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Challenge Micromass' big probability challenge

  1. May 20, 2016 #1

    micromass

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    Probability theory is very nice. It contains many questions which are very easy to state, but not so easily solved. Let's see if you can solve these questions.

    • For an answer to count, not only the answer must be given but also a detailed explanation.
    • Any use of outside sources is allowed, but do not look up the question directly. For example, it is ok to go check probability books, but it is not allowed to google the exact question.
    • If you previously encountered this statement and remember the solution, then you cannot participate in this particular statement.
    • All mathematical methods are allowed.
    • Please reference every source you use.

    Here you go:

    1. SOLVED BY BiGyElLoWhAt, PeroK The universe is about ##432,043,200,000,000,000## seconds old. Assume that from the very start of the universe, a solitary monk was flipping a coin over and over again. He flips a coin every second. A run of order ##n## are ##n## consecutive heads in a row, or tails in a row. On average, what is the largest run that the monk has had?

    2. SOLVED BY ProfuselyQuarky Every cereal box contains one toy. There are in total ##20## different toys. How much boxes of cereal do you expect to buy - on average - in order to collect all the toys.

    3. REMOVED

    4. SOLVED BY PeroK A closet contains ##100## pairs of shoes. I choose at random ##8## shoes. What is the probability that I will have selected exactly one complete pair?

    5. SOLVED BY mfb In a town of ##n+1## inhabitants. A person tells a rumor to two distinct numbers, the "first generation". These repeat the performance and generally, each person tells a rumor to two people at random without regard to the past development. Find the probability that the generation ##r## will not contain the person who started the rumor. Find the median of this distribution assuming ##n## large.

    6. SOLVED BY andrewkirk Find the probability that in a random arrangment of ##52## cards no two aces are adjacent.

    7. SOLVED BY Fightfish, mfb In a ballot, candidate Trump scores ##304## votes and Clinton scores ##216## votes. When the votes are counted, find the probability that throughout the counting, there are always more votes for Trump than Clinton.

    8. SOLVED BY thephystudent I have a stick of ##1## meter. I break this stick in ##3## pieces in such a way that every point of the stick has as much chance of being a break point. Find the probability that I can combine the three pieces into an obtuse triangle.

    9. SOLVED BY Charles Link, PeroK In a room of ##1000## people. How many people do you expect there to be so that nobody else in the room shares their birthday?

    10. REMOVED

    Thank you all for participating! I hope many of you have fun with this! Don't hesitate to post any feedback in the thread!

    More information:
    1. gato-docs.its.txstate.edu/mathworks/DistributionOfLongestRun.pdf
    2. http://www.math.uah.edu/stat/urn/Coupon.html
    3. REMOVED
    4. Feller "An introduction to probability theory and its applications Vol1" Chapter II "Elements of Combinatorial Analysis" Exercise 26
    5. Feller "An introduction to probability theory and its applications Vol1" Chapter II "Elements of Combinatorial Analysis"
    6. Feller "An introduction to probability theory and its applications Vol1" Chapter II "Elements of Combinatorial Analysis" Exercise 42
      Solution: Look at ##48## non-aces. These determine ##49## gaps. Choose ##4## gaps.
    7. Feller "An introduction to probability theory and its applications Vol1" Chapter III "Fluctuations in coin tossing and random walks"
    8. http://www.math.uah.edu/stat/buffon/Triangles.html
    9. Invented on my own
    10. REMOVED
     
    Last edited: Jun 1, 2016
  2. jcsd
  3. May 20, 2016 #2

    BiGyElLoWhAt

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    So here's my thinking (not a huge probability guy, so I might be missing something, but I think this makes sense):

    We want to find the number of consecutive heads/tails such that the probability is 1 in 4.320432x10^17. The probability of flipping 1 heads is 1/2, the probability of two is (1/2)^2, and the probability of n heads is (1/2)^n. So we have (1/2)^n = 1/4.320432x10^(17)
    or
    ##2^n = 4.320432E17##
    ##\text{log}_2(2^n) = \text{log}_2(4.320432E17) ## which gives us about ##n=58.584## which honestly seems a little low to me. But that's my guess, anyways.
     
  4. May 20, 2016 #3

    PeroK

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    For number 4:

    To have exactly one complete pair, you need one pair and six odd shoes. The pair is equally likely to be any two shoes, from the 1st and 2nd to the 7th and 8th. There are, therefore, ##\binom{8}{2} = 28## equally likely ways to get a pair.

    The probability that the 1st and 2nd shoes are the pair is:

    ##p_{12} = \frac{100}{100} \times \frac{1}{99} \times \frac{98}{98} \times \frac{96}{97} \times \frac{94}{96} \times \frac{92}{95} \times \frac{90}{94} \times \frac{88}{93}##

    Hence, the total probablity that there is exactly one pair is:

    ##p = 28 \times p_{12} = 0.24##
     
  5. May 20, 2016 #4

    micromass

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    Are you sure you're not overcounting? For example, if the shoes have colors, are you sure you're distinguishing between (blue, blue, red, purple, green, yellow) and (blue, blue, red, purple, yellow, green) ? In either case, I'm unsure how you found your ##p_{12}## probability. And are you distinguishing between left and right shoes? The first two shoes the same and left shoes is different from the first two shoes the same and right shoes.
     
  6. May 20, 2016 #5

    micromass

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    That's a nice approximation. Unless anybody can provide a more accurate result, I'll accept your solution.
     
  7. May 20, 2016 #6

    PeroK

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    Sorry, I did 100 shoes (50 pairs), not 100 pairs! Out of time today to fix things now.
     
  8. May 20, 2016 #7
    Just a guess by doing division, would number 4 be 0.16 or 1.6% chance of there being a whole pair? I'm probably wrong but you never know unless you try.
     
  9. May 20, 2016 #8
    I don't understand what question 9 is asking, can someone elaborate?
     
  10. May 20, 2016 #9

    micromass

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    For example in a room with ##5## people, you have the following birthdays (June 13, June 28, May 1, June 13, May 1). In this case, only the second person has a birthday that isn't shared with anybody else. So in this particular situation, the answer is ##1##. I'm looking for the average of all these situations.
     
  11. May 20, 2016 #10

    micromass

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    That's not the correct answer.
     
  12. May 20, 2016 #11
    I figured as much.
     
  13. May 20, 2016 #12

    micromass

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    You were close though!
     
  14. May 20, 2016 #13
    In question number three are the balls numbered in order or are they each given a random number? If the former is the case then my answer for that question would be 213 since the last ball (the highest number of all five balls) was 213. If the latter is the case then my answer is five, since we know their to be at least five balls since you only picked five from the box.
     
    Last edited: May 20, 2016
  15. May 20, 2016 #14
    Wait... how close, because my first answer was 0.24 or 2.4%.
     
  16. May 20, 2016 #15

    ShayanJ

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    I haven't thought about it myself and just tried to understand BiGy's reasoning. In the process, I realized it should be 1 in ##\frac{4.320432 \times 10^{17}}{n} ##, which makes the answer equal to almost 53.
     
  17. May 20, 2016 #16

    micromass

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    The balls are numbered in order. So if there are ##n## balls, then there are numbered with ##\{1,...,n\}##.

    Can you give a reasoning for why you think ##213## is the best answer?
     
  18. May 20, 2016 #17

    micromass

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    Incorrect. In either case, you should give a reasoning.
     
  19. May 20, 2016 #18
    Because the highest number ball from the box you picked from was labeled as ball number 213, which means their are at least a total of 213 balls in the box.
     
  20. May 20, 2016 #19
    For the cereals question I suppose , as the probability of getting a type of toy is 1/20, we have to buy 400 cereal boxes.I'm not so sure but a try anyways.
     
  21. May 20, 2016 #20

    micromass

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    Indeed, there are at least ##213##. But what makes ##213## the optimal number of balls in the box? Sure, saying ##213## is the safest options, it's definitely a minimum number of balls. But what is most likely?
     
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