Probability In Rolling Two dices A dice is rolled again and again till a total of 5 or 7 is obtained. What will be the chance that a total of 5 comes before a total of 7 ?
 A: I assume that you are throwing $2$ dice again and again, until the
sum of them is $5$ or $7$. Let $X$ be the sum that occurs at your
first throwing and let $E$ denote the event that $5$ becomes before
$7$. Then:
$P\left(E\right)=P\left(E\mid X=5\right)P\left(X=5\right)+P\left(E\mid X=7\right)P\left(X=7\right)+P\left(E\mid X\notin\left\{ 5,7\right\} \right)P\left(X\notin\left\{ 5,7\right\} \right)$
$X\notin\left\{ 5,7\right\} $ means that the first result is not
relevant, hence $P\left(E\mid X\notin\left\{ 5,7\right\} \right)=P\left(E\right)$
and:
$P\left(E\right)=1\times\frac{4}{36}+0\times\frac{6}{36}+P\left(E\right)\times\frac{26}{36}=\frac{1}{9}+P\left(E\right)\times\frac{13}{18}$
This leads to $P\left(E\right)=\frac{2}{5}$.
A: Probability of 5 = 4/36 = 1/9    Probability of 7 = 6/36 = 1/6
Probability of no result = 1 - (1/9 + 1/6)
                     = 13/18

Let p = the probability that we will roll a total of 5 before we roll 
a total of 7.
If we roll the two dice, we can win with probability 1/9, lose with 
probability 1/6, or return to start with probability 13/18,
and so  p = 1.(1/9) + 0.(1/6) + p(13/18)
 p[1 - 13/18] = 1/9

      p(5/18) = 1/9

            p = 2/5

A: P(5) = 1/9, P(7) = 1/6
P(5 nor 7) = 26/36 = 13/18
Required probability 
= 1/9 + (26/36)(1/9) + (26/36)^2 + (26/36)^3 + ..... (infinity)
= 2/5
= sum of probabilities that we get a sum of 5 in first throw, second throw etc.
