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1. Another hard probability problem
"A dice it thrown until the sum of the outcomes is greater than 300. Find, approximately, the probability that at least 80 experiments are required."

I have the suspicion that if
X = number of throws required until sum>300

then X has normal distribution with mean 85 and ... some variance

How could I prove that?
2. Re: Another hard probability problem
A sum of independent variables tends to the normal distribution - is certainly the CLT

Call the random variable for the nth throw . Work out the mean and variance of each , realise that they are independent, ....
3. Re: Another hard probability problem
It is not obvious what the distribution that "number of throws required until sum>300" is going to look like.

On the other hand, the variable "sum of the first 79 experiments" is fairly standard.

So you would be better off considering p(sum of first 79 <= 300).
4. Re: Another hard probability problem
(Original post by DFranklin)
It is not obvious what the distribution that "number of throws required until sum>300" is going to look like.

On the other hand, the variable "sum of the first 79 experiments" is fairly standard.

So you would be better off considering p(sum of first 79 <= 300).
Is this correct?
P(more than 80 experiments needed to get sum > 300) = P(sum <= 300 in exactly 79 experiments)

I know this:
S|N = sum of N dices thrown ~ Normal (mean = , s.d. = )

So:
P(sum <= 300 in exactly 79 exps)
= P ( S <= 300 | N= 79)
= P ( Z <=
= P ( Z <= 1.5481)

The thing is, i've simulated the experiment and I don't get 0.93, I get 0.89 or 0.90...
5. Re: Another hard probability problem
Its good that you are checking your answer with simulation, this is a clever thing to do and definitely a nice habit to get into.

However, you probably didnt do enough simulations, if the number is too low then the randomness means you wont get an accurate result:

Code:
> sims <- 10000000; x <- numeric(sims)
> for (s in 1:sims) {x[s] <- sum(sample(1:6,79,replace=TRUE))}
> sum(x<=300)/sims
[1] 0.9430214
(R code, takes about a minute to run)

since your answer deviates slightly from that, then unless your math is wrong (I didnt check) the problem will be that you are using an asymptotic approximation which is going to be slightly off.

edit: P(Z < 1.5481) = 0.9392009, which rounds to 0.94, not 0.93
Last edited by poohat; 23-07-2012 at 03:50.
6. Re: Another hard probability problem
(Original post by poohat)
Its good that you are checking your answer with simulation, this is a clever thing to do and definitely a nice habit to get into.

However, you probably didnt do enough simulations, if the number is too low then the randomness means you wont get an accurate result:

Code:
> sims <- 10000000; x <- numeric(sims)
> for (s in 1:sims) {x[s] <- sum(sample(1:6,79,replace=TRUE))}
> sum(x<=300)/sims
[1] 0.9430214
(R code, takes about a minute to run)

since your answer deviates slightly from that, then unless your math is wrong (I didnt check) the problem will be that you are using an asymptotic approximation which is going to be slightly off.

edit: P(Z < 1.5481) = 0.9392009, which rounds to 0.94, not 0.93
Ah, there has to be a problem with my simulation then.
7. Re: Another hard probability problem
lifes too short for probability problems