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France — Being able to speak the local language is also a major factor for expat parents. Belgium — The country regained its position from the ranking due to it showing "improvements across the board, but most notably in regard to the availability of childcare and education. Australia — Scored well across the board but it hit the number one spot for available leisure activities for children, which boosted it into the top Norway — The nation jumped to 6th from 17th this year due a huge improvement in expat parents rating childcare options.

Sweden — The country fell from 3rd to 5th this year as a result of a decrease in the Quality of Education subcategory. Austria — The country fell from first place last year, to 4th place this year, due to a poor showing in the Family Well-Being and Quality of Education sub-indexes. The 13 cheapest countries to live in the world for expats.

These are the 19 best countries for raising a family, as voted by expats These are the 19 best countries for raising a family, as voted by expats When moving abroad, especially when you have a You are logged into Facebook Social: You are logged in with Google Social: Registration on or use of this site constitutes acceptance of our Terms of Service , Privacy Policy , and Cookies Policy. Phooey, this isn't at all a mathematical puzzle. I didn't immediately understand his reasoning. But if all families are enumerated 1,2,3, Viewed this way, the rule for stopping when the first B is reached is clearly a red herring!

And clearly the proportion of boys and girls will be equal. At least asymptotically, with probability 1, by the Strong Law of Large Numbers. On the other hand, this does not work for each possible stopping rule.

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Say we're back to the usual assumption of each birth having an equal chance of being a boy or girl. Then the ratio of girls to boys in the population will clearly be greater than 2. There has been some heated? I thought, for my own amusement, that I would do some Monte-Carlo experiments with a "plausible" model involving Pilgrims traveling to the New World. However, in thinking about possible models, I came up with the following issue. Suppose we assume that:.

Under these assumptions, it follows that if there were N male Pilgrims in the first settlement, then, at any given time, there are at most 3N male Pilgrims. Moreover, the probability of the settlement dying out over several generations because all the children are girls is non-zero. By Kolmogorov's zero-one law overkill , it follows that almost surely the Settlement will die out, and not become the kick-ass country it may well have been.

Perhaps I am missing something here but it seems quite intuitive to me that it has to be Think of it as coin tosses: As the resulting distribution is completely independent of the stopping time the proportion will in the limit always converge to the original distribution which is supposed to be That is because the resulting stochastic process is markovian and a martingale Or put another way: It doesn't matter when you stop tossing, the outcome will always add up to because the coin doesn't have a memory.

It is a little bit like trying to invest in the stock market and getting out every time you are in the plus sitting out negative phases and after that beginning all over again. This seems like a clever infallible strategy - alas it doesn't work and you will stay at zero in the long run here minus transaction costs of course BTW: This reminds me of some friends of ours who desperately wanted a boy - now they have three girls and stopped "trying" The correct answer has nothing to do with the number of families.

This is a very tricky problem, and many people fall into the trap of trying to average each possible fraction of girls, weighted only by the probability of that outcome. But in fact they would need to be weighted also by the size of the population, if that strategy is used to find the answer. Google's reasoning is perfectly correct, but here is another route to the same result. We just find the expected number of boys and the expected number of girls for one family.

The number of boys is obviously 1 for any outcome of the form G n B , and so its expectation is 1. This sum is perhaps surprisingly also 1, which is easy to verify. And so on average, the population will have an equal number of children of each sex. I will agree that Google's phrasing could have been more precise. But that is the case with virtually any math problem that is phrased as a problem in the real world, and I believe Google's intended problem is sufficiently clear that there is no real value in debating all its possible meanings.

My way to look it is a bit different.

So the ratio remains equal.

The possibile sequence and the respective probabilty of new born can be:. After Summing it up. Total Number of Boys is: Which also, give 1 on summing up. Maybe this is off topic since it is applied statistics. But why not look at the problem as "What is the expected number of failure X till there is as success? So, the expected number of failure girls, sorry! We have infinite countable number of families.

These are the 19 best countries for raising a family, as voted by expats

Each family stops to have children when the first boy appears. It is well defined and converges as. The answer is simply the same as in the coin tossing experiment, and there is a one word proof of it, Martingale, which gives This also shows why killing babies is wrong. Home Questions Tags Users Unanswered. In a country in which people only want boys [closed] Ask Question.

Starting to solve the problem for myself I got that part of girls can be calculated with following series: The official solution is: This one caused quite the debate, but we figured it out following these steps: Imagine you have 10 couples who have 10 babies. Where the truth is? Caicedo Jun 27 '12 at 2: As is almost always the case with these sorts of things it's not that "the usual explanation is wrong" or anything, it's simply that the question is ambiguously stated.

The answer to the question "what is the proportion? The question should perhaps say something like "what is the average proportion" and this is ambiguous already. Imagine for example there were just 1 boy-girl family, and they produced offspring until they got a boy and then stopped. Now people will say "well that's not what the question meant" and that's exactly my point. These are visibly going to be different.

Europe's young adults living with parents - a country by country breakdown

There may be other ambiguities too. It's always the same with these sorts of questions. I'm tempted to vote to close as not a real question. Based on the answers and questions people have had about this question, it seems that it is still interesting. I am not voting to close and I don't think others should either.

I think that sometimes an acceptable answer to a mathematical question is that the question is actually ill posed. When that happens and it is not obvious, it is interesting. Clearly this question is confusing, and it is a subtle point that the question is ambiguous. If I had asked the question and you had posted your replies to Tom as an answer, I would have accepted it. The question is "interesting" in the same way that the Monty Hall problem is "interesting", and the question is generating a lot of noise, just as Monty Hall always does.

I will happily accept that you are highlighting a different issue with the question. This is really nice! I voted to close because this is an old chestnut, but you have found new life in it. You've changed the question by assuming that the population i. The difference in answers between the original problem and the completed-families problem is the "bias" you calculated.

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Jul 9 '10 at 0: T, you are right that the bias depends on the formalization of the problem, but not that its existence depends on this particular one. If you assume that some families have not stopped reproducing, there is still a bias. I do not see a reasonable way to interpret the problem so that the population size is fixed or does not depend on the sexes of the children, but feel free to point one out.

I just wanted to say, that you assume that all children of one family are born instantaneous with the last child a boy. Dec 20 '10 at There were some very interesting and thoughtful comments on your post. However, since each argument continues until the participants finally agree, I expected that the fraction of thoughtful comments would be just over half.

Sadly this seemed not to be the case Tom Leinster 19k 4 75 It's funny that you mention that. I just discussed that puzzle in the StoxPoker. I learned of it on the TwoPlusTwo. I also posted a variant in the ProjectEuler forums. For the history of this problem, you could try asking Peter Winkler at Dartmouth , who calls the bottom card of the deck the "Predestination Card. It is not so clear to me why the answer is 0. So I understand that the dealer could have pulled out any card form the remaining deck and the game is equivalent.

However, if a lot of black cards had been pulled already from the deck, then the number of black cards remaining in the deck would be low so it would be sensible to say any card pulled from the deck would have a high probability of being red. I think you are biased toward the favourable scenario where you've just seen X say 5 black cards in a row and are thinking you should yell 'now'. You are dicounting the equallly probable scenario where you start by seeing X red cards in a row which is equally likely. For every favourable scenario there is an equally unfavourable one to counter-balance the average.

So when you start playing, you might as well take your chances and yell 'now'. You need another assumption. So there are really two questions here: I guess there is some ambiguity about what number is meant by a proportion A: A is the complement so that computing E[A: B] is essentially the same as E[B: If you interpret the proportion A: B] can be 1 while E[B: A] is not 1, and may not exist.

Your answer is more complete, and I took the lazy man's approach: For those who still don't get it, it might help to consider this ultrasimplified example: Jan 4 '11 at 5: Even if it is made crystal clear that average of fractions is not equal to the fraction of average, this does not entitle anybody to chose the wrong number. The question asks about the ratio of boys to girls, not about the ratio of expected boys to expected girls.

You ask why the ratio should interest anyone. I suggest you address that query to the person who posed the question, not the people who answered it. Since it's been well established elsewhere that you're not the least bit interested in the question that was posed, or any of the interesting subsidiary questions that it raises, but only in blustering and hurling insults, I won't be responding to any followups. The question asks about the ratio of boys and girls and not about the mean value of averages in families.

My own treatment and my recognition of this fact should show anybody that I am very interested in this question. I attribute your insulting manner to the fact that you recognize to have lost against Lubos Motl who, as a Harvard string theorist, is certainly not less than you able to understand the simple error made by Douglas and accepted by you. Vipul Naik 3, 1 22 Thorny 1, 8 9. You answered the false question. And as Thorny says and as I also few minutes ago found myself: It is completely irrelevant which couple decides to cease fire and which will continue.

Therefore the independent variables will remain equal within the statistical margin. TonyK 1, 13 My commenter also adds the following in my opinion, quite insightful remarks: Even for independent variables, the expected value of a ratio is not equal to the ratio of the expected values. The expected value of a product of uncorrelated variables is the product of the expected values, though.

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This is one of the most important keys to understanding this problem, I believe. And this is why I suggested the Taylor series to expand the ratio about its mean. I also think it is a little easier to find the expected proportion of boys because the random part G only appears in the denominator. Steven, that is incorrect.

It isn't convex, as simple calculations demonstrate.