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EventHorizon

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Everything posted by EventHorizon

  1. Hmm....seems even that may not be solvable. What I had is simply an upper bound. Looks like this thread is done.
  2. Alright...question 3 does need to be simplified. Let S(x,m) be the number of people you need to gather before there is a group of x people all with the same emotion, where m is the number of emotions. (new 3) Assume S(x,2) is known for all x. Give a recursive function for S(x,m).
  3. Yup, you got 1 and 2. Very interesting that this is Ramsey Theory. Thanks for pointing that out. I adapted this from a question in a math class I took quite a few years ago (actually, question 1 is the exact question). I thought I had a method to solve question 3, but it must break down somewhere. I didn't really put much effort into part 3 yet, though it is surprising to me that it is an unsolved question....
  4. But I suppose, EH is looking for a polynomial solution. Is that so? Or is the problem to calculate probability for two specific cards? But that would be too simple, not EH-like problem. Very nice. I'll consider this one solved. Though here's an equation that will blow your mind So that just leaves question 2...
  5. Now that this one is solved....why not solve a related one I posted about a week ago... Question 3
  6. that's right. So, what are some of the answers?
  7. yup, there's the answer to the first question. It's fitting that you got it first. Any ideas on the next two?
  8. For 2, I said "nope...." when I should have said, you can get a better estimate...and to use spoilers
  9. I have a smaller number than that for number 2.
  10. I think my labeling the cards a and a+1 made the problem a little confusing. There is no wrap around, and a is not a specific card, just one that has the card with the number one greater right after it. Look at my example for n=3 in my previous post.
  11. Largaroth: Remember, it is a probability. This means it needs to be between 0 and 1. Avalanche5x: You answer fails for n=3. the probability is 1/2, and you have 1/3.
  12. 1)correct 2)nope 3)nope Though I haven't yet worked out the solutions for 2 and 3 completely.
  13. nope. For n=3, it is 1/2, but your equation gives 1/3.
  14. Please use spoilers, so others can look over posts without seeing answers.
  15. n, m, and x are all variables
  16. There are n people in a room. Between each pair of people, they either hate each other or like each other (e.g., the feeling is always mutual). (1) What is the minimum number of people, n, such that you are guaranteed to have either a group of three that all like each other or a group of three that all hate each other? (2) Now we add in the ability to be indifferent (not like or hate). What must n be such that there is guaranteed to be a group of three that all like each other, all hate each other, or are all indifferent towards each other? (3) Now say there are m different emotions. How many people must there be such that there is guaranteed to be a group of x people that all have the same emotion towards each other?
  17. 1) look it up. (though I did include the information needed in the spoiler) 2) nope.... 3) Yes, n is a variable.
  18. I'll start with extensions of Prof. Templeton's Birthday and SSN puzzles... (1) How many Plutonians would you need to gather before there is a 50:50 chance that two of them have the same birthday? (2) Planet Gplex takes a googolplex (10^(10^100)) of its days to orbit its sun. Approximately how many Gplexians would you need to gather before there is a 50:50 chance that two of them have the same birthday? Prove that your estimate is a good one. (yeah, I'm trying to make direct computation impossible, and yes I have an answer/proof) And now, a somewhat related original puzzle... (3) You are given a stack of cards numbered 1 to n, and their order is randomized. What is the probability that two consecutively numbered cards (lets say, a and a+1) are found consecutively (a immediately before a+1) somewhere in the stack?
  19. oops....forgot part 2
  20. EventHorizon

    Either that, or they end up endlessly running clockwise in the smallest circle they can manage (crazy point horses...). But theoretically, yes, they do meet....as long as their speed does not decrease below a given positive threshold (the path itself has finite length).
  21. EventHorizon

    Their paths will not be on circular arcs. Their paths will be smooth and will be arced, just not along the path of a circle.
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