bubbled

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About bubbled

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  1. Take the last chip

    I think have the answer to the second part: Spoiler Turkey (T) should go first unless the starting chips are in the set of losing (L) starting conditions. L = {L1 = 2, L2 = 3, L3 = L1 + L2, L4 = L2 + L3, L5 = L3 + L4....}. So it's the Fibonacci sequence without the 1, 1 at the start as those numbers don't make sense given the problem. My logic is a bit convoluted, but it's the only way I've managed to get anywhere. Basically, the key numbers are the ones in L. There are two ways to win. 1. Ideally, on your first play, take as many chips as necessary to leave a number in L. However, you can only do this if the number of chips you take is strictly less than 1/2 of the number of chips you leave. For example, if your starting chips is 7, you can take 2, and leave the homunculus (H) with 5, and since 2 < (5/2) and 5 is in L, you win. However, you lose on 8, because if you take 3 leaving 5, H can win by taking the rest of the chips. So you must take either 1 or 2, but either way, H will leave you with 5 next, and you lose. 2. You also win if the (starting chips) - (the largest member of L that's less than starting chips) is not in L. For example, let's look at 20 chips to start. The key number is 13, as that's highest number in L that is less than 20. Ideally, we'd like to get right down to 13, but if we take 7, H wins by taking the rest of the chips. But, we are in luck as 20 - 13 = 7, and 7 in not in L. This means we can play the game as though the starting chips were 7. Once we take the last chip above 13, we will leave H with 13 and win. If the number of starting chips gets really large as in the problem, the only way I can think solving it is recursively. The idea is, you need to keep leaving H with various numbers in L, all the way down the ladder. To do that, you may need to examine how you win from smaller numbers and build up. In example 2 above, I know that 7 is a winner. I also know that 5 is a loser, so I'd take 2, leaving 18 which is 5 more than 13. H can take any number from 1-4, but none of those will get him to 13. The most interesting is 1, leaving T with 17. 17 is 4 more than 13, so all still good. We know 3 is in L, so T takes 1, leaving H with 16, which is 3 more than 13. H can take 1 or 2, but either way, T's next play will leave H with 13. Next, T just needs get to the next stop which is 8. He will repeat the above process, stopping at 11 and/or 10 to get there. Then he'll go from 8 to 5, 5 to 3, and then win.      
  2. Bus Schedule

    Is it OK assume exactly 4 busses per hour, but arriving at completely random times? I can simulate that. Clearly, if the city had the busses arrive at exactly 15 minute intervals, then wait times would average around 7.5 minutes. And no one would ever wait longer than 15 minutes.
  3. Take the last chip

    I think I figured out the first question: Spoiler Define N (even number) to be the number of chips in the pot. As long as N is not equal to a number  2^n (n = integer), then Turkey (T) should go first. Otherwise go second. If N is odd, the game is trivial. T takes 1 on the first move and wins. If N is not equal to 2^n, then search for the smallest number 2^a where (N / 2^a) is an odd number. Start by taking 2^a from N. If the homunculus (H), takes an odd number next, the game is over. T takes 1 next and will win. So, to keep things interesting, H must take an even number next. (N - 2^a) will be divisible by all numbers 2^b where b <= a. And (N - 2^a) / 2^b will be an even number. Therefore, if H takes 2^b next, (N - 2^a - 2^b) will be divisible by 2^b and will be odd. So, whatever even number, 2^b, H takes from (N - 2^a), T should next just take 2^b. H can now take 2^c where c <= b, and T continues with the same strategy above until he wins. If N = 2^n, then N / 2^a (a <= n) will be even. So no matter what even number H takes from N, T can go second and take 2^a next, and continue the strategy above to win. In the specific game with 43546758343209876 chips in the pot, T should go first and take 4. 43546758343209876 - 4 = 43546758343209872. That number is divisible by 2 and 4, and those divisions result in an even number. So, no matter whether H takes 4 or 2 next, he will leave an odd number of 4's or 2's for T. Here's one concrete example. If N is 20: T takes 4, H 4, T 4, H 4, and T wins with 4. Or T takes 4, H 4, T 4, H 2, T 2, H 2, T wins with 2. Or T takes 4, H 2, T 2, H 2, T 2, H 2, T 2, H 2, T wins with 2.  
  4. Counterfeit Coins

    Agree with Ed on first part. Can you clarify the second question?
  5. Who is taller, A or B?

    Here are my results:
  6. I just ran one more 100M simulation and captured a lot more info.  Probability that a new random point will fall in the largest arc: 0.61104318 Mean of the arc length that captures the random point: 0.499978733972 Mean of the sums of the squared arc lengths: 0.500018084204 Median length of the longest arc: 0.591768881555 Histogram of the shortest arc lengths: Histogram of the middle arc lengths: Histogram of the longest arc lengths: I might be the only one, but I find these graphs to be interesting. Anytime you have two or more random variables and are trying to say things about their relationships to each other, things get pretty non-intuitive.       
  7. I think it's interesting that the sum of the squares of the means doesn't match the mean of the sum of the squares.  Maybe that shouldn't be interesting?
  8. OK, so I continued by using bonanova's idea of picking the arcs and then spinning the wheel. I used the same method as above (picking two random points) and creating three arcs. I then pick a random new point and find the length of the arc it falls in. I store those lengths in a separate array and find the mean length. With 100,000,000 million trials, I get 0.500034371498. This it what we'd expect. But I'm confused as to why the sum of the squares doesn't add up to .5 as well. It must be something to do with the distribution of the actual lengths of the arcs. Maybe because when the longest arc is bigger than the mean of the other longest arcs, the random point will fall in it more often? More simulations necessary to answer that question.  
  9. I ran a simulation of 100,000,000 trials using bonanova's method (I think) on a unit circumference. I pick two random numbers, P1 and P2,  between 0 and 1. If P1 < P2, then my three lengths are L1 = P1, L2 = P2 - P1, and L3 = 1 - P2. If P2 < P1, L1 = P2, L2 = P1 - P2, and L3 = 1 - P1. I sort the three lengths and put them into a matrix where the first column consists of all the shortest lengths, the second column is the middle length and the third column is the longest length. I then calculate the mean of the various columns. I get this result: mean of longest arcs = 0.611101750112 mean of middle arcs = 0.277769774084 mean of shortest arcs = 0.111128475804 This matches bonanova's results pretty closely and those lengths add up to 1, as expected. It would seem to me that the probability that a particular point falls in the longest arc should be 0.6111... And the middle arc 0.277... and shortest arc 0111... Which means that the average length of the arc that a random point (or fixed point) falls in should be the sum of the squares of the mean arc lengths. However, the sum of the squares is only 0.462950934519. I would expect it to be .5, given the analysis from the previous problem. I'm a bit confused. Thoughts? 
  10. cutting a circle to find a point

    I can't disagree with any of this. I was completely wrong and the fact that my answer matched with the correct answer blinded me to the fact that I was introducing an extra point into the problem. I now realize that you have to be very careful with how exactly you apply random conditions to problems like these.
  11. Find The Three Fastest

    At a track meet, you have 25 runners for a 100 meter event and only 5 lanes. You've forgotten all of your timing devices at home. Assume that all runners {R1, R2,...R25} run the event in different times from each other, but an individual runner RN, will run the event in the same time each time he/she runs. Your job is to figure out who is the 1st, 2nd and 3rd fastest runners. What is the minimum number of heats needed to make your determination?