# Probability of Sum of Dice Rolls for Positive Integer "m

• Abomb23
In summary, the problem involves finding the probability of getting a sum of "m" after "n" rolls of a six-sided die, where "n" is less than or equal to "m" and both are positive integers. The formula for this probability is P(n) = (r1^n + r2^n + r3^n + r4^n + r5^n)/7 + 2/7, where r1, r2, r3, r4, r5 are the roots of the equation 6x^5 + 5x^4 + 4x^3 + 3x^2 + 2x + 1 = 0. As "n" approaches infinity, the probability
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 ?

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.

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)

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.

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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

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

CRGreathouse said:
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

Did the problem sheet really say "a dice"?

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

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.

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

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## What is the probability of getting a sum of "m" when rolling two dice?

The probability of getting a sum of "m" when rolling two dice is (6-m+1)/36. This is because there are a total of 36 possible outcomes when rolling two dice, and out of those, (6-m+1) outcomes will result in a sum of "m".

## What is the most common sum when rolling two dice?

The most common sum when rolling two dice is 7. This is because there are more ways to get a sum of 7 (6 possible combinations) compared to any other sum.

## What is the probability of getting a sum of "m" when rolling three dice?

The probability of getting a sum of "m" when rolling three dice is (21-m)/216. This is because there are a total of 216 possible outcomes when rolling three dice, and out of those, (21-m) outcomes will result in a sum of "m".

## What is the probability of getting a sum of "m" when rolling four dice?

The probability of getting a sum of "m" when rolling four dice is (56-m)/1296. This is because there are a total of 1296 possible outcomes when rolling four dice, and out of those, (56-m) outcomes will result in a sum of "m".

## How does the probability of getting a sum of "m" change as the number of dice increases?

As the number of dice increases, the probability of getting a sum of "m" decreases. This is because with more dice, there are more possible outcomes and therefore, the chances of getting a specific sum becomes smaller.

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