# Probability- sum of n tosses of a fair die

## Homework Statement

You roll a fair die. Let Sn be the sum of the first n tosses. what is the probability of having Sn divisible by 5 ?

a) n is infinite

## The Attempt at a Solution

there are also cases that n is finite, but I believe if I know how the infinite case works, I will be able to figure out the finite case.
I managed to solve it by using the counting method for n=2 and n=3... but couldn't find a real solution with some pattern.

Would appreciate any help.

Thanks,
Ron.

a bump for those who didn't see the question :\

D H
Staff Emeritus
a bump for those who didn't see the question :\
You have the template in your original post but you did not follow it. You need to show some work and some of the relevant equations before you are going to get much help at this site.

"Bump" does not constitute showing some work.

You have the template in your original post but you did not follow it. You need to show some work and some of the relevant equations before you are going to get much help at this site.

"Bump" does not constitute showing some work.

Actually, this is not my first post here and I always provide any information I have.
I couldn't relate it to any topic so I couldn't provide any related equations...

sorry ...

uart

## Homework Statement

You roll a fair die. Let Sn be the sum of the first n tosses. what is the probability of having Sn divisible by 5 ?

a) n is infinite

## The Attempt at a Solution

there are also cases that n is finite, but I believe if I know how the infinite case works, I will be able to figure out the finite case.
I managed to solve it by using the counting method for n=2 and n=3... but couldn't find a real solution with some pattern.

Would appreciate any help.

Thanks,
Ron.

Hi Ron, did you get any further with this?

Yes it's difficult to find a "number pattern" as in a specific function of "n". I couldn't find one. Never the less it is still possible to guess the asymptotic behavior, and when you think about it the result is somewhat obvious.