Statistical Analysis   of  the number of Rounds until you have at least ONE    dollar

                     In a  Casino with a Certain Roulette



                              By Shalosh B. Ekhad



Once upon a time there was a casino with a strange roulette, where there were,

    4, possible outcomes as follows.



                                Regarding losing



                   You lose, 1, dollars with probability, 1/4

                   You lose, 2, dollars with probability, 1/4



                               Regarding winning



                   You win, 1, dollars with probability, 1/4

                   You win, 3, dollars with probability, 1/4

You start with a capital of 0 dollars, and have unlimited credit. Your goal \
    in life is to gain a POSITIVE amount, and

                    you quit as soon as you reach that goal.

                Note that the expected gain in ONE round is, 1/4

        Since it is positive, sooner or later you will reach your goal.

The probability generating function, let's call it f=f(t), for the random va\
    riable "duration of game"

in other words, the formal power series whose coefficient of, t^n is the pro\
    bability of ending after n rounds satisfies the

                         following algebraic equation



 6  10       6       5   9        6       5       4   8
t  f   + (7 t  - 12 t ) f  + (22 t  - 80 t  + 48 t ) f

             6        5        4       3   7
     + (102 t  - 260 t  + 368 t  - 64 t ) f

             6         5         4        3   6
     + (303 t  - 1156 t  + 1088 t  - 768 t ) f

             6         5         4         3        2   5
     + (642 t  - 2676 t  + 5088 t  - 2752 t  + 768 t ) f

             6         5          4          3         2   4
     + (843 t  - 5644 t  + 10432 t  - 10816 t  + 3840 t ) f

              6         5          4          3          2            3
     + (1230 t  - 5820 t  + 16176 t  - 18688 t  + 11264 t  - 3072 t) f

             6         5          4          3          2            2
     + (528 t  - 4988 t  + 13824 t  - 24000 t  + 20224 t  - 6144 t) f

             6         5          4          3          2
     + (227 t  - 2880 t  + 11008 t  - 18176 t  + 19200 t  - 13312 t + 4096) f

            6         5         4         3         2
     + 191 t  - 1060 t  + 3408 t  - 6656 t  + 6144 t  - 2048 t = 0



                             and in Maple notation



t^6*f^10+(7*t^6-12*t^5)*f^9+(22*t^6-80*t^5+48*t^4)*f^8+(102*t^6-260*t^5+368*t^4
-64*t^3)*f^7+(303*t^6-1156*t^5+1088*t^4-768*t^3)*f^6+(642*t^6-2676*t^5+5088*t^4
-2752*t^3+768*t^2)*f^5+(843*t^6-5644*t^5+10432*t^4-10816*t^3+3840*t^2)*f^4+(
1230*t^6-5820*t^5+16176*t^4-18688*t^3+11264*t^2-3072*t)*f^3+(528*t^6-4988*t^5+
13824*t^4-24000*t^3+20224*t^2-6144*t)*f^2+(227*t^6-2880*t^5+11008*t^4-18176*t^3
+19200*t^2-13312*t+4096)*f+191*t^6-1060*t^5+3408*t^4-6656*t^3+6144*t^2-2048*t =
0


                     459.09115960443116165, did not make it

          This ends this paper, since the parameter, 0.2, is too small