Results 1 to 8 of 8

Math Help - Game involving probability

  1. #1
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10

    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 ).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432
    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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    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 07:23 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member DivideBy0's Avatar
    Joined
    Mar 2007
    From
    Melbourne, Australia
    Posts
    432
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    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).
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: January 21st 2011, 12:47 PM
  2. Probability Game
    Posted in the Statistics Forum
    Replies: 0
    Last Post: June 10th 2008, 01:48 PM
  3. Some probability in the game Big 2
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: November 15th 2007, 11:44 PM
  4. probability game
    Posted in the Statistics Forum
    Replies: 2
    Last Post: January 5th 2007, 07:14 PM
  5. Please!!help!! probability game
    Posted in the Statistics Forum
    Replies: 3
    Last Post: January 5th 2007, 11:02 AM

Search Tags


/mathhelpforum @mathhelpforum