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Random walk?

  1. Jul 11, 2012 #1
    In a gambling game , you can win 1$ in each round with probability 0.6 or lose 2$ in probability 0.4. suppose you start with 100$.
    find the probability that after 10 rounds you have between 93 to 107 dollars.

    i am not sure how to start the solution,

  2. jcsd
  3. Jul 11, 2012 #2


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    Well, one obvious, though not necessarily efficient, method is to determine all possible "walks" that end up between 93 and 107. A bit more efficient method might be to enumerate walks that put you outside that interval since there appear to be fewer of those. The probability you are inside the interval is, of course, 1 minus the probability you are outside.

    Note that the order in which you win or lose is irrelevant.
  4. Jul 11, 2012 #3

    Ray Vickson

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    It might be a bit easier to simplify the "randomness" as follows: assume that in each play you always start by losing $2; then you gain an additional $0 with probability 0.4 or an additional $3 with probability 0.6. In 10 rounds you will have [itex]D = 100 - 20 + 3S[/itex] dollars, where [itex]S = \sum_{i=1}^{10} X_i,[/itex] and the [itex]X_i[/itex] are iid random variables with distribution P{X=0} = 4/10, P{X=1} = 6/10. What limits on S give D between 93 and 107?

    Last edited: Jul 11, 2012
  5. Jul 11, 2012 #4


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    No, I count fewer possibilities that put the expected gain between -7 and +7 inclusive. That's between 1 and 5 losses inclusive, which is 5 possibilities.

    The complementary set would have 6 possibilities (0 losses and 6 to 10 inclusive). That's more work.
  6. Jul 11, 2012 #5


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    First work out how many losses will yield an expected gain between -7 and +7 inclusive. This can be done by solving a simple pair of simultaneous equations (or one equation and one inequality).

    Then treat it as a Binomial probability problem and just sum the relevant probabilities.
  7. Jul 13, 2012 #6
    so S is a binomal variable with n=10 , p=0.6 and i left to find P([itex]5\leq S \leq 9[/itex]) ?
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