Probability- sum of n tosses of a fair die

In summary: If you multiply both sides by 5/6 you get Sn = 5*6*(n+1) which simplifies to Sn = 25. So the probability of a die with a sum of 1 through 5 being divisible by 25 is about 1 in 6.
  • #1
Roni1985
201
0

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


Homework Equations





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.
 
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  • #2
a bump for those who didn't see the question :\
 
  • #3
Roni1985 said:
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.
 
  • #4
D H said:
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 ...
 
  • #5
Roni1985 said:

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


Homework Equations





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.
 

1. What is the probability of getting a specific number when rolling a fair die n times?

The probability of getting a specific number when rolling a fair die is 1/6 for each roll. Therefore, the probability of getting that number n times in a row is (1/6)^n or 1/6^n.

2. What is the probability of getting a certain sum when rolling a die n times?

The probability of getting a certain sum when rolling a die n times can be calculated by dividing the number of ways to get that sum by the total number of possible outcomes. For example, the probability of getting a sum of 7 when rolling a die twice is 6/36 or 1/6.

3. How does the number of rolls affect the probability of getting a specific sum?

The more rolls you have, the higher the probability of getting a specific sum. This is because the more rolls you have, the more chances you have to get that sum. However, the probability of getting a specific sum does not increase proportionally with the number of rolls. It is still dependent on the total number of possible outcomes.

4. What is the expected value when rolling a die n times?

The expected value when rolling a die n times is the sum of all possible outcomes divided by the total number of possible outcomes. For a fair die, the expected value would be (1+2+3+4+5+6)/6 or 3.5. This means that on average, we can expect to get a sum of 3.5 when rolling a die once.

5. How does the probability change when using multiple dice instead of one?

The probability changes when using multiple dice instead of one because there are more possible outcomes. For example, when rolling two dice, the probability of getting a sum of 7 increases to 1/6, compared to 1/11 when rolling just one die. This is because there are more combinations of numbers that can add up to 7 when using two dice.

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