Greetings to all.
I started an MSc that includes a course in statistics. Full disclosure: my bachelor’s had no courses of statics and it is in biology.
So, the professor was trying to explain the Bayes theorem and conditional probability through the following example.
“A friend of yours invites you over. He says he has 2 children. When you go over, a child opens the door for you and it is a boy. What is the probability that the other child is a boy as well.”
The math say the probability the other child is a boy is increased the moment we learn that one of the kids is a boy. Which i cannot wrap my head around, assuming that each birth is a separate event (the fact that a boy was born does not affect the result of the other birth), and the result of each birth can be a boy or a girl with 50/50 chance.
I get that “math says so” but… Could someone please explain? thank you
In: 77
LOL a coworker tried for *years* to explain this exact thing to me…I finally got it, but also immediately forgot it (I didn’t need it, we’re just nerds.)
Let’s say no child answers the door. Options are BB/BG/GB/GG so 1/4 both boys, 1/4 both girls and 1/2 one of each.
Now a boy answers the door. GG is now eliminated. So 1/3 chance both are boys.
Edit: it’s like the Let’s Make A Deal problem
Think of it another way:
The professor tells you that you will be visited by four people. Each one will either give you a dollar or give you nothing. What are the chances of getting at least $2 in the end? Basically, you need two of the four people to give you a dollar.
Now, the first one visits you and gives you a dollar. What are the chances now that you will have at least $2 in the end? You only need one of the next three to give you a dollar. The chances of getting to $2 are higher than before, because you’re halfway there.
Each visitor’s decision is independent, but your chances of having $2 in the end is conditioned by how much you have in your hand at each point. You can’t predict the choice the visitor will make, but the number of possible outcomes changes – i.e. you can’t end up with no dollars once you already have $1.
Before you went there, if someone asked you to guess the genders of the two kids, you’d naturally think, it could be anything: any one of BB, BG, GB, GG. You have one in four chances of being right.
But when a boy opens the door, you can certainly eliminate GG as a possibility. But it could still be one of the other choices, so you have one in three chances of being right. What has happened is that you have gained information, and based on that information (“conditional to that information”), your guesswork has improved (“higher probability”). More information equates to less uncertainty, which improves the odds you place on that event.
Remember that the theory of probability has long been associated with gambling and insurance, which is tied to how much money you are willing to wager on something. The higher the probability, the less risky a wager it becomes.
Edit: this is wrong, the chances the other kid is a boy is 50%.
Imagine there are 100 families with 2 kids.
25 have BB
25 have BG
25 have GB
25 have GG
So you have a 25% chance of visiting a family with 2 boys.
Now imagine you know the family has 1 boy.
Your possible families are
25 with BB
25 with BG
25 with GB
So before it was 25/100 (1/4) chance of happening to visit a BB family but now it’s 25/75 (1/3)
> What is the probability that the other child is a boy as well.
The problem is that the phrasing question is mis leading. In a certain sense, you can still answer that with 0.5 because of independence as you mentioned before.
The actual question is:
* Given the 1st kid is a boy, what is the probability that it is a BB situation?
Before we answer that question, let me ask another simpler question:
* What is the probability that it is a GG situation (without knowing anything else)?
Obviously the answer is 1/4
* What is the probability that it is a GG situation GIVEN that the 1st kid is a boy?
I hope it is obvious why here, the answer is zero, not 0.5 . It cannot ever be a GG situation because we already know that the 1st kid is a boy.
It is in this context why the P(BB|BX) is not 1/2.
It seems that your main confusion is why information about one child gives information on the other.
The important thing to note it that you’re not given information about one child in particular. You’re given information about BOTH children as a whole: at least one of them is a boy. That’s why it gives you additional information about the the combined identity of BOTH children.
If the question was instead “Given that the YOUNGER child is a boy…” Then you don’t have any additional information about the elder one, they still have 50/50 of being boy or girl.
I don’t think your prof is applying the theorem correctly. Let’s do the math P(the unknown child is male “A”|the known child is male “B”)=P(A and B)/P(B). Naively: P(.5×.5)/P(.5)=.5 After learning that one child is a boy: P(.5×1)/1=.5 either way the math works out as a coin flip
This problem sounds like someone told you “my first child is a boy. What is the probability that my second child is also a boy?”. If that were the case, the information is completely independent from the question and the answer would be one-half.
But what you actually learned was “at least one of my two children is a boy”. That statement gives you “partial* information about *both* children.
Take it to an extreme example. Imagine that your friend has 1000 children. The door opens and 999 boys are standing there. What are the chances that your friend had 999 boys and 1 girl AND by pure chance the girl is not there? Seems more likely that he had all sons and one is missing.
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