I was asked the following question on a practice test:
If two dice are rolled until their sum is seven, or they have been rolled twice, let the random variable
Nbe equal to the number of rolls. Find the expectation ofN.a)$\frac{7}{2}$ b)$\frac{5}{12}$ c)$\frac{11}{6}$ d)$\frac{5}{6}$ e)$\frac{5}{36}$
Right away you can kind of tell that c must be the solution because intuitively the expectation should be close to but less than $2$, but I did the calculation anyway.
I thought about it this way. The probability that the number of rolls is $1$ is equal to the probability that the sum of the dice is equal to $7$ on the first roll, which is $\frac{7}{36}$. Then, since we can have no more than $2$ rolls, the probability that the number of rolls is $2$ must be $\frac{29}{36}$. In other words,
$$P(N=1)=\frac{7}{36}$$ $$P(N=2)=\frac{29}{36}$$
So then the expectation of $N$ is
$$ E[N]=1\times\frac{7}{36}+2\times\frac{29}{36} =\frac{65}{36} $$
This is freakishly close to the intuitive answer, c, but not quite right. Did I make a mistake, or did the test writer just forget to say they just wanted to the closest solution?
You are right in everything except that there are only 6 ways to roll a 7.