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Two dice are rolled at a time, for many time until either A or B wins. A wins if we get two consecutive 7s and B wins if we get one 6 at any time. what is the probability of A winning the game??

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when you say "one 6", do you mean that one of the dice has a 6 on it, or that both dice add to 6 at any one time? –  Robert Mastragostino Jun 4 '12 at 13:03
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up vote 7 down vote accepted

Let $p$ be the probability that $A$ wins. Let $h$ be the probability that $A$ wins given that neither $A$ nor $B$ has yet won, and a $7$ has just been thrown, so that $A$ is "hot."

We break the calculation into cases, depending on what happens on the first throw. There are two ways that $A$ can (ultimately) win. Either (i) the first throw is a $7$ (probability $1/6$), making $A$ hot, in which case the probability of a win is $h$, or (ii) the first throw is something other than $7$ or $6$ (probability $25/36)$), in which case for all practical purposes the game starts again. In that case, the probability $A$ ultimately wins is $p$. We thus obtain the equation $$p=\frac{1}{6}h +\frac{25}{36}p.\tag{$1$}$$

If $A$ is hot, then $A$ wins if (i) the next throw is $7$, or (ii) the next throw is something other than a $6$ or $7$, but $A$ ultimately wins. In case (ii), the probability of an ultimate win for $A$ is $p$. Thus $$h=\frac{1}{6}+\frac{25}{36}p.\tag{$2$}$$

We have two linear equations in two unknowns, and it is not hard to solve for $p$. From $(1)$ we obtain $11p/6$. Substituting in $(2)$, we find that $p=6/41$.

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@ AndreNicolas can you please explain this part bit more clearly - "If A is hot, then A wins if (i) the next throw is 7, or (ii) the next throw is something other than a 6 or 7, but A ultimately wins. In case (ii), the probability of an ultimate win for A is p." –  Archie.bpgc Jun 6 '12 at 9:07
    
how you got the equation for h –  Archie.bpgc Jun 6 '12 at 9:08
    
I will assume you know the meaning of $h$. So given just got a $7$. How can $A$ win? Two ways. (i) On next throw, get $7$. Then game over, $A$ won. Probability of this is $1/6$; (ii) next throw, get neither $7$ nor $6$, but $A$ ultimately wins. Probability of neither $7$ nor $6$ is $25/36$. If get neither $7$ nor $6$, then for all practical purposes it's a new game, probability $A$ wins ultimately is $p$. So $h=1/6 +(25/36)p$. –  André Nicolas Jun 6 '12 at 11:19
    
what do you mean by "but A ultimately wins" in the (ii) step?? –  Archie.bpgc Jun 6 '12 at 11:52
    
Simply $A$ wins. Suppose $A$ has not won or lost, then there is a $7$. Then on next throw (i) If there is $7$, then $A$ wins or (ii) neither $7$ nor $6$ or (iii) $6$ (but then $A$ loses). Prob. of neither $7$ nor $6$ is $25/36$. If get neither $7$ nor $6$, then everything that has happened so far is irrelevant, the probability $A$ wins is just the same as it was at the beginning of the game, namely $p$. –  André Nicolas Jun 6 '12 at 12:00
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Let $p_A$ and $p_B$ the winning probabilities for $A$ and $B$, and $p_6$ and $p_7$ the probabilities to roll 6 and 7.

Now, by regarding the different possibilities for the first roll (and if one starts with 7, also the second roll), we find:

$$p_A=p_6 \cdot 0 + p_7 (p_6 \cdot 0 +p_7+(1-p_6-p_7)p_A) + (1-p_6-p_7)p_A.$$

This is a linear equation for $p_A$.

We have $p_6=\frac5{36}$, $p_7=\frac{6}{36}$, and therefore:

$$p_A=\frac{p_7^2}{p_7p_6+p_7^2+p_6} =\frac{6}{41}.$$

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Note that this is indeed just the comparison between getting a single 6 and getting two 7s, because the two events do not overlap. But my above approach will also work for more complicated scenarios. –  Phira Jun 4 '12 at 13:11
    
This is not quite correct. The second term $p_7(p_7+p_A)$ overcounts the event that the second roll is a 7. It needs to be $p_7(p_7+(1-p_7-p_6)p_A)$. –  Zander Jun 4 '12 at 13:23
    
@Zander Thank you. –  Phira Jun 4 '12 at 18:36
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