Twinzen

Members
  • Content count

    90
  • Joined

  • Last visited

About Twinzen

  • Rank
    Mini-Member
  1. I don't know that he's actually made me shed tears, but I find Carl Sagan and his "billions and billions" quite moving sometimes.
  2. This is absolutely true, and has the interesting implication that were an "intelligent being" (by human standards) to emerge in a couple of billion years from now, which might be very likely given the size of the universe, they would probably have to conclude that the universe is static and that it consists of only their own galaxy.
  3. I lol'd
  4. Noo! "Go 6 it!"
  5. Absolutely, it is a good idea. Indeed it is the best strategy in the long run. The point is that we use the first 36% of the dataset as a representative subset. Also the a posteriori odds are either 100% or 0%. My odds are a priori. If you were "the poor devil" facing this situation, you would be well adviced to use my strategy, as there is nothing better. You need to discard a certain amount of datapoints just to get a representative subset that will make the standard deviation small enough to go by. If you discard too few, the standard deviation stays so large that you can't make good judgment where to stop. If you discard too many, your sd will be smaller but you risk rejecting the right maiden for every new datapoint you collect. This problem is about finding the balance between the two. That being said, I am glad I don't face this dilemma irl. :p Edit: My apologies. Looks like I quoted you out of context. In a searching algorithm over an unsorted dataset you have no choice but to iterate over all the elements, and choose the biggest one you find. By using my "method" you will only be right in ~37% of all cases. However, your life is at stake, so you need to take a risk, but try to minimize it. By iterating over all maidens like if it were a searching algorithm you would have to choose the last one, and the chance of being right would be 1/100 as you correctly point out. The good thing about a searching algorithm is that you can go back and select a datapoint a posteriori, but in this riddle you can't go back to an earlier maiden. A searching algorithm under the condition that it can't go back would indeed have to use my strategy, and only yield a 37% success rate. Not relevant for computer science, I agree.
  6. Yes. If you have 100 maidens lined up the chance of the right one being "to the right" of the first 36 is 64%. That was all I was saying. The first chance you take is discarding the first 36 as the right one, the second chance you take is assuming the first maiden who presents a higher dowry than the first 36 is the right one, if any such exists, whatever the probability is. With hindsight I can say that the probability of that is 58% since I know the product probability to be around 37%. Thanks for the interesting problem Hellbent.
  7. I will demonstrate why your argument fails. Say there were only 5 maidens, and your strategy was to hear the 2 first maidens out, then pick the first of the 3 last ones who gives a higher number than the 2 first. I will use the numbers 1-5 to represent the maidens' dowries, where 1 is the lowest and 5 is the highest, though they are only representations of the real values. Line up the maidens, and start interrogating them. There are 120 (5!) ways in which they can be lined up, and I will demonstrate that by using my strategy the chance of winning is higher than 1/5, and by analogy, the 100 maiden problem has a winning chance of more than 1/100: The 5 is first: 5****, you lose (24 permutations) The 5 is second: *5***, you lose (24 permutations) The 5 is third: **5**, you win (24 permutations) The 5 is last: You only win if the 3rd and 4th digits are less than any of two first, that is, if 4 is among the 2 first: 34215 34125 43215 43125 24135 42135 24315 42315 14235 41235 14325 41325 12 wins, the rest are losses (4!-12 = 12) we have determined that you will win in at least 36/120 scenarios, which is 30%. The real chance is even bigger since there are winning scenarios where the 5 shows up as the 4th digit, but I don't need to calculate it as I have shown that this strategy gives a winning chance higher than 20% (1/5). I realized that the solution I gave earlier in this thread actually is analoguous to using standard deviations and statistical inference, as I use the first 36 values as a representative subset. Can you do better than my strategy? Probably, but you need to determine where to stop, how many standard deviations from the mean f.ex? This is subtle statistics that are out of my expertise, so I will wait for OP to post his answer. I will be surprised if it is much better than my strategy yielding ~37.1%.
  8. Cool, reminds me of some av arhitecture.
  9. This is the point I would contest. Knowing the highest dowry you have heard up till a certain point would make you know for certain whether some of them NOT could be the right one, as they might present lower dowries. This makes it safe to exclude some. My argument is, take a chance that it is none of the 36 first maidens (64% chance of being right) then pick the first one who presents a dowry higher than any of the 36 first. (You could still be wrong of course, but I think the chance of success using this strategy is much better than 1/100.) Anyway, I might be wrong in my analysis, but I can't see where exactly the problem lies.
  10. I might need to revise my original assertion. When using standard deviation on a dataset you assume there is rhyme and reason to it, but in this problem the dowry doesn't necessarily have an upper bound a priori. As an educated mathematician within the more theoretical branches, I can't claim to be an expert on numerics/statistics, but I have a tentative solution based on experimentation. (Using C++, which means my random number generator won't be very good, but I did the experiment 1 million times per number so it should be in the ballpark of the real solution.) Basically, my method is as follows; Ask the first N maidens what their dowry is, then iterate over the rest of the maidens, asking each one what their dowry is, and as soon as someone claims to have a higher one than any of the first N maidens, choose her. Which value for N gives the highest probability of success? Turns out it is somewhere around 35-37. I'll go with 36. In other words, had I beed faced with this dilemma, I would ask the first 36 maidens what their dowry is, then I'd choose the first maiden among the rest who had a higher dowry than any of the first 36. Chance of survival would then be 37.1% or so. Is there a better solution?
  11. This is a very interesting problem. I think the way to solve it is by statistical methods involving calculating the standard deviation and average for each new maiden you are presented with, and then choosing the first maiden whose dowry far exceeds it. The problem is that there is a lot of uncertainty in the beginning, so you better hope the right maiden is not among the first ones. I will try to solve this rigoruously today.
  12. I asked god after sacrificing a kitten to him and he revealed to me in his infinite wisdom that you making a map for this project was part of his infinitely wise plan.
  13. Very professional looking textures. If I didn't fail as a mapper I would definitely make stuff using these.