DominO123 Posted April 30, 2004 Report Share Posted April 30, 2004 (edited) Fad, I understand how you are looking at it and I think it's right. The way I did it: 2 feet on the short side and 2 feet on the long side of the board cancel. On the long side, we have 4 feet of board left (18 lb) on average 4 feet from the fulcrum. On the short side we have 48 lbs of weight at 1.5 feet from fulcrum. Since 18 * 4 = 48 * 1.5, everything is balanced. Of course its right. I was first doing it like you. But then, I said why involve each sides center of gravity for the board? Let imagine a board without a mass, and the board with a mass of 36 pound putting it on the one without one. Where would be the center of its gravity? In its center of course. Then the distance from it to the fulcrum(I just learned that word ) would be 2. The mass of the board being 36 pound, I multiply it by its center of gravity. The other side being what you found. Actually, I find my method the "purest" way to resolve the solution. Edited April 30, 2004 by Fadix Quote Link to comment Share on other sites More sharing options...
DominO123 Posted April 30, 2004 Report Share Posted April 30, 2004 Sorry Domino, I guess I am not understanding what the 2 * 36 is? 36 is the mass of the entire board, and 2 being the distance between the center of gravity of the entire board and the fulcrum. Once you understand how I did it, I think I will impress you. Quote Link to comment Share on other sites More sharing options...
Azat Posted April 30, 2004 Report Share Posted April 30, 2004 Fadiushka, i see how you did it. I don't care what Sip says about you, you are a smart man. Quote Link to comment Share on other sites More sharing options...
Sasun Posted July 30, 2004 Report Share Posted July 30, 2004 There is a village where couples live. There are equal number of men and women. The couples have children, the aim is to have a boy. As soon as a couple has a boy they stop having children. It is assumed that boys and girls are born with equal probabilities, 50-50 chance. The village grows in this way, the boys and girls grow up, make couples and follow the same principle the aim being to have one boy and stop. They will not stop having children until they have a boy. Question: assuming infinite number of generations, what would be the proportion of men and women in this village? Would there be more women or more men? Whatever the answer is prove it. Quote Link to comment Share on other sites More sharing options...
Azat Posted July 30, 2004 Report Share Posted July 30, 2004 over time 0 assuming the boys do not cheat Quote Link to comment Share on other sites More sharing options...
Sasun Posted July 30, 2004 Report Share Posted July 30, 2004 I think (intuitively) they will be 50-50 in the long run but I don't know how to prove it formally. Quote Link to comment Share on other sites More sharing options...
Sip Posted July 30, 2004 Report Share Posted July 30, 2004 (edited) It won't be 50/50 ... I just woke up and can't focus but I suspect the ratio of boys to girls will be 1 to 2. If you want, I'll post the solution after you guys had a chance to think more. Edited July 30, 2004 by Seapahn Quote Link to comment Share on other sites More sharing options...
DominO123 Posted July 30, 2004 Report Share Posted July 30, 2004 It won't be 50/50 ... I just woke up and can't focus but I suspect the ratio of boys to girls will be 1 to 2. If you want, I'll post the solution after you guys had a chance to think more. I disagree, and I have seen your solution before you deleted it. Your solution supposes that it adventages boys, like if a selective randomness will change the ramndomnity in this cases. It does not. It is not because you stop with boys that it means you advantage boys... because the only reason that you continue having kids supposes that you had girls, which in this cases gives for the next children the same chances to be a girl or a boy... for instance, if first you had a girl, the next time if you had a boy you'll stop... stopping there does not advantage boys, because you had a girl previously... in the same token, if you had a boy in the first shot, elsewhere another might have a girl etc... That's just my opinion, and I think you need a good argument to make me change. Quote Link to comment Share on other sites More sharing options...
Sip Posted July 30, 2004 Report Share Posted July 30, 2004 (edited) Alright, since you saw it already, here it is: Here are ALL the possible ways a couple can have children (assuming they'll have at least one): boy girl, boy girl, girl, girl, boy girl, girl, girl, girl, boy .... Basically in the form of (girl)^k, boy where k>=0 The probability of each is (1/2)^k * 1/2 = (1/2)^(k+1) I think the ratio of boys:girls will be: k : ( sum over k (from 1 to infinity) of [k*(1/2)^k] = 2 ) So the ratio boys:girls will be 1:2. Edited July 30, 2004 by Seapahn Quote Link to comment Share on other sites More sharing options...
Sasun Posted July 30, 2004 Report Share Posted July 30, 2004 Just to clarify, one man cannot marry with more than one woman (although this probably would not change the outcome). And the question is about both kids and adults. My reasoning is that the numbers of men and women are determined by births. Births are independent of each other, this classifies as a Bernoulli process, like tossing coins. Boy is a success of trial, and girl is failure. Even if the probabilities were different it would not matter. Suppose there is an experiment with infinite number of people lined up to toss coins, they are trying to get heads (success trial). Every time a person gets heads he/she stops tossing and the next person takes the coin to toss. Same goes for everyone. Now, all we can see is that there is tossing of coins experiment going. One person's likelihood of success is independent of another person's history of tossing, since each tossing is independent of any other tossings. So we can say that numbers of success and failure are equal in the long run, just like one person was tossing coins all the time without stopping. Now replace "tossing" with "giving birth", I think these are very similar experiments. All that we should care is about birhts, one person's behavior upon success or failure should not affect the other births. Overall, equal number of men and women are born. The chances are there are some single women at all times who never have children, but as they don't live forever their numbers are in the long run infinitely smaller than the number of existing couples and children with (approximately) equal number of both sexes. To sum up, in my reasoning the key is independence of trials (births). Quote Link to comment Share on other sites More sharing options...
Sasun Posted July 30, 2004 Report Share Posted July 30, 2004 (edited) Alright, since you saw it already, here it is: Here are ALL the possible ways a couple can have children (assuming they'll have at least one): boy girl, boy girl, girl, girl, boy girl, girl, girl, girl, boy .... Basically in the form of (girl)^k, boy where k>=0 The probability of each is (1/2)^k * 1/2 = (1/2)^(k+1) I think the ratio of boys:girls will be: k : ( sum over k (from 1 to infinity) of [k*(1/2)^k] = 2 ) So the ratio boys:girls will be 1:2. I see, but half of the time the experiment wil end in just one boy (the first line in the pyramid). Is that included in your calculations? Edited July 30, 2004 by Sasun Quote Link to comment Share on other sites More sharing options...
Sip Posted July 30, 2004 Report Share Posted July 30, 2004 Sasun, independence of tossings is a different issue than the relative counts of success vs failures. The problem here is asking you to count the ratio of successful flips to the unsuccessful ones. The mistake you make by combining all tossing parties into 1 tossing party is in effectively reducing the number of unsuccessful tosses! In other words, it's true that one person tossing a coin ad infinitum tosses is the same number of times as n people tossing a coin ad infinitum. However, you CANNOT take a ratio in this case (the infinity over infinity problem). What I did above, was to count the number of tails vs heads that each person is expected to get in the long run. Now since all of them are doing this experiment independently, the ratio overall would stay the same as that of the expectation for one experiment. Quote Link to comment Share on other sites More sharing options...
Sip Posted July 30, 2004 Report Share Posted July 30, 2004 (edited) I see, but half of the time the experiment wil end in just one boy (the first line in the pyramid). Is that included in your calculations? The experiment will ALWAYS end in one boy. 1/2 will have 0 girls, 1 boy 1/4 will have 1 girl, 1 boy 1/8 will have 2 girls, 1 boy 1/16 will have 3 girls, 1 boy ... So when we add up the series for girls, we get 2 girls!!! Note that you can also add up the series for the boys and you'll get 1 as expected: Boys: 1/2 + 1/4 + 1/8 + 1/16 ... = 1 Edited July 30, 2004 by Seapahn Quote Link to comment Share on other sites More sharing options...
Sasun Posted July 30, 2004 Report Share Posted July 30, 2004 However, you CANNOT take a ratio in this case (the infinity over infinity problem). Of course, we are assuming large numbers as infinity, and what I mean is that the ratio will be a number close to 1. At this point I am undecided, let me think a little more. Quote Link to comment Share on other sites More sharing options...
DominO123 Posted July 30, 2004 Report Share Posted July 30, 2004 (edited) The experiment will ALWAYS end in one boy. 1/2 will have 0 girls, 1 boy 1/4 will have 1 girl, 1 boy 1/8 will have 2 girls, 1 boy 1/16 will have 3 girls, 1 boy ... So when we add up the series for girls, we get 2 girls!!! Note that you can also add up the series for the boys and you'll get 1 as expected: Boys: 1/2 + 1/4 + 1/8 + 1/16 ... = 1 There is a problem with your method, how can you find a total of higher than 100% in a probabilistic calculation that when adding them we should find 100% ? Furthermore, if we were to do the same addintion for girls... we should find for girls. 1/2 + 1/4 + 1/4 + 1/4 + 1/4 etc... Edited July 30, 2004 by Fadix Quote Link to comment Share on other sites More sharing options...
DominO123 Posted July 30, 2004 Report Share Posted July 30, 2004 Alright, since you saw it already, here it is: Here are ALL the possible ways a couple can have children (assuming they'll have at least one): boy girl, boy girl, girl, girl, boy girl, girl, girl, girl, boy .... Basically in the form of (girl)^k, boy where k>=0 The probability of each is (1/2)^k * 1/2 = (1/2)^(k+1) I think the ratio of boys:girls will be: k : ( sum over k (from 1 to infinity) of [k*(1/2)^k] = 2 ) So the ratio boys:girls will be 1:2. I sweared having seen in your prior calculation the other way around... so forget my first reply(not the above). Quote Link to comment Share on other sites More sharing options...
DominO123 Posted July 30, 2004 Report Share Posted July 30, 2004 (edited) After thinking about it again, I entirly disagree with you, more so than before... you can not place boys and girls like that, the way yo try to find the proportions is against the question that state there is 50/50 chances and this is regardless of what sex of the previous chidren... It change nothing absolutly nothing when and where you stop the randomnity... Edited July 30, 2004 by Fadix Quote Link to comment Share on other sites More sharing options...
Twilight Bark Posted July 30, 2004 Report Share Posted July 30, 2004 k : ( sum over k (from 1 to infinity) of [k*(1/2)^k] = 2 ) I only have about a minute, so forgive me if I am off, but ... I think the sum should be of (k-1)/(2^k) rather than k/(2^k). And that would make the sum of girls 1 as well is my guess. So maybe Sasun's intuition was right. Cute problem. TB PS: Math and science is a good refuge when the "soft" issues start hurting your mind and heart. Quote Link to comment Share on other sites More sharing options...
Sip Posted July 30, 2004 Report Share Posted July 30, 2004 By the way, if anyone would be kind enough to (re)teach me why the series n/2^n sums to 2, I'd be very grateful. I couldn't remember how to do it so I just simplified it in mathcad and it gave me 2. Quote Link to comment Share on other sites More sharing options...
Sip Posted July 30, 2004 Report Share Posted July 30, 2004 I only have about a minute, so forgive me if I am off, but ... I think the sum should be of (k-1)/(2^k) rather than k/(2^k). And that would make the sum of girls 1 as well is my guess. So maybe Sasun's intuition was right. Cute problem. TB Actually I think that was the most complicated part of the problem. I may well have made a mistake with the k starting from 0 or 1 and the k-1's. Quote Link to comment Share on other sites More sharing options...
Sasun Posted July 30, 2004 Report Share Posted July 30, 2004 By the way, if anyone would be kind enough to (re)teach me why the series n/2^n sums to 2, I'd be very grateful. I couldn't remember how to do it so I just simplified it in mathcad and it gave me 2. Here you go: 1+1/2+1/4+1/8+1?16....=1+S 1+S=1+1/2(1+1/2+1/4+1/8+...)=1+1/2(1+S) S=1/2(1+S) 2S=1+S Therefore S=1 (and the sum is 2) Quote Link to comment Share on other sites More sharing options...
Sip Posted July 30, 2004 Report Share Posted July 30, 2004 (edited) 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!) The way I did it above, the number of boys would be (k+1), thus also 2. Edited July 30, 2004 by Seapahn Quote Link to comment Share on other sites More sharing options...
Sasun 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!) Well, you said you just woke up So you should never solve mathematical problems when you just wake up Quote Link to comment Share on other sites More sharing options...
Sip Posted July 30, 2004 Report Share Posted July 30, 2004 Believe it or not, I sometimes seems to think better in my sleep. Quote Link to comment Share on other sites More sharing options...
Azat 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 Edited July 30, 2004 by Azat Quote Link to comment Share on other sites More sharing options...
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