Reply With QuoteExcept that it isn't as we know that she has one daughter which rules out the boy-boy possibility. Before we knew this the odds were 1/4 of two girls. With the information that she has one daughter this reduced to 1/3.Originally Posted by Pebble
Matt I can't agree,
The probability of boy or girl offspring is dependent on genetics not naming algorhythms.
The probability of two girls remains 1/4 irrespective of their names or order of naming.
Except that it isn't as we know that she has one daughter which rules out the boy-boy possibility. Before we knew this the odds were 1/4 of two girls. With the information that she has one daughter this reduced to 1/3.
I had it all...but I threw it away. Then I found it again, so that's alright.
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In both cases the chances are 1 in 2.
One child is a girl, and it doesn't matter what her name is.
The other is either a boy or a girl...or a hermaphrodite.
Yes but my comments were specifically directed at the possibilities drawn up by Matt which included Boy-Boy.
So to be more specific Mendelian Inheritance requires a 1/2 ratio of the sexes (accepted reality is slightly different) so for 2 children we end up with 1/4 BB and 1/4GG what ever their names. Accepted that if you exclude the BB group the remainder have 1/3 ratios left.
No, no, no. In a family of two, the possibilities are B-B, B-G, G-B, G-G. We are told that one is a girl, so we can omit B-B. We are then left with three equally likely possibilities B-G, G-B, G-G, one of which gives us 2 girls. So, given that one is a girl, the probability of two girls is 1 in 3.
What is the difference between B-G and G-B ?
Both are distinct possibilities. The first child could be a Boy or Girl - and, quite separately, the second child could be a Boy or Girl. So, for both kids, the 4 (2X2) distinct and equally likely possibilities are Boy-Boy, Boy-Girl, Girl-Boy, Girl-Girl.
But there is nothing in the question that requires a distinction in the order of birth.
If the order is important the possibilities are
G and G, G and B, B and G, B and B
otherwise
G and G, B or G, B and B
Mr Smoketoomuch,
take two coins. Toss them both. The are three possible outcomes.
both heads,
both tails
one of each.
If you repeat this over and over recording the results you will see that "one of each" comes up twice as often (half the time) as either both heads (quarter of the time) or both tails (also quarter of the time)
You'll also see that out of all the times that there's at least one head, (three quarters of the time) in approximately one third of those instances (one quarter of the time) there will also be a second head.
Don't take my word for it, see for yourself.
This paradox was raised in New Scientist recently. Apparently the answer comes from the fact that by including a specific additional detail, like naming a child, you exclude certain possibilities, thereby reducing the portion of the population to whom the question applies. So instead of considering the whole population of parents, where the odds remain 1 in 3, you are looking at a particulat subset, where the odds now approach 1 in 2.
Just because it appeared in that illustrious publication doesn't make it so. Anything beginning "apparently" looks more like something expounded by that fount of all knowledge "the bloke in the pub" than a mathematical truth. Show me the working step by step and I'll consider it.
I'm not personally claiming anything. I just noticed that the same problem had been discussed elsewhere and passed on their conclusion. I can see, however, how by introducing an additional condition you reduce the number of people to whom the problem applies, thereby changing the odds. I'm not interested in working out the new odds, I am happy with seeing the principle.
The problem I have with questions like this is that in order to reach an answer you need to access information that is not contained within the question.
For example if we change the first example too...
Jane has two balls, one of which is red. What is the probability that the other is red.
..then the question becomes unanswerable because we do not know how many colours there are or how they are chosen.
By making the subject of the question children we automatically know there are two choices of gender and we also know that one has to come before the other (ie order is important as even with the case of twins one has to come out first).
It therefore follows that to HAVE (the have is very important because here we are making a prediction before the event) two children (one of which is a girl) there are three possibilities...
B then G
G then B
G then G
...so the probability of two girls is 1/3. This however is not the situation presented to us. In the question the two children already exist so the possibilities become...
B then G
G then B
G1 then G2
G2 then G1
...so the probability of two girls is 1/2.
Therefore, in conclusion the answer to both questions is 1/2.
skb
I don't see the logic of your last bit. We know there is at least one girl which rules out the B - B possibility. Whether G1 comes before or after G2 doesn't matter, surely?
Replace the boy girl with a coin toss - heads & tails. The question asks "What is the probability of getting two tails (girls) when one of the results is a tail (girl)?"
So you can get:
HT
TH
TT
with HH excluded as we know there is at least one tail.
The TT result is still 1/3 whether T(1) is first or T(2), isn't it?
Naming one of the tails as "Emma" should make no difference as we are still looking for two tails, not a specific order of tails, wouldn't it?
If we were looking for two tails with "Emma" first, then we would have:
HT(Emma)
T(Emma)H
T(Emma)T
TT(Emma)
which would be 1/4.
Wouldn't it?
I had it all...but I threw it away. Then I found it again, so that's alright.
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No, as we cannot substitute the children for coin tosses as G1 =/= G2 whereas T = T.
But the order is important. Rather than think about it as Bs and Gs think of it as drawing two names from a hat that contains three names, two are girls names and one is a boy’s name (there can’t be two boys names as BB has been excluded as a possibility). Let’s call those names Emma, Sue and Steve. If you were to draw two names from the hat then the possible combinations are:
Emma then Sue
Sue then Emma
Steve then Sue
Sue then Steve
Steve then Emma
Emma then Steve
We are however given a further stipulation in that one of the names is not only a girl but is also Emma, we can therefore drop any of the choices which don’t contain Emma to leave us with...
Emma then Sue
Sue then Emma
Steve then Emma
Emma then Steve
...which gives the probability of having two girls as 1/2.
skb
Last edited by skbuncks; 16th July 2010 at 03:25 PM.
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