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

CaptainBlack

Well-known member
Jan 26, 2012
890
"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 \(\approx 93.6\%\)

Please check the arithmetic etc.

CB
 
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