A and B are playing a game, the outcome of which is determined by throwing 2 dice. If we get sum of 7 then A wins, if we get 2 equal digits then B wins, else we throw the dice over and over again, until someone wins.
What is the expected of number of games they have to play, until someone wins 2 consecutive games?
Well, the probability of A or B to win is 0.5. Let $X$ be the number of games A and B have to play, until someone wins 2 consecutive games.
I think $X$ obeys negative binomial distribution, when $X \sim \mathcal{NB}(0.5,2)$, then $$\mathbb{E}[X]=r/p=2/0.5=4.$$
But,I'm not sure..
Best Answer
I interpret this as: There is a chance of 1/6 that A wins a game, 1/6 that B wins a game, and 4/6 that none wins. drhab's answer is mostly right except the probabilities are not quite correct. So we have the equations
$~\mathsf EX = 1 + \tfrac 46\,\mathsf EX + \tfrac 26\,\mathsf EY~$ or $~\mathsf EX = 3 + \mathsf EY$
$~\mathsf EY = 1 + \tfrac 16(0)+ \tfrac 16\,\mathsf EY + \tfrac 46\,\mathsf EX~$ or $~5~\mathsf EY = 6 + 4\,\mathsf EX~$ or $~\mathsf EY = 1.2 + 0.8\,\mathsf EX$.
Substitute 2nd equation in the first:
$~\mathsf EX = 3 + 1.2 + 0.8\,\mathsf EX~$ or $~0.2\,\mathsf EX = 4.2~$ or $~\mathsf EX = 21~$.
$~\mathsf EX$ is the expected count of games to be played given nobody won the previous one.
$~\mathsf EY$ is the expected count of games to be played given that somebody won the prior one.