Probability of Sum of Dice Rolls for Positive Integer "m

[Abomb23]

I ran into this next problem and I am having hard time getting a final answer for it for every m :


A dice is rolled and summed over and over , What is the probability That the sum will be "m" , "m" is a positive integer


My problem starts after the numer 6 , as I start to loose options , as I can't use 1 dice for higher numbers then 7 , same goes for 2 dices after 13 and so on.

Any1 got a soultion ?

 

[radou]

It's about a pair of dices, right? Try to write down all the sums that can occur, and for each of these sums, write down the number of combinations which give these sums.

 

[Abomb23]

Nope , It can be endless number of rolls but using same dice all over again.

So let's say that p(1) = 1/6 as there is only 1 way to get it , by getting 1 in the first roll.

then
P(2) = 1/6 + 1/36 as u got 2 options now , either u get 2 in first roll or 2 times one with 2 rolls.

and so on , but again after 7 it gets harder as u can't get a sum of 7+ with 1 dice , so I'm trying to find the overall function to get to ANY number m (again positive integer)

 

[mattmns]

Do you have the original wording of the problem?

In my opinion if you are rolling one die an unlimited number of times, and summing the values after each roll, then p(m) = 0 for all m.

There is no way to get m = 1 since once you roll twice you are already past 1, same for 2, 3, 4, etc.

edit... I just re-read your last post, and you say that you can't get a sum of 7+ with 1 die. So are we just rolling the die an unlimited number of times, and looking at the probability that the last roll is some number? The exercise seems to be very vague, could you please elaborate a little.

 

[Abomb23]

I copied the Question from the sheet I'm trying to solve so That's the way i understood what to do :

for any number "m" I need to find the the entire field of possible answer without surpesing it (thus p(m)!=0) , so let's say I want to get to m=3 .

then

p(3) = 1*(1/6) + 2*(1/6^2) + 1*(1/6_^3) = something

so in this case there are a total of 4 "true" final outcomes I can get that will give me a sum of 3 :

1) I will roll the die once and get a 3 > 1/6
2) I will roll the die 1st time and get a 2 , then roll again and get a 1 > 1/36
3) i will roll the die 1st and get a 1 , then roll again and get a 2 > 1/36
4) I will roll the die 1st and get a 1 , roll again and get a 1 , roll again and get a 1 > 1/216

so in reality I am not really rolling infinite rolls, as in "worse" case I will have to roll the die m times to get to m (getting 1 in each roll).

The problem again is when passing 6n+1 numbers (n=1,2...) like 7 and 13 etc... as I'm starting to lose options as I can't get a sum of 7 with 1 die , sam goes to 13 with 2 dices.

Hope that clears it up a bit :D

 

[CRGreathouse]

Let each p_i be chosen with equal probability from {1, 2, 3, 4, 5, 6}, and let S_n=p_1+p_2+\cdots+p_n. P_n is then the probability that S_i=n for some i. The question is: What is P_n, and how is it found, for n > 6?

P_1 = 1 = 1/6
P_2 = 1/6 + 1/36
P_3 = 1/6 + 2/36 + 1/216

 

[Abomb23]

Let each p_i be chosen with equal probability from {1, 2, 3, 4, 5, 6}, and let S_n=p_1+p_2+\cdots+p_n. P_n is then the probability that S_i=n for some i. The question is: What is P_n, and how is it found, for n > 6?

P_1 = 1 = 1/6
P_2 = 1/6 + 1/36
P_3 = 1/6 + 2/36 + 1/216

Yes , this is what i meant (or I think the question meant).

I already know (from working with some numbers) the Pn = 0.28 = 2/7 (more or less)

But the problem is that I need to make a general formule / Function that will compute that

 

[HallsofIvy]

Did the problem sheet really say "a dice"?

Dice is already plural. It is "a die" and "several dice".

 

[ssd]

What I feel is, you want the probability of getting a sum of m after n rolls of a six faced unbiased die, n <= m <= 6n. Obviously the total number of possible situations arising on the top face of the die is = 6^n. Let N= the number of favourable possibilities such that the sum is m.
Consider (x + x^2 + x^3 +...+ x^6)^n. Now the coefficient of x^m in the expression gives the number of ways in which a sum of m can be obtained. Therefore N= coeff. of x^m in {x(x^6-1)/(x-1)}^n = coeff. of x^(m-n) in {(x^6-1)/(x-1)}^n.
Then N/6^n is the required probability.

 

[balakrishnan_v]

The answer is
P(\text{getting n})=\frac{r_1^n+r_2^n+r_3^n+r_4^n+r_5^n}{7} +\frac{2}{7}where r_1,r_2,r_3,r_4,r_5 are the roots of the equation
6x^5+5x^4+4x^3+3x^2+2x+1=0

The roots are:
r_1 = 0.29419455636014125+0.66836709744330092i
r_2=0.29419455636014125000+-0.66836709744330092000i
r_3=-0.67033204760309673000+0.00000000000000000000i
r_4=-0.37569519922525918000+0.57017516101141252000i
r_5=-0.37569519922525918000+-0.57017516101141252000i
eg
P(getting 1)=1/6
P(getting 2)=7/36
P(getting 3)=49/216
P(getting 4)=343/1296
P(getting 5)=2401/7776
P(getting 6)=16807/46656
P(getting 7)=70993/279936
P(getting 8)=450295/1679616
P(getting 9)=2825473/10077696
P(getting 10)=17492167/60466176
P(getting 11)=106442161/362797056
P(getting 12)=633074071/2176782336
P(getting 13)=3647371105/13060694016
P(getting 14)=22219348327/78364164096
P(getting 15)=134526474769/470184984576
P(getting 16)=809860055095/2821109907456
P(getting 17)=4852905842113/16926659444736
P(getting 18)=29004175431175/101559956668416
P(getting 19)=173492524161649/609359740010496
P(getting 20)=1044275922856663/3656158440062976

and it approaches 2/7 as n->infinity