Probability problem. Sum of a 3-side die.?

Question

Hi,

I've looking in the internet but I can't find a straight answer.

The problem is this:

Suppose we have a 3-side die (faces marked 0, 1, and 2).

Are there general formulas to answer questions such as:

a) "After throwing the die 100 times, get the probability of the sum being 120"?
b) "After throwing the die 100 times, get the probability of the sum being greater or equal to 120"?

Any help is appreciated.

Thanks for answering.

Dr D:
I don't follow your reasoning. I'm a complete novice in prob, so I don't see why Var(X) = 2/3 or how do you go from "P(1.195 < Ybar < 1.205)" to "P(2.388 < Z < 2.511)".

The sum (120) and the number of trials (100) are just an example, I would like to know if there is a general formula.

kain2396:
The 0 doesn't need to be there. The faces can be 1, 2, and 3. However, I don't see the problem with a face 0. After all, the experiment is finite, meaning that if I roll the die (faces 0, 1, and 2) 200 times, the sum will be between 0 and 400. For this case, I think that P(sum=0) = 1 / 3^200.

In summary, what I'm asking is if there are formulas to find P(sum=x) and P(sum>=x) for any "x" between 0 and 2n, where "n" is the number of trials.

Thanks for your time Dr D.

I read about the normal function, normalization, etc, and now I see what you are doing. Yet, I still have some "novice" questions:

1. Why "Var(Ybar) = Var(Y) / 100^2 = 2/300"?
The book defines the sample variance as sum_i(x_i - media)^2/n
2. Is this a more or less exact method compared to counting combinations?
I ask because for 100 trials, for example, though the maximum sum is 200 I think that P(sum > 200) is not exactly 0.

Answer 1

I would do this using the normal distribution.
X = score on one throw
follows a uniform discrete distribution, such that
E(X) = 1, Var(X) = 2/3
See the link below for details.

Let Y = ∑ X1 + X2 + ... + X100
E(Y) = 100
Var(Y) = 200/3

Ybar = mean value of Y
E(Ybar) = 1
Var(Ybar) = Var(Y) / 100^2 = 2/300

Ybar ~ N(1, 2/300)
We are required to find the P(Y = 120)
= P(119.5 < Y < 120.5) ---------- continuity correction
= P(1.195 < Ybar < 1.205)
= P(2.388 < Z < 2.511)
= 0.0024 from normal tables

P(Y >= 120) = P(Y > 119.5)
= P(Ybar > 1.195)
= P(Z > 2.511)
= 0.006

*EDIT*
This is a pretty advanced question for someone who is a novice in stats.
The variance of a uniform discrete distribution is (n^2 -1 )/12 where n = no. of options, in this case 3. Check the first link below.

The other part about going from Ybar to Z is called the standardization of the normal variable. If you've studied the normal distribution, you will have seen this. If you haven't, then explaining is not going to help. You need to study the normal distribution and go through the problem step by step.

You would also need to know about hte Central Limit Theorem, which basically says that if n is large (100), then the mean of the distribution tends to a normal distribution.

*EDIT*
There is a difference between sample variance and population variance. The formula you just gave shows you how to calculate the sample variance of a few given data. In the case of rolling a die, we are calculating the variance from first principles rather than from observed data. I didn't do that here, but I'm sure you could find the derivation on the internet somewhere. The wikipedia link just says it's (n^2 - 1) / 12.

There are some general variance formulas.
If Var(X) = a, Var(Y) = b
Var(X + Y) = a + b
Also Var(10X) = 10^2 * a
So that's why when I added X 100 times, I multiplied the variance by 100. But when I divided this by 100, I divided the variance by 100^2.

And also note that the normal approximation is an approximation. Yes P(Ybar > 2) is not exactly zero, but it is very close to it. This would translate to find the p value of 12.25 in the normal tables.

Answer 2

Something seems strange here. Both of those questions have a problem. You could, in theory, roll the dice and get zero infinite number of times. This, to me, would seem to make the set of rolls that add up to any number be a set which is countably infinite. Take a certain example:

{2,2,2,2,2,2,2,2,2,2} = 20

Now put a zero in front of it. You can do that as many times as you like and get a new unique set in the probability space.

Make sure the face 0 is supposed to be there. You might be able to do this with a continuous distribution, but I can't imagine what that might be. Perhaps the multinomial distribution, but that's not continuous.

Sorry I can't help more, but I'm slightly suspicious.