A male friend of yours wants to test if you are ok at Math and logical thinking, he says I am one of two children of my parents, and asks you to find out the probability that his sibling is a sister, he carefully also states he wants you to assume a sex ratio of 100 boys to 100 girls and that they were not twins.
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$\begingroup$ Agree with you that it may in spirit be similar in nature, but due to the changed wordings (though by mistake) its not asking for the same info, so it may not be a duplicate $\endgroup$– skvCommented Oct 23, 2014 at 11:03
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$\begingroup$ I don't think the sibling part of the question is relevant; someone asks you what the probability is for a certain person is, the answer is going to be 50%. You made it clear that loopholes like twins aren't to be considered, so the family structure adds nothing to the mystery! $\endgroup$– TheRubberDuckCommented Oct 23, 2014 at 13:05
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$\begingroup$ @EnvisionAndDevelop yes this is a question gone wrong :) so I wanted to delete it, but then I got serious warnings, so I just left it there undecided $\endgroup$– skvCommented Oct 23, 2014 at 13:16
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1$\begingroup$ I flagged as a potential duplicate not because it's literally the same question, but because the answer to this question is contained within the larger discussion of this type of question going on in the answers there. A number of variant cases are examined in the quest to find the right wording, and I believe this is one of them. In fact, this could be an answer to that question! $\endgroup$– YamikuronueCommented Oct 23, 2014 at 14:14
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$\begingroup$ I agree, I considered many words myself without finding a duplicate, I have voted for the close myself, lets see $\endgroup$– skvCommented Oct 23, 2014 at 14:35
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3 Answers
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The accepted answer has a subtle error. If we randomly select a 2-sibling family with at least one male, the odds of the other being female are indeed 2/3. If we randomly select a male who is part of a 2 person family, the odds of them having a male sibling are 1/2.
The three, equally probable male-containing families are
MF FM MM
There are four males listed, and 2 of them have male siblings.
(Edit to remove irrelevant female only family)
Edit again: Note that if you were to randomly select 2-child families with at least one male instead, (for example, by selecting a group of fathers and asking for those with 2 children and at least 1 male) you'd get 2/3 of them having a girl
answered Oct 23, 2014 at 8:09
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$\begingroup$ Can you tell me why FF is a probability in this case, I mean you know one of them is a male, so FF cannot be a possibility right $\endgroup$– skvCommented Oct 23, 2014 at 8:13
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$\begingroup$ I meant the four equally probable families before you select for males. You can reduce it to three equally probable male-containing families; there are still 4 males, 2 of whom have a brother. $\endgroup$ Commented Oct 23, 2014 at 8:19
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3$\begingroup$ @skv Nope, MM family contains two such cases. In other words, chance that you befriend male from this family is greater then of any single one of the other families because there are simply more males to possibly befriend. $\endgroup$– CthulhuCommented Oct 23, 2014 at 8:25
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4$\begingroup$ @skv, but you have 2 instances of John1 (sibling to Janie, sibling to John2). If you meet John1 there is a 50% chance his sibling is male. There are 2 instances of John2 (sibling to Janie, sibling to John1). Again, there is a 50% chance that John2's sibling is male. $\endgroup$ Commented Oct 23, 2014 at 8:57
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1$\begingroup$ Agreed, mistake in wording the question :) $\endgroup$– skvCommented Oct 23, 2014 at 9:38
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The fact that the first child among two is a boy has no effect on the sex of the second because the draw "sex of the first child" and "sex of the second child" are independant.
The probability of the other sibling being a girl is equal to the probability for any child picked at random to be a girl.
The chance is 1/2.
Another form of the question would be:
You play heads or tails -with a balanced coin, no loophole- twice.
The first time you play you get heads, what is the chance for the second time you play for it to get tails ?
It is 1/2. The chance for you to get tails is not increased by the fact that you had heads previously.
As soon as we know that the first child is a male, FM and FF become impossible, leaving MF and MM.
EDIT: I see that I still am not the accepted answer. And after discussing with my friend I can see why.
My answer is counter intuitive because in real life people tend to apply the Law of Large Numbers -or law of averages- to their everyday / low amount of draws routines. For instance:
You are at the casino, and play roulette.
You place your bets on red and lose 3 times.
You apply the law of average and believe that now your chance to win by betting red is very high. But this reasoning is irrational. The roulette has no memory of the previous rolls. Your chances by betting red or black are still the same.
Let me make another attempt at making you see that Skywalker's answer is flawed.
A group of 4 male dudes come to you at a party and tell you that they have a 5th friend of which they ask you to guess the gender.
According to the logic used in his argument the chance for the 5th friend to be a girl is 4/5 = 0.80 = 80% because the possible combinations are
MMMMM MMMMF MMMFM MMFMM MFMMM FMMMM
The fact that the 4 other dudes are males bring no knowledge of the gender of the 5th.
The chance in that case is 1/2 too.
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$\begingroup$ Your line of thinking is sound, except for that fact that in the question as stated is that we don't know the first child is male... we just know *a* child is male which means that FM, MF, and MM are all possibilities. $\endgroup$– MordredCommented Oct 23, 2014 at 22:39
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$\begingroup$ @Mordred In this case, the fact that I named it "the first" is just for a matter of identification. I can rephrase it like that: A and B are 2 people. A is male and asks the chance for B to be a female. The possibilities are (A = M and B = M) (A = M and B = F) (A = F and B = M) (A = F and B = F) but because A is male, then the only possibilities are (A = M and B = M) (A = M and B = F) That is 1/2 chance for B to be female. $\endgroup$ Commented Oct 24, 2014 at 3:47
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1$\begingroup$ @TheNaturalTanuki, I don't see any disagreement. In the "4 dudes at a party" example, there is a 50% chance of the final person being female. If, however, you told someone "round up five people" and afterwards asked "are there at least 4 dudes?", the chance of the final person being female is 5/6. The distinction here is between selecting people and asking about their group versus selecting groups and asking about their people. There will be twice as many (random) 2-person groups with 1 female present as none. There will be 5 times as many 5-person groups with 1 female as none. $\endgroup$ Commented Oct 24, 2014 at 10:57
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$\begingroup$ @frodoskywalker But it isn't round up 2 people. The gender of the first person is preselected as male so there is only one variable remaining. With friend = F and sibling = S possible conditions: F=Male,S=Male (possible); F=Male,S=Female (possible); F=Female,S=Male (impossible); F=Female,S=Female (impossible). In this case half of the possible 2 person groups are eliminated by the fact that we know the identity of one. $\endgroup$– MylesCommented Oct 24, 2014 at 13:42
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$\begingroup$ @Myles I think the confusion is that frodoskywalker's answer actually answers the question both ways, for selecting the group, and selecting the people. On further reading I think he and TheNaturalTanuki both agree. $\endgroup$– MordredCommented Oct 24, 2014 at 15:56
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it doesn't state in the question that ether of the offspring are male or female, it says a male friend of yours asks you(the reader) a question... etc. the variables you are speaking of(as far as the existing sex of the alive child, not the child to come into play in the question) will change depending on the sex of the reader, also still does not change the outcome of the end result, for the odds reset for each child born, even if the children born are all male so far, it does not mean that there is a higher or lower likelihood of the next being male or female, it is still a 50/50 shot, mathematically... while scientifically speaking if the father is predisposed to produce one sex more than the other then there would be a 2/3 likelihood that the child in question would be born male or female.
