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.

  • $\begingroup$ Is player 1 the same person as player A? Also, what have you tried? You tagged this as markov-chains, so you probably know about states and transition matrices. What states can this game be in? What's the transition probability between them? $\endgroup$ – Arthur Feb 21 '18 at 10:09
  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. $\endgroup$ – José Carlos Santos Feb 21 '18 at 10:10
  • $\begingroup$ As you are new to the site: people here tend to not respond well (or at all) to questions like this that look like routine homework problems and which show no effort. What have you tried? Where are you getting stuck? $\endgroup$ – lulu Feb 21 '18 at 10:29
  • $\begingroup$ sorry about that, and yes - I'm new to this site. player A = player 1. $\endgroup$ – Don Juan Feb 21 '18 at 10:48
  • $\begingroup$ i've tried a conditional probability, to answer part a. defined A as the chance to ger 6, B to get 2 or 4 and C to get odd number. then i ended up with p(A/A) + p(A/B) = 1/3. obviously I'm missing something here. $\endgroup$ – Don Juan Feb 21 '18 at 10:54

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 $x=7$.

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  • $\begingroup$ thank you very much! why did you give an equal chance to win(p) for going through even or odd steps? $\endgroup$ – Don Juan Feb 21 '18 at 21:37
  • $\begingroup$ The first roll brings us to $e$ or $o$ with equal probability. $\endgroup$ – Christian Blatter Feb 22 '18 at 8:18

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