1. ## Hard probability question

"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?

Also, I calculated the probability using another method and I got 0.765 as a result, but when I simulated the experiment I got 0.89 as a result ...

2. ## Re: Hard probability question

I doubt it's normal. We know that between 50 and 300 experiments are required, and the expected number of experiments is about 300/3.5, 85.7.

Edit: Between 51 and 301 (if you are considering "strictly greater than 300").

3. ## Re: Hard probability question

Originally Posted by smalltalk
"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?

Also, I calculated the probability using another method and I got 0.765 as a result, but when I simulated the experiment I got 0.89 as a result ...
Hi Smalltalk,

Let's say the outcome of one die throw is $X_i$. Then you want to find $\Pr(\sum_{i=1}^{80} X_i \le 300)$. For brevity, let's say $Y = \sum_{i=1}^{80} X_i$.

Y is approximately normally distributed, by the Central Limit Theorem. You need to compute the mean of Y (it's not 85), and you need to find its standard deviation before you can look up the sought-for probability in a table.

Hint: $var(Y) = \sum_{i=1}^{80} var(X_i)$.
Can you find the variance of $X_i$?