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Two Interesting Probability Questions

  1. Mar 19, 2009 #1
    p is probability that a man will success in a single shot
    n is number of shots man is going to make

    1)what is probability that a man will score at least one series of at least k successive successful shots (in the interval of n)?

    2)what is probability that a difference between his unsuccessful shots at any time (for the n={1,..,n}) and his successful shots (also score in that same time/for that same value of n) will never be grater that d ?

    Hope you understand
  2. jcsd
  3. Mar 19, 2009 #2


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    Welcome to PF!

    Hi Nikolaj! Welcome to PF! :wink:

    Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
  4. Mar 19, 2009 #3
    This looks like a homework question :rolleyes:
  5. Mar 19, 2009 #4
    I don't know even how to begin, so if you could direct me, maybe some good article on internet that would help me to solve this, don't really know math terminology in English so don't know how to find it. :confused:
  6. Mar 21, 2009 #5
    well, for part 1 you may want to start by using the binomial distribution and fiddling with it to get to part 2
  7. Mar 22, 2009 #6
    Both problems look non-trivial. For the first you'll probably need to form an auxiliary variable p(n,i) with the probability of the i most recent shots being successful, then set up a recurrence formula. Similarly for the second.

  8. Mar 31, 2009 #7
    Are you still working on this problem? You should take bpet's advice about using a recurrence relation. Here's how I would start on the first problem:

    [tex]\pi_{i,j}[/tex] is the probability of getting a series of k successful shots given that the last k-i shots have been successful (he needs i more to win right now) and he has j shots left. This probability satisfies the recurrence

    [tex] \pi_{i,j} = p \pi_{i-1,j-1} + (1-p)\pi_{k,j-1}[/tex]

    [tex]\pi_{0,j} = 1[/tex] for j = 0, 1, 2, 3, ......

    [tex]\pi_{i,0} = 0[/tex] for i = 1, 2, 3, ......

    which can be seen by conditioning on whether the next shot is successful or not. The next shot is successful with probability p, in which case he will need i-1 more consecutive successes to win, and will have j-1 chances left. The next shot is unsuccessful with probability 1-p, in which case he will need k consecutive successes to win, and will have j-1 shots left. If you can solve the recurrence, then [tex]\pi_{k,n}[/tex] is the probability you are looking for.

    The second problem sounds like a random walk of length n, and you are looking for the probability that it stays between -d and d. The random walk is called biased if p is not 1/2.

    BTW: if you don't need a general formula in terms of k, n, and p, but only need a way to compute the answer, there is a much easier way to do this using matrices. For the first problem, there are k+1 states: 0, 1, 2, ..., k. The man is in state i when the last i shots have been successful. If he gets to state k at some point during the series of n shots, he wins, and remains in state k for all later shots. Set up a matrix P where the element P_ij is the probability of going from state i to state j. For k = 3 the matrix would look like this

    P =

    q p 0 0
    q 0 p 0
    q 0 0 p
    0 0 0 1

    where q = 1-p

    Then you would first figure out the probability that the man does not reach state k during the n shots. If n = 10, for example, then you raise the matrix P to the 10th power and sum the first three elements of the first row of the result. This is the probability that, starting in state 0, the man is in state 0, 1, or 2 after 10 shots. Call this probability x. Then the probability that he does reach state 3 (that is, he does have a series of 3 consecutive successful shots) will be 1-x.
    Last edited: Mar 31, 2009
  9. Apr 1, 2009 #8


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    Hi Nikolaj! :smile:

    just woke up … :zzz:

    This is a very difficult question.

    Let's see how far you can get, before we have to jump in and help you.

    Try 1) first …

    you probably know that the main difficulty is to avoid counting things twice …

    so the trick is not to count the number of ways of getting k successes starting with the mth shot, but of getting one failure and then k successes starting with the (m-1)th shot :wink:
  10. Apr 1, 2009 #9
    Hi Adeimantus, I had given up - it was too difficult, but thanks for pointing this out, I will study it and try again to solve this. Thanks for your help, and your too tiny-tim. As I sad I will try again when I would have free time, and post it if have some difficulties.
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