Pair of fair die are rolled (independently I hope) infinitely. Find probability sum of 5 appears before sum of 7.
2 approaches:
- $$P(\text{sum of 5 appears before sum of 7})$$
$$= P(\text{roll 1 is 5})$$
$$+ P(\text{roll 2 is 5, roll 1 is not 7})$$
$$+ P(\text{roll 3 is 5, roll 1,2 are not 7})$$
$$+ P(\text{roll 4 is 5, roll 1,2,3 are not 7})$$
$$+ \ldots$$
- $$P(\text{sum of 5 appears before sum of 7})$$
$$= P(\text{roll 1 is 5})$$
$$+ P(\text{roll 2 is 5, roll 1 is not 7}, \ \color{red}{\text{roll 1 is not 5}})$$
$$+ P(\text{roll 3 is 5, roll 1,2 are not 7}, \ \color{red}{\text{roll 1,2 are not 5}})$$
$$+ P(\text{roll 4 is 5, roll 1,2,3 are not 7}, \ \color{red}{\text{roll 1,2,3 are not 5}})$$
$$+ \ldots$$
Which if any is right?
Mathematically:
Let $n = 1,2,…$
Let $A_n$ be probability that sum of 5 appears on roll $n$
Let $B_n$ be probability that sum of 7 appears on roll $n$
Let $B_0^C = \Omega$
Observe that $A_n$ and $B_n$ are disjoint. Hence $A_n \subseteq B_n^C$
Approach 1 gives:
$$\sum_{n=1}^{\infty} P(A_n \cap \bigcap_{m=0}^{n} B_m^C)$$
$$\sum_{n=1}^{\infty} P(A_n \cap \bigcap_{m=0}^{\color{red}{n-1}} B_m^C)$$
$$ = \frac{4}{36} \sum_{n=1}^{\infty} (\frac{30}{36})^{n-1}$$
Approach 2 gives:
$$\sum_{n=1}^{\infty} P(A_n \cap \bigcap_{m=0}^{n} B_m^C \color{red}{\cap \bigcap_{m=0}^{n-1} A_m^C})$$
$$\sum_{n=1}^{\infty} P(A_n \cap \bigcap_{m=0}^{\color{red}{n-1}} B_m^C \color{red}{\cap \bigcap_{m=0}^{n-1} A_m^C})$$
$$ = \frac{4}{36} \sum_{n=1}^{\infty} (\frac{30}{36})^{n-1} \color{red}{(\frac{32}{36})^{n-1}}$$
What is the weakest independence assumption we need to make?
For approach 1 it seems that we need to assume independence of
$$A_n, B_1, B_2, …, B_{n-1}$$.
For approach 2 it seems that we need to assume independence of
$$A_1, A_2, …, A_n, B_1, B_2, …, B_{n-1}$$.
Is that right?
Best Answer
A simple approach would be to "split the world" into two events:
Since these are complementary events, the probability that either one of them will occur is $1$.
Now, the probability of a sum of $5$ is $\color\red{1/9}$, and the probability of a sum of $7$ is $\color\green{1/6}$.
Therefore: