Probability of Sum of Dice Rolls 10-15 Inclusive

[theloathedone]

Homework Statement

There is a 5-sided fair dice with faces showing 1, 2, 3, 4, 5 dots. The dice is rolled four times. Find the probability of getting a total score between 10 and 15 (inclusive).

Homework Equations

If X is the score from one roll, then it follows a discrete uniform distribution with parameter 5.

The Attempt at a Solution

I'm really clueless as to how to start. I can only think of writing out all the possible sums and finding the probability. But that seems rather tedious, is there a better way?

 

[Ray Vickson]

Homework Statement

There is a 5-sided fair dice

"dice" = plural
"die" = singular
So, you have one die.

with faces showing 1, 2, 3, 4, 5 dots. The dice is rolled four times. Find the probability of getting a total score between 10 and 15 (inclusive).

Homework Equations

If X is the score from one roll, then it follows a discrete uniform distribution with parameter 5.

The Attempt at a Solution

I'm really clueless as to how to start. I can only think of writing out all the possible sums and finding the probability. But that seems rather tedious, is there a better way?

Can you at least show your calculations in the easiest cases, with total = 4 and total = 5? (What I mean is: when you say you can write it out---well do just that: write it out.) That would prove that you at least know what to do using a brute-force method. After that, I would be willing to show you some tricks (actually, standard methods) that make the computations for total = 10, 11, 12, ... much, much easier.

R.G. Vickson

 

[theloathedone]

Ok, what I did was to draw out two tables. The first one is here:

1st Dice 1 2 3 4 5
2nd
1 2 3 4 5 6
2 3 4 5 6 7
3 4 5 6 7 8
4 5 6 7 8 9
5 6 7 8 9 10

If Y = X₁ + X₂
So P(Y = 2) = 1/25 = P(Y=10), P(Y=3) = 2/25 = P(Y = 9), P(Y=4) = 3/25 = P(Y=8), P(Y=5) = 4/25 = P(Y=7), P(Y=6) = 5/25.

Then I drew out another table, since the total T = Y₁ + Y₂, where the row and column headers are from 2 to 10, and I wrote the probability next to each of them. Then I found all the possible sums and added it all up one by one. The answer I got was 433/625 = 0.6928

But that's really tedious and I hope to learn a better way :)

 

[Ray Vickson]

Ok, what I did was to draw out two tables. The first one is here:

1st Dice 1 2 3 4 5
2nd
1 2 3 4 5 6
2 3 4 5 6 7
3 4 5 6 7 8
4 5 6 7 8 9
5 6 7 8 9 10

"Die", not "dice"!

If Y = X₁ + X₂
So P(Y = 2) = 1/25 = P(Y=10), P(Y=3) = 2/25 = P(Y = 9), P(Y=4) = 3/25 = P(Y=8), P(Y=5) = 4/25 = P(Y=7), P(Y=6) = 5/25.

Then I drew out another table, since the total T = Y₁ + Y₂, where the row and column headers are from 2 to 10, and I wrote the probability next to each of them. Then I found all the possible sums and added it all up one by one. The answer I got was 433/625 = 0.6928

But that's really tedious and I hope to learn a better way :)

OK, but *all* methods are a bit tedious in this type of problem. One more-or-less standard method for getting the sum of k independent, identically discrete random variables, each with distribution P{X=0} = p0, P{X=1} = p1, ..., P{X=r} = pr, with p0, p1, ..., pr >= 0 and sum{p_j, j=0..r} = 1, is to use the so-called *moment-generating function*, q(z) = p0 + p1*z + p2*z^2 + ... + pr*z^r. This is a polynomial in z of degree r. Now, it is fairly easy to show that the generating function of a sum of independent random variables is the product of their generating functions, so the generating function for S = X1 + X2 + X3 + X4 is Q(z) = q(z)^4. In your case, q(z) = (1/5)f(z), where f(z) = z + z^2 + z^3 + z^4 + z^5, so Q(z) = (1/5^4)(z + z^2 + z^3 + z^4 + z^5)^4. If we expand this out, the coefficient of z^i is the probability that S=i.

This is still tedious, but we can approach it recursively: we have a product f(z)*[f(z)*f(z)*f(z)]. Each factor contributes to z^i. In fact, if we have z^j1 from factor 1, z^j2 from factor 2, etc., the total is z^(j1+j2+j3+j4), so we need j1+j2+j3+j4=i. The coefficient of z^i is the total number of combinations {j1,j2,j3,j4} that sum to i; call it N4. How can we get N4? Well, if the first factor contributes z^j the remaining three must contribute i-j, and the number of these is N3[i-j]. Each j makes such a contribution, so the total is N4 = sum{N3[i-j], i=1..5} for i >= 8, while N4 = sum{N3[i-j], j=1..min(5,i-3)} for i = 4, 5,6,7. So, if we know N3[k] for k=3,...,15 we can get all the N4.

Similarly, N3 = sum{N2[i-j],j=1..min(5,i-2)} for i = 3, ..., 15, etc. These are most straightforward if we just let N1 = 0 for i <= 0 and for i >= 6. Then N2 = sum{N1[i-j] ,=1..5} holds for all i. Then we let N2=0 for i <= 1 and for i >= 11. This let's us write N3 = sum{N2[i-j]. j=1..5} for all i, etc.

To do this by hand is reasonably straightforward but somewhat tedious, although less tedious than other methods. It can all be done effortlessly in a spreadsheet, for example.

RGV

 

[LCKurtz]

I would sure like to see a picture of this "5-sided fair die".

 

[theloathedone]

I would sure like to see a picture of this "5-sided fair die".

Haha me too.

Thanks Ray for the help, I'll try to understand it :)

 

[DaveC426913]

Ray, please be advised: we do not do homework for people here. We guide them the answer.

I would sure like to see a picture of this "5-sided fair die".

I was thinking the same thing but it's not hard to do. A fair die does not have to be a regular polyhedron.
You can make a fair die of any number from 2 to theoretically infinite.

Start with a prism with the requisite number of sides and just round off the ends. The die can only come to rest in one of five equally likely configurations.