Question:
In rolling a pair of a fair dice, what is the probability that a sum of $7 $is rolled before a sum of $8$ is rolled?
I've seen the answer from other's question that I can solve it like this:
Suppose A is the event that '$7$ is rolled'. $B$ is the event that '8 is rolled'.
$$P(A)=\frac{1}{6}$$
$$P(B)=\frac{5}{36}$$
According to total law of probability,
$P(A$ before $B)=P(A$ before$B|A$ happens first$)\cdot P(A$ happens first$)+P(A $before $B|B$ happens first$)\cdot P(B$ happens first$)+P(A$ before $B|A$ happens first$)\cdot P($neither A nor B happens first$)$
Suppose $P(A$ before $B)=r$
We can get an equation $$r=1\cdot\frac{1}{6}+0\cdot\frac{5}{36}+r\cdot(1-\frac{1}{6}-\frac{5}{36})$$
Therefore
$$r=\frac{6}{11}$$
But can I also solve this problem using geometric series?
Here's my solution:
Let $C$ be the event '$7$ is rolled at $n$-th trial' and $D$ be the event '$8$ is rolled after $n$-th trial'.
Hence
$P(A$ before $B)=\sum_{n=1}^{\infty}(\frac{5}{6})^{n-1}\cdot\frac{1}{6}\cdot(\frac{31}{36})^{n}=0.508$
So which of this two methods should I use to figure out the probability?
Best Answer
Just look at the first time that a seven or an eight is rolled and neglect everything else.
For that situation you must calculate the probability that a seven shows up.
So actually to be calculated is the probability: $$\mathsf P(\text{seven rolled}\mid\text{seven rolled or eight rolled})=\frac{\frac6{36}}{\frac6{36}+\frac5{36}}=\frac6{11}$$