A fair 20-sided die is rolled repeatedly, until a gambler decides to stop. The gambler receives the amount shown on the die when the gambler stops. The gambler decides in advance to roll the die until a value of m or greater is obtained, and then stop (where m is a fixed integer with 1 < m <20).
(a) What is the expected number of rolls (simplify)?
Let $N$ denote the number of rolls until the first $m$ or a greater number appears. $N$ may take values greater or equal than $1$.
For the sake of simplicity, let $m=18$.
The probability that $N=1$ equals the probability that the first roll gives either $18$ or $19$ or $20$, That is
$$P(N=1)=\frac3{20}.$$
The probability that $N=2$ equals the probability that the first roll is neither $18$ nor $19$ nor $20$ and the third one is. So
$$P(N=2)=\left(1-\frac3{20}\right)\frac3{20}.$$
Perhaps, it is clear now that
$$P(N=k)= \left(1-\frac3{20}\right)^{k-1}\frac3{20}.$$
There remain a few easy questions:
What kind of distribution is this?
How to calculate the expectation?
Finally, how to generalize the result from $18$ to an arbitrary number between $1$ and $20$?
Hint:
Let $X$ denote the number of rolls and let $A$ denote the event that at first throw a value $\geq m$ is obtained.
Then:$$\mathbb EX=\mathbb E[X\mid A]P(A)+\mathbb E[X\mid A^{\complement}]P(A^{\complement})$$ Work this out and a linear equation arises allowing you to find $\mathbb EX$.
