Science1 min ago
Sex
165 Answers
Following on from the entertaining Dice and Socks threads earlier this week I've now found the problem about the sex of children.
One version goes: "You know that Mr. Smith has two children and that at least one of them is a boy. What is the probability that both children are boys?"
Thoughts please?
One version goes: "You know that Mr. Smith has two children and that at least one of them is a boy. What is the probability that both children are boys?"
Thoughts please?
Answers
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It's worth considering a separate example that is similar in nature.
"You know that Mr Smith has two pets and that at least one of them is a cat, and the other is either a cat or a dog. What is the probability that the other is also a cat?"
There's no answer to this one, but it ought to be obvious why: we have no idea what the probabilities are for various events. The point is that we have to calculate. You can't just list options and assume that they have equal probability. Here he could have a cat and a dog, or two cats, but obviously this isn't necessarily going to be a 50-50 choice just because there are two options.
The same logic is true here. Mr Smith could have a boy or a girl, or two boys, and this isn't automatically a 50-50 choice either. You're conned into thinking it is because we know the probability of any one child being a boy or a girl is 1/2. But, as has been seen, the probability of both being a boy is less than the probability of one boy and one girl -- because we have to calculate, and when we do we see that one is twice as likely as the other (for there are two distinct ways it can have happened).
Every time people try to solve a maths problem without actually doing any maths, they're going to find it a lot harder than by actually calculating.
"You know that Mr Smith has two pets and that at least one of them is a cat, and the other is either a cat or a dog. What is the probability that the other is also a cat?"
There's no answer to this one, but it ought to be obvious why: we have no idea what the probabilities are for various events. The point is that we have to calculate. You can't just list options and assume that they have equal probability. Here he could have a cat and a dog, or two cats, but obviously this isn't necessarily going to be a 50-50 choice just because there are two options.
The same logic is true here. Mr Smith could have a boy or a girl, or two boys, and this isn't automatically a 50-50 choice either. You're conned into thinking it is because we know the probability of any one child being a boy or a girl is 1/2. But, as has been seen, the probability of both being a boy is less than the probability of one boy and one girl -- because we have to calculate, and when we do we see that one is twice as likely as the other (for there are two distinct ways it can have happened).
Every time people try to solve a maths problem without actually doing any maths, they're going to find it a lot harder than by actually calculating.
That's understandable. The trick then is to stop trying to rely on intuition, and to actually check what's going on.
I'd also say that if you start introducing ideas of how you find out that one is a boy, it starts to cloud the issue. I was just drawing the probability tree for this one out and was starting to wonder if I might have to backtrack in somewhat humiliating fashion, before realising that I don't, because the manner in which we have discovered the knowledge is an event whose probability isn't known and you could start throwing all sorts at it. What if the other child is never seen with the father anyway? Perhaps at that time of day he/ she is always busy doing tennis or something, or maybe you met at a football pitch in which the other sibling has no interest, etc etc...
The only part of the probability tree about which anything can be said is the part where it's established that there are four possible two-child families, all of which are equally likely. In practice the answer will almost certainly not be exactly 1/3 but, as the question is worded, that is the answer you should give.
I'd also say that if you start introducing ideas of how you find out that one is a boy, it starts to cloud the issue. I was just drawing the probability tree for this one out and was starting to wonder if I might have to backtrack in somewhat humiliating fashion, before realising that I don't, because the manner in which we have discovered the knowledge is an event whose probability isn't known and you could start throwing all sorts at it. What if the other child is never seen with the father anyway? Perhaps at that time of day he/ she is always busy doing tennis or something, or maybe you met at a football pitch in which the other sibling has no interest, etc etc...
The only part of the probability tree about which anything can be said is the part where it's established that there are four possible two-child families, all of which are equally likely. In practice the answer will almost certainly not be exactly 1/3 but, as the question is worded, that is the answer you should give.
It's not that the probability is altered exactly -- it's just re-evaluated in the light of new information that is excluding one possibility.
Thus, P(two boys given two children) is still 1/4, but we're now being asked to evaluate P(two boys given definitely at least one boy out of two children). This at least can be a different number since it's a different event we're considering, and the new calculation gives P(two boys given at least one) = (1/4) /(3/4) = 1/3, where the division by three quarters represents the reweighted space of possible outcomes from 4 (GG BG GB BB) to just three (GB BG BB).
Providing new evidence generally is going to lead to a change in your answer. It is surely obvious that if you are aware that one of two children is a boy they can't both be girls -- the probability of this event is clearly 0.
Thus, P(two boys given two children) is still 1/4, but we're now being asked to evaluate P(two boys given definitely at least one boy out of two children). This at least can be a different number since it's a different event we're considering, and the new calculation gives P(two boys given at least one) = (1/4) /(3/4) = 1/3, where the division by three quarters represents the reweighted space of possible outcomes from 4 (GG BG GB BB) to just three (GB BG BB).
Providing new evidence generally is going to lead to a change in your answer. It is surely obvious that if you are aware that one of two children is a boy they can't both be girls -- the probability of this event is clearly 0.
This is getting silly!
Once again, factor, this question does not ask about the possibility of a second child being the same sex as the (given) first. It asks what are the chances of a couple with two children having two boys. Introducing "the first born is male" provides more information than is provided in the original question.
This conundrum arises because information is being provided in the question which actually has no bearing on, or adds any useful information to solving the problem. If the question was "what is the likelihood of a couple with two children having two boys?" I don't think anybody here disagrees that the answer is one in four. This question says "What is the likelihood of a couple having two children of the same sex where at least one of them is a boy?" (sounds clumsy but that's what it really means). As has been adequately demonstrated, the answer to this is one chance in three. All the phrase "at least one is a boy" does is to constrain the results to eliminate the G-G possibility. It does nothing to alter the overall number of possible combinations. The information that one is a boy does nothing to alter the overall chances of them both being boys as one in four. If we are to ignore the G-G possibility then there are only three options left of which B-B is just one.
Since we're putting our money where our mouths are I take it, talbot, that since you believe the true odds of the other child also being a boy are Evens you would be perfectly willing to offer me the same odds that the other child is a girl. If so I (and probably many others) will bite your hand off. In fact, in a long sequence of such bets you would lose twice as often as you will win.
The "Monty Hall" problem which I mentioned earlier is similar in nature as it provides information about an event which appears to modify the probability of an overall event, but in fact does not. To avoid hijacking factor's question I'll pose it separately.
Once again, factor, this question does not ask about the possibility of a second child being the same sex as the (given) first. It asks what are the chances of a couple with two children having two boys. Introducing "the first born is male" provides more information than is provided in the original question.
This conundrum arises because information is being provided in the question which actually has no bearing on, or adds any useful information to solving the problem. If the question was "what is the likelihood of a couple with two children having two boys?" I don't think anybody here disagrees that the answer is one in four. This question says "What is the likelihood of a couple having two children of the same sex where at least one of them is a boy?" (sounds clumsy but that's what it really means). As has been adequately demonstrated, the answer to this is one chance in three. All the phrase "at least one is a boy" does is to constrain the results to eliminate the G-G possibility. It does nothing to alter the overall number of possible combinations. The information that one is a boy does nothing to alter the overall chances of them both being boys as one in four. If we are to ignore the G-G possibility then there are only three options left of which B-B is just one.
Since we're putting our money where our mouths are I take it, talbot, that since you believe the true odds of the other child also being a boy are Evens you would be perfectly willing to offer me the same odds that the other child is a girl. If so I (and probably many others) will bite your hand off. In fact, in a long sequence of such bets you would lose twice as often as you will win.
The "Monty Hall" problem which I mentioned earlier is similar in nature as it provides information about an event which appears to modify the probability of an overall event, but in fact does not. To avoid hijacking factor's question I'll pose it separately.
Okay- don't all the theoretical probabilities that we drew up about the sexes of 2 children before we knew anything (that you can have BG,GB,BB,BG) become irrelevant once we know something about one of the children. Why should the sex of one child have any bearing on the sex of the other .
I start with a simple tree diagram like Prudie did showing BB 1/4, Bg, 1/4, gB 1/4, GG 1/4, where the first letter denotes the first child I see and the second the other one.. Once I see the boy I can cross of two branches leaving just BB and BG. Since I already know the first I see is a boy there is a 50-50 chance the other is a boy.
Isn't there?
I start with a simple tree diagram like Prudie did showing BB 1/4, Bg, 1/4, gB 1/4, GG 1/4, where the first letter denotes the first child I see and the second the other one.. Once I see the boy I can cross of two branches leaving just BB and BG. Since I already know the first I see is a boy there is a 50-50 chance the other is a boy.
Isn't there?
