Sasun Posted July 30, 2004 Report Share Posted July 30, 2004 (edited) But you guys are failing to realize that over time there are going to be couples who will never have a boy, thus eventually the boy population is going to be reduced to 0 Why not? They will (theoretically) give as many births as needed just to have one boy. Edited July 30, 2004 by Sasun Quote Link to comment Share on other sites More sharing options...
Azat Posted July 30, 2004 Report Share Posted July 30, 2004 theoretically yes, but there ARE going to be couples who cant. After 20th child(all girls) they can't have kids anymore Quote Link to comment Share on other sites More sharing options...
Sip Posted July 30, 2004 Report Share Posted July 30, 2004 Well, theoretically, it's possible for a couple to give birth to n girls, for any n. But as n approaches infinity, the chance of that happening approaches 0 VERY quickly according to 1/2 ^ n. So even though it is possible (with non zero chance) that all offspring are girls for all practical purposes, when you average it out over all possible outcomes that can happen, you get a 1 to 1 ratio. Quote Link to comment Share on other sites More sharing options...
Twilight Bark Posted July 30, 2004 Report Share Posted July 30, 2004 Well, sasun, remind me never to argume against your intuition. The ratio is 1:1 after all since I messed up with the k=1 vs k=0 (even though I made sure to check like 50 times!)Â The way I did it above, the number of boys would be (k+1), thus also 2. Your k is "number of girls", and my k is "number of children". In your case the sum should start from k=0 and should sum the term k/(2^(k+1)). And in my notation the girls' sum should start from k=1 and should sum the term (k-1)/(2^k). But in either case, the exponent of 2 should be one more than the factor in front of it. And both the boys' sum and girls' sum are 1, not 2. Right? Quote Link to comment Share on other sites More sharing options...
Twilight Bark Posted July 30, 2004 Report Share Posted July 30, 2004 Oh, and the fertility rate would be 2 per couple on average, assuming that people can have an arbitrarily large numbers of children! But even then if we take into account non-fertile couples, the fertility rate would be less than two per couple, and the village would eventually disappear. But the last one standing would probably be female, given the longer life expectancy. So, even theoretically, the probable outcome is a devastated village, with 100% female population (of one). Probably not the intended result of the boy-loving population! There may be a more general lesson in there somewhere, and beyond the girl-boy issue. Quote Link to comment Share on other sites More sharing options...
Sip Posted July 30, 2004 Report Share Posted July 30, 2004 (edited) Alright, let me try again and be as rigorous as possible with a uniform notation: k=1: boy k=2: girl boy k=3: gir girl boy k=4: girl girl girl boy .... Giving a general form for the probabilty of (girl)^(k-1) boy where k=1,2,3, ... Counting boys we have: sum over k=1...inf of [ (1/2)^k ] = 1 Counting girls we have: sum over k=1...inf of [ (k-1)*(1/2)^k ] = 1 Thus the ratio is 1:1 Edited July 30, 2004 by Seapahn Quote Link to comment Share on other sites More sharing options...
Sip Posted July 30, 2004 Report Share Posted July 30, 2004 And just to convince myself of the fertility rate:Â Number of children = sum over k=1...inf of [ k * (1/2)^k ] = 2 Â But wouldn't infertile couples be averaged out against the infinitely fertile couples? Quote Link to comment Share on other sites More sharing options...
Twilight Bark Posted July 30, 2004 Report Share Posted July 30, 2004 But wouldn't infertile couples be averaged out against the infinitely fertile couples? The number 2 is obtained by assuming every couple had one ore more (including infinity) children. Any infertile couples would reduce the overall fertility rate below 2, in spite of the herculean efforts of some rare couples! Sorry. Quote Link to comment Share on other sites More sharing options...
Sip Posted July 30, 2004 Report Share Posted July 30, 2004 (edited) That makes sense. Now you got me thinking ... I wonder if having infertile couples would have any impact on the ratio ... it seems like there shouldn't be any impact. Also, an observation here is that the "herculean" couples would produce basically exclusively a very very large number of girls. Kind of interesting how they are supposed to balance the single boys that are born!!!! Because any family that has more than 1 child produces at least one girl for the boy they get. A very neat problem in deed. Edited July 30, 2004 by Seapahn Quote Link to comment Share on other sites More sharing options...
vava Posted July 31, 2004 Report Share Posted July 31, 2004 Also, an observation here is that the "herculean" couples would produce basically exclusively a very very large number of girls. Kind of interesting how they are supposed to balance the single boys that are born!!!! Because any family that has more than 1 child produces at least one girl for the boy they get.  That's balanced out by half the total number of couples that have a boy first. Great problem guys - too bad I saw it too late! Quote Link to comment Share on other sites More sharing options...
Sasun Posted July 31, 2004 Report Share Posted July 31, 2004 Thank you guys for suggesting the right direction. I thought I would recap and generalize the proof. Let's say, Prob(Boy)=p Prob(Girl)=q p+q=1 Expected number of boys and girls from each couple: Number(Boy)=p+pq+pq^2+pq^3+pq^4+.....=p[sum over k=1...inf of (q^k)]= =p/(1-q)=1 (this result is not surprising since the couple is exactly doing everything to get 1 boy) Number(Girl)=pq+2pq^2+3pq^3+4pq^4+.....=p[sum over k=1...inf of (kq^k)]= =pq/(1-q)^2=pq/p^2=q/p It must be emphasized that the events "Boy" and "Girl" are independent, therefore we multiply their probabilities above. Total expected number of kids from each couple is  Number(Kids)=1+q/p . If boys had better odds to be born (p>q) we have a fertility rate<2 and the population will decline. If p2 and the population will grow. In our case p=q=1/2, so the population will remain the same. In either case, the ratio between boys and girls is Number(Boy)/Number(Girl)=p/q, and as it was proved before, this ratio is one assuming equal odds for boys and girls. Hmm... I like it, let's do more proofs like this Quote Link to comment Share on other sites More sharing options...
DominO123 Posted July 31, 2004 Report Share Posted July 31, 2004 Seems that that question has gone far while I was too occupied to answer back. Quote Link to comment Share on other sites More sharing options...
Sip Posted July 31, 2004 Report Share Posted July 31, 2004 Good job sasun! I like the p/q ratio conclusion. I didn't see it before but now it's all clear. Quote Link to comment Share on other sites More sharing options...
Boghos Posted August 18, 2004 Report Share Posted August 18, 2004 (edited) Using 1,3,4 and 6 and only the four basic operations get to 24. Use each number only once. Edited August 18, 2004 by Boghos Quote Link to comment Share on other sites More sharing options...
vava Posted August 18, 2004 Report Share Posted August 18, 2004 Can we only use each number once? Quote Link to comment Share on other sites More sharing options...
vava Posted August 18, 2004 Report Share Posted August 18, 2004 Can we only use each number once? Nevermind - what would be the point? I'm a bone head. Quote Link to comment Share on other sites More sharing options...
Boghos Posted August 18, 2004 Report Share Posted August 18, 2004 Fair question and I fixed it. Quote Link to comment Share on other sites More sharing options...
vava Posted August 18, 2004 Report Share Posted August 18, 2004 (edited) 6/(1-(3/4))=24Â Edited August 18, 2004 by vava Quote Link to comment Share on other sites More sharing options...
ED Posted August 18, 2004 Report Share Posted August 18, 2004 4+6x2+1+3=24? Quote Link to comment Share on other sites More sharing options...
vava Posted August 18, 2004 Report Share Posted August 18, 2004 (edited) 4+6x2+1+3=24? doesn't that make 20? Or did you forget the brackets, maybe... Edited August 18, 2004 by vava Quote Link to comment Share on other sites More sharing options...
Sip Posted August 18, 2004 Report Share Posted August 18, 2004 Can't use 2 Quote Link to comment Share on other sites More sharing options...
ED Posted August 18, 2004 Report Share Posted August 18, 2004 No vava its 4+6 times 2 =20 +1+3=24 Â anyways it was worth the shot Quote Link to comment Share on other sites More sharing options...
Azat Posted August 18, 2004 Report Share Posted August 18, 2004 25-1=24*Â *Based on Edwards answer- if he can use 2 then I want to use 25 Quote Link to comment Share on other sites More sharing options...
ED Posted August 18, 2004 Report Share Posted August 18, 2004 25-1=24*Â *Based on Edwards answer- if he can use 2 then I want to use 25 LMAO Azat Quote Link to comment Share on other sites More sharing options...
Boghos Posted August 18, 2004 Report Share Posted August 18, 2004 (edited) Bravo vava, quick solution for a simple but nonetheless elegant problem. Edited August 18, 2004 by Boghos Quote Link to comment Share on other sites More sharing options...
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