we roll a fair dice over and over again until one of the players wins. player A wins when in the end of an even sequence of 2 numbers he gets 4, i.e 24,44,64. player B wins when he gets 3 consequtive odd numbers.
a. what is the winning probability of player A? b. how many rounds we expect the game to last? c. how many rounds we expect the game to last, if we know that player 1 had won.
Thanks in advance.
Apart from the start and end positions we distinguish $3$ intermediate states of the game, namely $e$: last throw was even; $o$: last throw was odd; $oo$: last two throws were odd.
(a) We denote by $p$ the overall probability that $A$ wins, and by $p_e$, $p_o$, $p_{oo}$ the conditional probability that $A$ wins, given that the game is in the indicated state. We then have the following equations: $$\eqalign{ p&={1\over2} p_e+{1\over2} p_o \cr p_e&={1\over6}+{1\over3} p_e+{1\over2} p_o\cr p_o&={1\over2}p_e+{1\over2} p_{oo}\cr p_{oo}&={1\over2}p_e\ .\cr}$$ Solving this system gives $p={1\over2}$.
Question (b) can be handled similarly: We denote by $x$ the expected total number of rounds, and by $x_e$, $x_o$, $x_{oo}$ the expected number of additional rounds when the game is in the indicated state. We then have the following equations: $$\eqalign{ x&=1+{1\over2} x_e+{1\over2} x_o \cr x_e&=1+{1\over3} x_e+{1\over2} x_o\cr x_o&=1+{1\over2}x_e+{1\over2} x_{oo}\cr x_{oo}&=1+{1\over2}x_e\ .\cr}$$ Solving this system gives $x=7$.