The algebraic equation satisfied by the Probability Generating Function of L\

    ife in a Casion where the stakes given by the Probability distribution,

    [[2, 1/10], [1, 1/10], [0, 1/10], [-1, 1/5], [-2, 1/2]]



                              By Shalosh B. Ekhad





Theorem:  Let a(n) be probability that you exit after EXACTLY n rounds in a \
    casino where in each turn

               The probability that you win, 2, dollars  is, 1/10

               The probability that you win, 1, dollars  is, 1/10

             The probability that you neither win nor lose is, 1/10

               The probability that you lose, 1, dollars  is, 1/5

               The probability that you lose, 2, dollars  is, 1/2



Let X(t) be the ordinary generating function of that sequence, in other word\
    s



                                   infinity
                                    -----
                                     \            n
                            X(t) =    )     a(n) t
                                     /
                                    -----
                                    n = 0



                    X(t) satisifies the algebraic equation



        3     6           3          2      5           3         2      4
1/1000 t  X(t)  + (1/250 t  - 1/100 t ) X(t)  + (1/100 t  - 1/25 t ) X(t)

       / 71   3          2        \     3   /61   3   23  2         \     2
     + |---- t  - 3/100 t  + 1/5 t| X(t)  + |--- t  - -- t  + 3/10 t| X(t)
       \1000                      /         \500      25            /

                                                 3       2
       /     477   3   43  2        \        63 t    57 t    7 t
     + |-1 + ---- t  - -- t  + 9/5 t| X(t) + ----- - ----- + --- = 0
       \     1000      50           /         200     50     10



                             and in Maple notation



1/1000*t^3*X(t)^6+(1/250*t^3-1/100*t^2)*X(t)^5+(1/100*t^3-1/25*t^2)*X(t)^4+(71/
1000*t^3-3/100*t^2+1/5*t)*X(t)^3+(61/500*t^3-23/25*t^2+3/10*t)*X(t)^2+(-1+477/
1000*t^3-43/50*t^2+9/5*t)*X(t)+63/200*t^3-57/50*t^2+7/10*t = 0


                  Or more usefully (for computing many terms)



               3     6           3          2      5
X(t) = 1/1000 t  X(t)  + (1/250 t  - 1/100 t ) X(t)

               3         2      4   / 71   3          2        \     3
     + (1/100 t  - 1/25 t ) X(t)  + |---- t  - 3/100 t  + 1/5 t| X(t)
                                    \1000                      /

                                                                              3
       /61   3   23  2         \     2   /477   3   43  2        \        63 t
     + |--- t  - -- t  + 3/10 t| X(t)  + |---- t  - -- t  + 9/5 t| X(t) + -----
       \500      25            /         \1000      50           /         200

           2
       57 t    7 t
     - ----- + ---
        50     10



                             and in Maple notation



X(t) = 1/1000*t^3*X(t)^6+(1/250*t^3-1/100*t^2)*X(t)^5+(1/100*t^3-1/25*t^2)*X(t)
^4+(71/1000*t^3-3/100*t^2+1/5*t)*X(t)^3+(61/500*t^3-23/25*t^2+3/10*t)*X(t)^2+(
477/1000*t^3-43/50*t^2+9/5*t)*X(t)+63/200*t^3-57/50*t^2+7/10*t


    The probability that you got broke  before, 1000, rounds is, 1.000000000



                        assuming that this is the case

The Conditional  expected number of rounds  in the casino, until being in th\

    e red is, 1.817279933



    The Conditonal standard-deviation of life in the casino is, 1.901530941

               The condintioanl scaled, 3, moment is, 4.443585189

               The condintioanl scaled, 4, moment is, 34.67914368

                          This took, 49.010, seconds.