# smalltalk's "Hard Probability Question" for another place.

#### CaptainBlack

##### Well-known member
"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 ... The sum after $$N$$ throws is approximately normal with mean $$\mu_N=N \mu_1$$ and with variance $$\sigma^2_N=N\sigma^2_1$$, where $$\mu_1=3.5$$ is the mean for a single throw, and $$\sigma^2_1=17.5$$ is the variance of a single throw.

Now the probability that the process finished on or before the $$N$$ throw is:

$p_{N}\approx P\left(\frac{N\mu_1-300.5}{\sqrt{N}\sigma_1}\right)$

where $$P(.)$$ is the cumulative Standard Normal distribution function.

So the probability that at least $$N+1$$ throws are necessary is $$1-p_N$$

This normal approximation method gives a probability of $$\approx 92.4\%$$ while simulation gives $$\approx 93.6\%$$

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