This is one of those questions that Intro to Probability teachers like to throw at their students: Suppose you approach someone who you know has two children, but you don’t know their children’s genders. You ask the person “Is one of your children a boy?” and they answer “Yes.” What is probability that the person’s other child is also a boy? (In other words, once you know that one of their children is a boy, what are the odds that they have two boys?)
1/2 The trouble is is that people know the chance of both being a boy is 1/4, but once they hear 1 is a boy they incorrectly conclude the second is now more likely to be a girl. Mathmaticaly: Biologicaly it's a 1/2 chance for each gender. For two boys, we multiply 1/2 and 1/2 for each boy and get the intuitive answer of 1/4 chance of having 2 boys. However, in this case, it is a 1/1 chance of the first one being a boy (we are absolutly certain he is) thus, when we multiply 1/1 by 1/2 and get the answer of 1/2 for the second one's probability of being male. -Andrew
The answer is 1/3. Here I am assuming the probability of any one child being a boy versus a girl is a fifty-fifty proposition. It's not; boys slightly outnumber girls at birth. But I'm ignoring that little fact. With two children, there are four equiprobable ordered events: <G,G>, <G,B>, <B,G>, and <B,B>. The knowledge that one of them is a boy eliminates the two daughers case. What's left are two cases with one boy and one girl, and one case with two boys. These are equiprobable, mutually exclusive events than span the space. Two of the three cases have a girl. The odds are 2/3 that the other child is a girl, 1/3 that the other child is a boy. This can also be solved with Baye's Law. It yields the same result. This can also be solved with Monte-Carlo techniques. These yield the same result (to within statistical error). I got a girl in 6661 out of 10000 cases and a boy in the other 3339 cases.
Could the probability not be restated in the following way? Assume a man is walking along and meets a high school friend. His friend is now married, and has his wife and son with him. The wife is also pregnant. The man knows one child is a boy, what is the probability that the yet-to-be born child will be a boy?
No, you don't know whether the first one was a boy or not. You only know that ONE of them was a boy. Before you receive any information, there are four equally likely possibilities: BB, BG, GB and GG. Since you know that one of them is a boy, that rules out the GG possibility. That leaves BB, BG and GB as (equiprobable) possibilities, so the answer is 1/3.
No, in this case you have extra information (that the first child was a boy). In the original problem, you only know that there is at least one boy, not whether he was the first-born or not. If you know the first-born is a boy, the odds that the second child is a boy are 1/2.
Perhaps you should read this article, it will explain why your logic is fallacious. -Wikipedia While you can plot them as ordered events, you must realize that both <G,G> and <G,B> are eliminated, because the first event of both of them is a girl, which we confirmed is false. It is a 50/50 chance. The sperm carries either an x or a y chromosome, and there are always en equal number of each created. Stitistically more males may be born, but this doesn't mean it isn't 50/50 (once again, Gamblers Fallacy.) Yep, exactly the same problem, 1 unknown with 2 possibilities, equal likelynes of each, and 1 known red hearring. -Andrew
The probability teacher's answer is 1/3, of course. Butl, in practice there's a considerable chance that the person would respond differently if they had two boys, eg: "Is one of your children a boy?" "No, they are both boys." This is a legitimate response - it assumes an implied "exactly" rather than an implied "at least one". This problem has a history of causing trouble in 'Net space, mainly because of the difficulty in contriving an unambiguous situation where you eliminate the possibility of two girls, without influencing the probability of two boys.
Ah yes sorry. (quadraphonics posted while I was typing.) However, since we are only concerned with the gender of the other boy, after the information, the events (G,B) and (B,G) become identical. If it were worded as "What is the probability that the second(or first) born child was a boy?" it would be 1/3. -Andrew
The two events are separate events. The mention of "the other child" is a linguistic trick which is (deliberately?) not mentioned in the OP. As soon as you think or say "the other child", you're falsely implying that a particular child has already been identified, which is not the case. We are only concerned with whether both children are boys.
'First-born' has nothing to do with it. The 'first-born' child could have been an unseen daughter at the movies. The only information I gave was that one child was a boy and the wife was pregnant. I gave less information than was in the original version.
No, that's a different problem. The OP deals with dependent events, your restatement deals with independent events. The events this child here is a boy and that unborn child is a boy are independent. The events at least one of these two children are boys and both of these two children are boys are dependent.
My usage of the term "first-born" means, here, first out of the two children under consideration. Telling us that the wife has had a boy before the other child in question is more info than the original version, where we don't know the ordering. This is why the original version gives a probability of 1/3, while your version gives a probability of 1/2. It is MORE information, and it is a different problem with a different outome.
Alright allow me to put it this way: The events (B,G) and (G,B) loose half their weight in this case; each is 25% likely to be true, while (B,B) is 50% likely. Let's break the problem down furthur: We incorporate the persons answer into the scenario, not just by eliminating the (G,G) possibility, but by taking which child she could be reffering to into account. She is either reffering to child 1 or child 2. The new events appear as follows: (B,G,1) (G,B,2) (B,B,1) (B,B,2) Events: (B,G,2), (G,B,1), (G,G,1) (G,G,2) are contradictory and are therefore imppossible. Now we see, there is a 1/2 chance of it being 2 boys and a 1/2 chance of there being a girl. The fact is, the gender of the childs are independant. -Andrew
It's more than that - the OP doesn't identify a child in any way. In order to answer the question "Do you have [at least one] boy?", the responder has to consider her two children as a set. The response could refer to either or both. In turn, the probability we want to calculate is a property of both children as a set, not just one of them.
That doesn't make any sense at all. Yes, that's the trap I referred to in my first response to the thread. In the idealised probability teacher's scenario, the person isn't referring to one of her children, she is referring to them both as a set. You can imagine that the person interprets the question in a different way (as I suggested), which would change the resulting answer. Your analysis implies that the person has listened to the question, then chosen one of her children at random, and said whether that chosen child is a boy or not. Ie you eliminate (B,G,2) and (G,B,1) because in those situations the person would have answered "No".
OK, let's restate the OP like this. You approach somone and ask if they have any children. He states 'I have a son and my wife is pregnant'. What is the probability the unborn child will be a boy?
