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Math Help - Game involving probability

  1. #1
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    Game involving probability

    Here is a nice probability problem I saw in class about a week ago. It involves playing a game.

    Consider a game where the player has initially d dollars. The player tosses an unfair die (or flips an unfair coin, or whatever) so that the probability of winning is p. If he wins he earns a dollar, if he loses he loses a dollar. This is played until one of two situations occurs: either he reachers a certain amount n or he is bankrupt i.e. he loses all his money. Given d,p,n find a formula that will give the probability of succes (meaning he reaches n).

    According to our textbook this is a classic going all the way back to Bernoulli (I just forgot which one it is ).
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    Senior Member DivideBy0's Avatar
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    Ok I'll have a crack at it, it might be wrong though

    Let random numbers x-y=n-d. It doesn't matter how many trials there are, but as long as the difference between the number of times a success (x) is flipped and a failure (y) is flipped is the same, then we will have increased by x-y = n-d to reach n.

    Then there are \ _{x+y}C_x ways to arrange the probabilities

    So

    Pr(Success)={{x+y} \choose x} p^x (1-p)^y

    Of course it requires to know 3 things:
    1. N
    2. D
    3. The number of trials.
    It also looks quite similar to hypergeometric distribution

    thanks for posting this problem :P
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by DivideBy0 View Post
    Ok I'll have a crack at it, it might be wrong though

    Let random numbers x-y=n-d. It doesn't matter how many trials there are, but as long as the difference between the number of times a success (x) is flipped and a failure (y) is flipped is the same, then we will have increased by x-y = n-d to reach n.

    Then there are \ _{x+y}C_x ways to arrange the probabilities

    So

    Pr(Success)={{x+y} \choose x} p^x (1-p)^y

    Of course it requires to know 3 things:
    1. N
    2. D
    3. The number of trials.
    It also looks quite similar to hypergeometric distribution

    thanks for posting this problem :P
    Except if y>d, you may have gone bust before winning n dollars.

    In fact if you don't worry about your fortune becoming negative you will
    be n dollars up at some time with probability 1 as long as p>0.

    RonL
    Last edited by CaptainBlack; October 17th 2007 at 06:23 AM.
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  4. #4
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Except if y>n, you may have gone bust before winning d dollars.
    I have an example here... i dunno, things are getting pretty weird...

    Say x + y = 10
    Take n = 2, d = 1

    Also x - y = n - d

    So x - y = 1

    Then x=11/2, y = 9/2, so now you have to do non-integer factorials! Is this always the case with y>n? I'm getting pretty bogged down with the maths here.
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    One way of analysing this is by stochastic matrices. You have a system with n states 0,1,2,...,n, in which 0 (the losing state) and n (the winning state) are absorbing states (once you get there, you stay there). The matrix of transition probabilities is

     \begin{array}{c|ccccccc}&n&0&1&2&3&\ldots&n-1\\ \hline<br />
n&1&0&0&0&0&\ldots&p\\ 0&0&1&q&0&0&\ldots&0\\ 1&0&0&0&q&0&\ldots&0\\ 2&0&0&p&0&q&\ldots&0\\  \vdots&\vdots&\vdots&\vdots&\ddots&\ddots&\ddots&\  vdots\\n-2&0&0&0&\ldots&\ldots&0&q\\ n-1&0&0&0&\ldots&\ldots&p&0\end{array},

    where q = 1-p. This matrix can be partitioned into blocks as follows:

    \begin{bmatrix}I&S\\0&R\end{bmatrix}, where I is the 2x2 identity matrix, S is the 2x(n-2) matrix \begin{bmatrix}0&0&\ldots&0&p\\ q&0&\ldots&0&0\end{bmatrix}, R is the (n-2)x(n-2) matrix

    \begin{bmatrix}0&q&0&\ldots&0\\ p&0&q&\ldots&0\\ \vdots&\ddots&\ddots&\ddots&\vdots\\0&\ldots&\ldot  s&0&q\\ 0&\ldots&0&p&0\end{bmatrix} and 0 is an (n-2)x2 matrix of zeros.

    The theory of Markov processes tells you that the probabilities of winning and losing, given that you started with d dollars, are given by the two entries in the d'th column of the matrix S(I-R)^{-1}.

    The only snag is that the inverse matrix is not easy to compute, even for a relatively small value of n (unless maybe you have access to a good computer algebra package, which I don't).
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    This problem is from my book is on chapter on conditional probability. The classical solution does not involve the use of Markov Chains or Stotaic Processes*.






    *)In fact, I have no idea what these things even are. I definitely heard of them but never used them.
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  7. #7
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    Hint: Consider the following probabilities P_i = \mbox{ probability of winning with }i \mbox{ dollars }. So P_0 = 0 and P_n =1 .
    (So you want to find P_d).
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    In case you are interested here is Bernoulli's elegant solution.

    Let P_k ( 0\leq k \leq n) be the probability that the player will win the game with k dollars. So P_0=0 because at zero dollars the game is over and he will not win. And P_n=1 because he already reached his desired amount.

    If the player gets wins on coin flip then his probability of winning is now P_{k+1}. If the player loses on a coin flip his probability is now P_{k-1}. Since the probability of getting a win on a die throw is p it and the probability of losing is (1-p) is means: P_k = pP_{k+1} + (1-p)P_{k-1}.

    This means, pP_k + (1-p)P_k = pP_{k+1} + (1-p)P_{k-1} thus P_{k+1} - P_k = \frac{1-p}{p} (P_k - P_{k-1}).

    Now we will list all these values from 1 to n:
    P_2 - P_1 = \frac{1-p}{p} (P_1 - P_0) = \frac{1-p}{p}P_1
    P_3 - P_2 = \frac{1-p}{p} (P_2 - P_1) = \frac{(1-p)^2}{q^2}(P_1 - P_0) = \frac{(1-p)^2}{p^2}P_1
    ...
    P_n - P_{n-1} = \frac{1-p}{p} (P_{n-1}-P_{n-2}) = \frac{(1-p)^{n-1}}{p^n} (P_1 - P_0) = \frac{(1-p)^{n-1}}{p^{n-1}}P_1

    We want to know P_d, that is what we are trying to find.

    So add the frist d-1 equations:
    P_d - P_1 = P_1 \sum_{j=1}^{d-1} \frac{(1-p)^j}{p^j}
    This is a geometric series.
    So the solution depends whether or not (1-p)/p=1 (i.e. p=1/2) or not.

    I am not going to go further because it gets a little messy but the algebra is straightforward.

    And finally to complete this problem we need to find P_1 be that is easy to find if we sum all these equations and use the fact that P_n = 1.
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