Expected number of rolls

ppouyan

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[h=1]A fair m-sided dice is rolled and summed until the sum is at least N. What is the expected number of rolls? In other words what is the number of rolls if we roll a m-sided dice and the sum of rolls become at least N.



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Just to clarify: The die is numbered from 1 through m? Thank you! ;)
 
A fair m-sided dice is rolled and summed until the sum is at least N. What is the expected number of rolls? In other words what is the number of rolls if we roll a m-sided dice and the sum of rolls become at least N.
I have some comments on this question.
First have you thought about which numbers \(\displaystyle M\) are possible for dice. To see what I mean read this.

If you want to use all possible natural numbers, I suggest using a model based upon the lottery forced-air machine but with replacement. Suppose you use \(\displaystyle M=10\), ten balls numbered 1 to 10 and you let \(\displaystyle N=8\). Then on the very first trial the \(\displaystyle 8,9,10-ball\) "is spit"; we done. To be done on the second ball, we must have \(\displaystyle (1,7),(1,8),(1,9),\text{ or }(1,10)\). Then we must count the ways to use 2: \(\displaystyle (2,6),(2,7),(2,8),(2,9)\text{ or }(2,10)\).
Then keep it up with 3, 4, 5,6, or 7 in the first place.
I hope that I have made my point about the difficulty that you raise with this question.
This is a very tedious counting problem.
 
A fair m-sided dice is rolled and summed until the sum is at least N. What is the expected number of rolls? In other words what is the number of rolls if we roll a m-sided dice and the sum of rolls become at least N.



What is the expected value of a single roll? On a six sided die it would be 3.5. So, to get 7 (N=7) you would expect to roll the die twice; for N=32 (or 33 or 34 or 35), you would expect to roll the die 10 times, etc.
 
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