25%)
One way to look at it is to simply eliminate the combination where there are two boys since we’re given that information…
Boy then Girl (25%)
Girl then Boy (25%)
Girl then Girl (25%)
so, now two of the three {equally liklely} combinations contains a boy. (therefore 2/3)
But some are quick to call foul here because order doesn’t matter.
So we can take order out of the equation, but not blindly; A boy followed by a girl may be equivalent to the vice-versa, but a B/G combination is still twice as likely as either G/G or B/B. The original breakdown becomes the following (order irrelevant):
Boy/Girl (50%)
Boy/Boy (25%)
Girl/Girl (25%)
(not as some might assume 33%/33%/33% respectively)
Eliminate Boy/Boy from the mix (per the information given) and once more we have 50/75 = 2/3 likelihood that the other child is a boy.
nuclearnewyears,
No, applying Bayes’ theorem, we need to first consider the probability within the entire population. Then we can do the math to get the new probability given known information. Taking the givens into account too early leads to errors.
P(x|y) = P(x^y) / P(y)
Or, the probability of x given y is equal to the probability of x AND y divided by the probability of y. You can’t assume that y is true, yet. Otherwise the denominator would always be 1.
Here’s another way to think about it for you 50%ers. (Feel free to substitute in your currency of choice.)
Your claim is that when I flip two coins and at least one is heads, there is a 50% chance that the overall outcome will be one heads one tails, and a 50% chance that it will be two heads. So I propose a bet. We flip two coins, and when at least one is heads, the bet comes into play. If the other coin is tails, you pay me $1. If the other coin is heads, I pay you $1.20, so that if you’re right you’ll tend to make money after a decent number of flips.
That means:
Two tails - Draw; no one wins
One heads one tails - I win $1
Two heads - You win $1.20
Would you take this bet?
You’re all wrong, says the philosopher, who was listening in on all this. You only know that on this particular trip through Ireland, your mind has brought to consciousness the percept, illusory or not, of a sheep, black on this side.
A 50% argument that the 66%ers can understand:
Lets assume the parent picked a child at random and revealed that it was a girl.
From 100 parents lets say:
25 have BB
50 have BG
25 have GG
Now… ONLY HALF of the BG parents will have said girl whereas ALL of the GG parents will have said it. So its 50% not 66%.
(You have to ask if there’s at least one girl to get 66%)
xooorx,
Nicely stated. Exactly the premise of my original 50% argument.
However, I switched to 67% because the purpose of this argument is to study Bayesian theory. If I were being practical, I’d be in the 100% camp.
Here’s another puzzle for you:
A certain king is often out fighting wars. He loves battle, but there is a problem: his kingdom is running low on men! So he thinks about it, and passes a new law: Women must stop having children after their first girl.
For he reasons thus: with this law, sure, there will be the occasional one-girl family, but there will also be boy-girl families, boy-boy-girl families, boy-boy-boy-girl families, and so on. Thus, the proportion of boys amongst children will increase.
So, the problem: calculate the boy:girl percentage in the kingdom after a sufficiently long period of time has passed (assuming 50-50 odds for each birth).