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)?

Freddy Chaney 2022-09-25 Answered
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 < m < 20).
(a) What is the expected number of rolls (simplify)?
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Answers (2)

Micah Hobbs
Answered 2022-09-26 Author has 8 answers
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:
E X = E [ X A ] P ( A ) + E [ X A ] P ( A ).
Work this out and a linear equation arises allowing you to find EX.
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memLosycecyjz
Answered 2022-09-27 Author has 2 answers
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 ( N = 1 ) = 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 ( N = 2 ) = ( 1 3 20 ) 3 20 .
Perhaps, it is clear now that P ( N = k ) = ( 1 3 20 ) k 1 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?
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