Accordingly, they continue reproducing until they have a boy. Or until the mother can't give birth anymore. To make things easier, suppose each mother is able to produce ten children.

In such a society, will there be more boys or more girls? (than in a standard society)

Now, a new behaviour has arisen. If parents have a girl as first-born, they kill her. Let us suppose that they can't do it for next-borns, so that they continue the above tactic of reproducing until a boy is born.

How does this change affect your answer?

Then every mother has 1 boy.

1/2 of all mothers have 0 girl, 1/4 of all mothers have 1 girls, 1/8 of all mothers have 2 girls....

Number of girls per mother is then sum(n=1 to infinity) (n-1)/2^n = 1.

In this scenario, there will be same number of girls than boys in this scenario.

If only 10 children are allowed (sum n to 10 instead of infinity), then there will be 1013/1024 girls per mother, and 1023/1024 boys per mother.

The ratio of boys to girls is then indeed 1023/1013 =~ 1.01, or about 1% more boys.

If parents kill their first-born girl (but not their second born girl), then half of all mothers kill a girl, and hence there will be 501/1024 girls per mother.

The ratio of boys to girls is then 1023/501 =~ 2.04.

Why do you think this problem is not that simple ?

Actually, I am trying to make a connection between this problem and the one I ask about in Blackjack/Effect of position.

Normally, whatever the bound (10 here), the ratio should be still 1 on 1. Or p if you don't assume equiprobability to begin with.

^{n}. But the conditional distribution is a truncated geometric: pq

^{n}/q

^{10}, n=0 to 10.

Yet there are some families with ten girls and no boy.

Quote:kubikulannThis is a country where culture makes parents prefer to have a boy than a girl.

Accordingly, they continue reproducing until they have a boy. Or until the mother can't give birth anymore. To make things easier, suppose each mother is able to produce ten children.

In such a society, will there be more boys or more girls?

I am going for a new record for the number of times in a day that I can point out that E(X) + E(Y) = E(X+Y).

It will be 50-50. And, no betting system can make money in a 0EV game.

Quote:Now, a new behaviour has arisen. If parents have a girl as first-born, they kill her. Let us suppose that they can't do it for next-borns, so that they continue the above tactic of reproducing until a boy is born.

How does this change affect your answer?

Ooooh, a loss rebate on your first bet, if it loses. Now we are +EV!

If we assume that mothers can have infinitely many children, the question requires very little arithmetic, so that's the one that I'm going to answer.

Each mother has 2 children on average. Each mother kills 1/2 of 1 girl on average. Half the girls are killed (!!!) We are left with 2 boys for every 1 girl.

Nature often gives a young mother a female because they are easier to take care of and more experienced mothers get males who roam around more and are adventuresome.

After major population shakeups, females tend to be valued more. Post Black Death women were not put on a pedestal, they were needed.

Interesting. Do you have a reference?Quote:FleaStiffNature often gives a young mother a female because they are easier to take care of and more experienced mothers get males who roam around more and are adventuresome.

What does it have to do with the problem here?Quote:AxiomOfChoiceI am going for a new record for the number of times in a day that I can point out that E(X) + E(Y) = E(X+Y).

Ditto?Quote:AxiomOfChoiceAnd, no betting system can make money in a 0EV game.

Proof?Quote:AxiomOfChoiceIt will be 50-50.

This kind of sentence costs points to my students. Shows a misunderstanding of the concept of average. "Each" and "average" are contradictory. The correct form is "Half the mothers, on average, kill one girl."Quote:AxiomOfChoiceEach mother kills 1/2 of 1 girl on average.

Quote:kubikulannWhat does it have to do with the problem here?

The expected chance of boy or girl doesn't change if you have 1 child or 100. Stopping early doesn't matter. Every birth is a fresh roll of the dice. Kinda neat way to look at the problem (that I've seen before).

Quote:Proof?

E(Boy) = 0.5 and E(Girl) = 0.5

Hence 50% boys, 50% girls. The only wrinkle is the capping out at 10 boys.

Half the families have no girls, a quarter one girl, an eighth two, etc., one in 1,024 nine girls and one in 1,024 ten girls. That makes (256+256+192+128+80+48+28+16+9+10)/1024 = 1023/1024. There are the same number.

It's the same sort of thinking behind betting systems, as becomes obvious when you put the mothers in a line: the mothers change, but the babies stay the same.

The second obviously changes things.

Putting the mothers in a line again, a girl born after a boy is killed, and a girl born after a boy and 10n girls is killed. The first takes out half the girls; the second, .5 * Σ (1/1024)^k (k starting at 1), which is equal to one girl in 2,046. Add it up and we get 1024/2046 girls killed, or 512/1023 killed, leaving 511 for every 1023 who would have been born, which since the birthrates were equal, means 511 for every 1023 boys.

Quote:kubikulannWhat does it have to do with the problem here?

The outcome of each birth is a random variable. The expectation of the sum is the sum of the expectations. In other words, 0.5 boys per birth.

Quote:kubikulannThis kind of sentence costs points to my students. Shows a misunderstanding of the concept of average. "Each" and "average" are contradictory. The correct form is "Half the mothers, on average, kill one girl."

I disagree with you. I'm guessing that English is not your first language?

Actually, I'm not sure that that excuses it. The expected number of girls killed per mother is 1/2. Expectation and average are the same.

Or, we can represent the number of girls killed by each mother by a different random variable. The expectation of each random variable is 1/2. Once again, "expectation" and "average" mean the same thing.

You clearly did not read or did not understand what I wrote. I'm guessing statistics is not your first language?Quote:AxiomOfChoiceI disagree with you. I'm guessing that English is not your first language?

Actually, I'm not sure that that excuses it. The expected number of girls killed per mother is 1/2. Expectation and average are the same.

Or, we can represent the number of girls killed by each mother by a different random variable. The expectation of each random variable is 1/2. Once again, "expectation" and "average" mean the same thing.

I said Each is the wrong term to use. You wrote "each mother", now you try to hide it by writing "each random variable". Naughty trick. Does not work with me.

I never contested that expectation and average are similar concepts.

(Note: they are different, though, because expectation is in probability and average is on a statistical sample. The expectation is a constant, while the average is a random variable.)

Again, you seem incapable of integrating the concept of conditional probability. It is not 0.5 boy when you have to allow for the case "no birth".Quote:AxiomOfChoiceThe outcome of each birth is a random variable. The expectation of the sum is the sum of the expectations. In other words, 0.5 boys per birth.

a × 0.5 + (1-a)× 0 is not 0.5 if a<1.

May you be more precise?Quote:thecesspitThe expected chance of boy or girl doesn't change

One talks of the expected number of something (a random variable), or of the chance of something happening (an event), but the "expected chance" is not meaningful (in your context).

It is not, since most people provide a wrong answer. Maybe I'm not getting the meaning of "obvious"? I thought it meant "evident", "trivial".Quote:24BingoThe answer to the first is pretty obvious, really...

;-) "obviously" again?Quote:24BingoThe second obviously changes things.

Your answer is correct for the special case of 50% chance of boys. Yet that assumption was introduced by a later poster, it is not in the original problem. Can you provide the answers for the general case of p : 1-p ?

Quote:kubikulannMay you be more precise?

One talks of the expected number of something (a random variable), or of the chance of something happening (an event), but the "expected chance" is not meaningful (in your context).

Okay drop the word 'expected'. I think you understood the point made well enough. Not being marked here for a final examination, I may make some slightly loose statements.

Quote:kubikulannYou clearly did not read or did not understand what I wrote. I'm guessing statistics is not your first language?Quote:AxiomOfChoiceI disagree with you. I'm guessing that English is not your first language?

Actually, I'm not sure that that excuses it. The expected number of girls killed per mother is 1/2. Expectation and average are the same.

Or, we can represent the number of girls killed by each mother by a different random variable. The expectation of each random variable is 1/2. Once again, "expectation" and "average" mean the same thing.

I said Each is the wrong term to use. You wrote "each mother", now you try to hide it by writing "each random variable". Naughty trick. Does not work with me.

In English, "each" is the right word to use in this case.

This is a word problem, and each mother is a word-representation of a random variable. I'm perfectly comfortable using "mother" and "random variable" interchangeably here.

Quote:I never contested that expectation and average are similar concepts.

(Note: they are different, though, because expectation is in probability and average is on a statistical sample. The expectation is a constant, while the average is a random variable.)

The observed average of a sample is a random variable, only because the observed results are themselves random variables. Average itself is not a random variable, and it is the same as expectation.

I thought that this problem was diffrent, but I did the calculation and he is he right 100%, on the first problem the ratio stays 1:1

He is the math calculation for the problem as stated (ie upto 10 children untill a boy is born)

Boys Girls

1 0 0,5

1 1 0,25

1 2 0,125

1 3 0,0625

1 4 0,03125

1 5 0,015625

1 6 0,0078125

1 7 0,00390625

1 8 0,001953125

1 9 0,000976563

0 10 0,000976563

Weighted Average Number of Boys = 0,999023438

Weighted Average Number of Girls = 0,999023438

Ratio 1:1

And for the second problem (first girl is killed)

Boys Girls

1 0 0,5

1 0 0,25

1 1 0,125

1 2 0,0625

1 3 0,03125

1 4 0,015625

1 5 0,0078125

1 6 0,00390625

1 7 0,001953125

1 8 0,000976563

0 9 0,000976563

Weighted Average Number of Boys = 0,999023438

Weighted Average Number of Girls = 0,499023438

Ratio 2,001956947:1

Not exactly 2 (I thing because we stop at 10)

OK. So you used brute force calculation for one example (p=50%).Quote:AceTwoI did the calculation and he is he right 100%, on the first problem the ratio stays 1:1

Can you provide the equational answer to the general problem (arbitrary p of boys)?

Does the ratio remain identical?

How does the ratio change in the killing case?

Quote:AceTwoNot exactly 2 (I thing because we stop at 10)

Yes, exactly, because we stop at 10. I specifically mentioned that I was solving a modified problem where we do not stop at 10, because it makes the problem much simpler (and I can solve it in my head instead of needing paper)

These problems are all essentially the same. You just add up expectations. They are beyond simple. I don't understand why people feel the need to make them difficult. It's like baccarat system players trying to use the fact that they change their betting conditionally to try to beat a negative expectation game. Just like Axel pointed out in another thread, they are taking the simplest possible game and making it as complicated as possible, with the same results. That is exactly what is going on here; people are taking the simplest possible problem and making it complicated.

Just add up expectations. That's it. It could not be simpler. If I can solve a problem in 10 seconds in my head using a simple math theorem, why would I go through pages and pages of calculations, just to come to the same solution (unless I make a mistake in the pages and pages of calculations, in which case the solution will be different)

Quote:FleaStiff

Nature often gives a young mother a female because they are easier to take care of and more experienced mothers get males who roam around more and are adventuresome.

Not true to humans.

In the human population the natural ratio of male to female born is 105:100. There is some scientific reason for this relating to the x and y chromosomes.

This is the ratio observed in western societies.

However the world male to female born is 107:100 and this is due to selective abortion of female featus in countries like China, India etc.

But because women live longer on average there are more women than men, the ratio of male to female alive being around 0,97 in the west.

Quote:AxiomOfChoice

These problems are all essentially the same. You just add up expectations. They are beyond simple. I don't understand why people feel the need to make them difficult. It's like baccarat system players trying to use the fact that they change their betting conditionally to try to beat a negative expectation game. Just like Axel pointed out in another thread, they are taking the simplest possible game and making it as complicated as possible, with the same results. That is exactly what is going on here; people are taking the simplest possible problem and making it complicated.

You are 100% right. The problem is like playing roulette and say that I would bet black untill it comes out and then stop. The EVs do not change.

For some reason I thought the problem was different and I thought that there was indeed a bias towards male and the ratio would be favouring males.

In my defence I can say, this is late in the hour where I am from and my quite tired.

Quote:kubikulannIt is not, since most people provide a wrong answer. Maybe I'm not getting the meaning of "obvious"? I thought it meant "evident", "trivial".

;-) "obviously" again?

Your answer is correct for the special case of 50% chance of boys. Yet that assumption was introduced by a later poster, it is not in the original problem. Can you provide the answers for the general case of p : 1-p ?

Well, that's pretty simple, too, except now the "mothers in a line" version is easier to work with than the brute force version.

Second case: again, every girl after a boy followed by 10n girls is killed. Let's allow n to be zero this time (probably should have done so the first time, really). We'll say p is the chance of a girl being born. (1-p)*Σp^(10n), this time starting at zero, adds up to (1-p)/(1-p^10) of a girl being killed, so p/(1-p^10) surviving girls for every boy.

(The girls before the first boy are killed differently, but since the expected number of such girls is fixed - at one - for any population size, they vanish in the long run.)

Quote:kubikulannIt is not, since most people provide a wrong answer. Maybe I'm not getting the meaning of "obvious"? I thought it meant "evident", "trivial".

;-) "obviously" again?

Obvious to someone who understands math.

Many people think that they can beat roulette or baccarat by varying their bets in certain ways. I would say that it is obvious that they are wrong, even though many people believe it. Go read the craps or baccarat forums for an example of the stuff that many people believe. Then realize that you are making exactly the same mistake that they are in the blackjack thread.

This is not the correct answer.Quote:24BingoFirst case: again, nothing changes but the mothers. The ratio stays the same.

Second case: again, every girl after a boy followed by 10n girls is killed. Let's allow n to be zero this time (probably should have done so the first time, really). We'll say p is the chance of a girl being born. (1-p)*Óp^(10n), this time starting at zero, adds up to (1-p)/(1-p^10) of a girl being killed, so p/(1-p^10) surviving girls for every boy.

(The girls before the first boy are killed differently, but since the expected number of such girls is fixed - at one - for any population size, they vanish in the long run.)

Note that with the p=0.50 case your answer gets 512/1023 surviving girls. Contradicts your previous answer.

Stop the harassment please. Your feeling superior does not give you the right to despise other people. Your (supposed) mathematical skills are matched by a lack of social skills.Quote:AxiomOfChoiceObvious to someone who understands math.

Many people think that they can beat roulette or baccarat by varying their bets in certain ways. I would say that it is obvious that they are wrong, even though many people believe it. Go read the craps or baccarat forums for an example of the stuff that many people believe. Then realize that you are making exactly the same mistake that they are in the blackjack thread.

Quote:kubikulannStop the harassment please. Your feeling superior does not give you the right to despise other people. Your (supposed) mathematical skills are matched by a lack of social skills.

You are the one who is posting questions, and then telling people that they are wrong when they give you the right answers, and then editing your posts when you realize that you are wrong.

Mango's initial answer to your question is spot on. The thread should have ended there. You said that there was something wrong with his solution, then edited that out of your post. The rest of this is you trying to figure out what's going on, and trying to make it more difficult than it really is, to save face.

Sad.

Quote:kubikulannI edited it to prevent bad feelings. His answer WAS wrong. Go reread it. He says the proportion changes in the first case.

Oh, I see. Yes I misunderstood what he was saying (he answered 3 problems, including one which you didn't ask, which threw me off) Admittedly I also answered a question which you didn't ask.

You do at least see how the question can be answered by repeatedly applying the identity that E(X+Y) = E(X) + E(Y), right? In this case, each birth is a random variable (representing number of boys born) which maps to 0 half the time and 1 half the time. Thus each birth has expectation of 1/2.

Total number of boys is E(X1 + X2 + X3 + ...) where each Xi is a birth.

By applying that identity, this is the same E(X1) + E(X2) + ... which is 1/2 + 1/2 + ... which is 1/2 of the number of births. Therefore the expected number of boys = 1/2 the number of births, regardless of how many births there are, or how you decide whether or not to have another birth. This is very similar to a betting system question where you decide to quit or not quit based on how you are doing in a session; it does not change the fact that your total expectation is the sum of the expectations of all your bets.

As for the 2nd question, we can apply the same identity. Number of surviving girls = Girls born - girls killed.

Expected number of surviving girls = E(girls born - girls killed) = E(girls born) - E(girls killed). There's that identity again.

E(girls killed) is easy -- again, we let Xi be the number of girls that the ith mother kills and use the same identity. So E(girls killed) = E(X1 + X2 + ... + Xn) = E(X1) + E(X2) + ... + E(Xn) = 1/2 + 1/2 + ... + 1/2 = 1/2 the number of mothers.

E(girls born) would be easy if it were not for the max of 10 children total -- it would be 1 per mother (using the identity again). The max of 10 lowers it slightly, since mothers no longer have an average of 2 children each. Instead, 1 in 1024 mothers stops after 10 girls. That group would normally average 12 children, so the average number of children per mother is 2 - 2/1024 = 1023/512.

So each mother has, on average, 1023/1024 boys and 1023/1024 girls, and kills 1/2 of a girl. That leaves us with 2047/2048 boys and 511/1024 girls for a ratio of 511:1023 girls:boys. IMO the max of 10 just makes the problem unnecessarily ugly. 1:2 is such a nicer answer.

Note that the only math that I did was repeatedly applying E(X+Y) = E(X) + E(Y). Over and over and over, just applying the same identity. It turns this into a simple arithmetic (addition and subtraction) problem.

The blackjack problem is the same.

I made a sign error in my second answer, putting down the number of girls killed where I meant to put down the survivors. The actual number of girls per boy is (1 - my answer).

Certainly. My solution is using that sort of argument. I take all the first-borns, all the second-born etc. In each layer, there must be a proportion p/q of boys and girls. So the total maintains the same proportion.Quote:AxiomOfChoiceYou do at least see how the question can be answered by repeatedly applying the identity that E(X+Y) = E(X) + E(Y), right?

Yes, because the sex of the next child is independent from the past. That is not exactly the same in the blackjack situation, because of the dependency on players' decisions.Quote:In this case, each birth is a random variable (representing number of boys born)

The general answer is p (1-qQuote:That leaves us with 2047/2048 boys and 511/1024 girls for a ratio of 511:1023 girls:boys. IMO the max of 10 just makes the problem unnecessarily ugly. 1:2 is such a nicer answer.

^{10}) boys for q²(1-q

^{9}) girls. (p being the prob of a boy)

With a max of two children (think China's policy), then the ratio is 1:3 (for a p=0.50). Also a nice answer, no?

Quote:In this case, each birth is a random variable (representing number of boys born)

Quote:Yes, because the sex of the next child is independent from the past. That is not exactly the same in the blackjack situation, because of the dependency on players' decisions.

No, no, no. NO!!!! This is the point; this is what I have been saying over and over again.

E(X+Y) = E(X) + E(Y). Always. Even if X and Y are dependent. Even if they are correlated. You can ignore the dependencies. That is what makes the question easy! It's what makes the hat question easy. If you pick your hat before I pick my hat, my probability of getting my hat is dependent on whether you got your hat, but it doesn't matter! You still have n people, and each E(Xn) = 1/n, so E(X1 + X2 + ... + Xn) = E(X1) + ... E(Xn) = 1, so the expected number of people to get their own hat back is 1. It doesn't matter than the Xn's are all dependent. That is the whole point of the identity. It's not particularly interesting for iid variables; the fact that it works for dependent, correlated, and different variables is what makes it so powerful! It's why I don't need to do a lot of work to figure out the answer to the hat problem, it's just n * 1/n = 1.

In other words, it's not just that E(2X) = 2E(X). That is boring and not that useful.

It's that E(X+Y) = E(X) + E(Y) for all X and Y. That is so much more powerful!

Everything you write here is perfectly correct.Quote:AxiomOfChoiceNo, no, no. NO!!!! This is the point; this is what I have been saying over and over again.

E(X+Y) = E(X) + E(Y). Always. Even if X and Y are dependent. Even if they are correlated. You can ignore the dependencies. That is what makes the question easy! It's what makes the hat question easy. If you pick your hat before I pick my hat, my probability of getting my hat is dependent on whether you got your hat, but it doesn't matter! You still have n people, and each E(Xn) = 1/n, so E(X1 + X2 + ... + Xn) = E(X1) + ... E(Xn) = 1, so the expected number of people to get their own hat back is 1. It doesn't matter than the Xn's are all dependent. That is the whole point of the identity. It's not particularly interesting for iid variables; the fact that it works for dependent, correlated, and different variables is what makes it so powerful! It's why I don't need to do a lot of work to figure out the answer to the hat problem, it's just n * 1/n = 1.

In other words, it's not just that E(2X) = 2E(X). That is boring and not that useful.

It's that E(X+Y) = E(X) + E(Y) for all X and Y. That is so much more powerful!

What I'm saying is that you cannot use this if your premises are E(x|a) and E(y|b). These cannot be added just like that, they have to be weighted. Also, E(x|b) and E(y|a) should be accounted for.

In another thread, you acknowledge this, but mix the above argument with another: that E(x|a) and E(y|b) are both zero, so the sum is zero, weighting or not. Well, that does not prevent the necessity of weighting, does it?

And, finally, the question was to ascertain that they were effectively zero in the problem under analysis. Which proved right under one assumption and false under another.

Quote:kubikulannI edited it to prevent bad feelings. His answer WAS wrong. Go reread it. He says the proportion changes in the first case.

Yes, the answer to the second problem was wrong. My fault, it was late in the evening, and I'm not a mathematician.

(a) A good mathematician should have pointed a changing of birth strategy would never result in any other distribution of boys vs. girls. His professioni is to provide answers of absolute truth, however abstract the proof may be.

(b) A good math teacher would have spotted the false argument, formula, or assumption. His profession is to teach and not to judge right or wrong without giving hint to the right direction.

While (a) has been achieved in this thread numerous time (and I feel ashamed of not seeing it even when fighting with sleep), there is no trace of (b) at all. Which, in my modest opinion, doesn't speak high for the "teacher".

Quote:kubikulannWhat I'm saying is that you cannot use this if your premises are E(x|a) and E(y|b). These cannot be added just like that, they have to be weighted.

Yes, but when the expectations are 0, the weightings become irrelevant.

You are basically saying, no, TC' is not TC + 0 + 0, it is TC + 0 + f(TC)*0, and telling me that I need to do a bunch of number crunching with some complicated function f before I multiply the result by 0.