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About bonanova

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    Retired Expert
  • Birthday November 3

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  • Location
    New York
  • Interests
    Music [performing and directing], photography.

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  1. Yikes! Really? It's 24 times more likely for a 3 to be drawn 5th than to be drawn 1st?
  2. I like phaze's solution. I would add just a tweak, as described in the second spoiler. Recap: Tweak.
  3. You are on the right track.
  4. So you're a writer, Bill. Care to share some of your work? Sure. Here are a few bits of it: FF?
  5. It would be more fun if there were five of us in a room working on it together. I've been distracted by other things lately but still interested.
  6. First thoughts (that are consistent with a better result than in the 3-person example):
  7. With a nod to jasen's recent and interesting puzzle, A traveler happened upon a village of huts, laid out as the circles in the picture below. The village's mayor explained to the traveler that the family living in each of the huts had an eldest son whose age was unique within the village. (No two eldest sons had the same age.) How interesting, replied the traveler. Tell me this: of all the male children here, what is the age of the very oldest? The mayor thought for a moment and replied, well I guess I could tell you that none of them are yet of voting age (21), and I guess you might be interested to hear that there are no gaps in their collective ages. But all of that wouldn't be enough information. I think it would be better for you to just knock on all the doors and ask. I don't have time for that, replied the traveler, and I'm really not that interested. Well, here's an interesting thing about our village, replied the mayor. You may have noticed, our huts are laid out so that many rows of 3, 4, or 5 huts cut across the entire village. Just ask at the huts along any of those rows. Add the ages that you hear, and divide the sum by two. That way you will learn the age of the oldest son in the village. And now, without knocking on any doors, you can learn it too.
  8. I've brought this up with site management to have replies ordered chronologically. I think that's the order now. It's confusing otherwise.
  9. Yeah, that's it. I have an online tree now ( with a few thousand names. I get inquiries and "matches" almost daily from people who descended from an uncle's sister-in-law of an 8th-generaltion direct ancestor. Soon you realize how interconnected the human family is. The gene pool is constantly being stirred. Even if you could map the whole thing you'd need a lot of color coding to discern in individual's tree. We'd all be different colors simultaneously, and a lot of "branches" would be tied into knots. (Rednecks all... )
  10. I get that. I thought it was strange that the question was introduced with an "overpopulation" flavor. (Maybe it was just a red herring.) My comment badly implied that yours wasn't responsive to the question; I meant to opine that the introduction wasn't all that germane. There are a couple of "paradoxes" when one considers family trees and such. For example we all have two parents, four grandparents, eight GGPs, sixteen GGGPs, and so on. So how is an exponentially increasing heritage as we go back in time consistent with a growing population going forward? Was there an original, parenting "couple" N generations back? If so, how could they singly play all the roles of my 2N progenitors at that time? And, if two kids/family going forward gives a stable population, how can it also produce an exponential personal progeny? I found Mitochondrial Eve an interesting read.
  11. Brute force Has anyone found a constructive solution?
  12. "... for a thousand years. How many descendants would you have?" But we're not enumerating the world's population. We're quantifying the part of it that comprises your descendants. Two children, four grandchildren, ...
  13. Give it a try:
  14. Doesn't each birth increase the population? What if, miraculously, no one will die? But the question does not hinge on when/whether people will die. The question asks: how many descendants will you have?
  15. @BMAD The three of us: you, I, and Logo, agree on the 5 4 1 5 3 = 18 assignment. I like your approach, where the solution can be constructed.