# 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)?

Discrete Probability - geometric and uniform
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).
(a) What is the expected number of rolls (simplify)?
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Micah Hobbs
Step 1
Let X denote the number of rolls and let A denote the event that at first throw a value ≥m is obtained.
Step 2
Then:
$\mathbb{E}X=\mathbb{E}\left[X\mid A\right]P\left(A\right)+\mathbb{E}\left[X\mid {A}^{\complement }\right]P\left({A}^{\complement }\right)$.
Work this out and a linear equation arises allowing you to find EX.
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Step 1
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 $P\left(N=1\right)=\frac{3}{20}.$
Step 2
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 $P\left(N=2\right)=\left(1-\frac{3}{20}\right)\frac{3}{20}.$
Perhaps, it is clear now that $P\left(N=k\right)={\left(1-\frac{3}{20}\right)}^{k-1}\frac{3}{20}.$
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?