My professor suggested that we first let $E_n$ denote the event that a $5$ occurs on the $n^{\text{th}}$ roll and no $5$ or $7$ occurs on the first $n-1$ rolls, then find $P(E_n)$ and take the sum $\sum_{n=1}^{\infty}P(E_n)$. He calculates $P(E_n)$ like this:
- The number of ways in which the first $n-1$ rolls will not sum to $5$ or $7$ is $26^{n-1}$
- The number of ways in which the $n^{\text{th}}$ roll can sum to $5$ is $4$
- The number of ways you can roll $n$ pairs of dice is $36^n$
- Therefore, $P(E_n) = \Large\frac{(26^{n-1})(4)}{36^n}$
What I find difficult to understand is that this method seems to be answering a different question altogether, namely:
A pair of dice is rolled $n$ times. Find the probability that a $5$ occurs on the last roll, and neither a $5$ nor a $7$ is rolled on the first $n-1$ rolls.
I agree that my professor's method for finding $P(E_n)$ works for this altered question because the sample space is the set of all sequences of dice pairs of length $n$. But for the original problem, isn't the sample space the set of all sequences of dice pairs that end with $5$ or $7$ while the rest of the terms are neither $5$ nor $7$? If so, the sample space would be infinite, so we could not simply divide by $36^n$ like we did in step $4$. Where am I going wrong?
Best Answer
As A.J. has commented, a final step of summing to infinity is missing.
If you carry out that step, you will arrive at an answer of $0.4$
However, you can avoid such a summation by using the method below:
Since a sum of $7$ is effectively barred by stipulating that a sum of $5$ must occur first,
Let $P$ be the ultimate probabilty that we get a sum of $5$, then either we get $5$ or are back to start without rolling either $5$ or $7$
Thus$\;\Large P = \frac4{36} + \frac{26}{36}P \Longrightarrow P= 0.4$