# Sum of dice Problem

1. Nov 28, 2006

### 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 ?

2. Nov 28, 2006

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.

3. Nov 28, 2006

### Abomb23

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

So lets 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)

4. Nov 28, 2006

### 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.

Last edited: Nov 28, 2006
5. Nov 28, 2006

### 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 lets 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

6. Nov 28, 2006

### 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

7. Nov 28, 2006

### Abomb23

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 genral formule / Function that will compute that

8. Nov 29, 2006

### HallsofIvy

Did the problem sheet really say "a dice"?

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

9. Dec 29, 2006

### 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.

Last edited: Dec 29, 2006
10. Jan 1, 2007

### balakrishnan_v

$$P(\text{getting n})=\frac{r_1^n+r_2^n+r_3^n+r_4^n+r_5^n}{7} +\frac{2}{7}$$

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

Last edited: Jan 1, 2007