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

[CaptainBlack]

"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>300then 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

 

[jvicens]

Hi CB,

Thank you for your forum post. Your suspicion is correct, the number of throws required until the sum is greater than 300 does indeed follow a normal distribution. To prove this, we can use the Central Limit Theorem which states that the sum of independent and identically distributed random variables will approach a normal distribution as the number of variables increases.

In this case, the number of throws (X) is the sum of the outcomes of each individual throw, which follows a discrete uniform distribution. As the number of throws increases, X will approach a normal distribution with mean 85 and variance 2975. This can be calculated using the formulas you provided: \(\mu_N=N \mu_1=85\) and \(\sigma^2_N=N\sigma^2_1=2975\).

To calculate the probability that at least 80 throws are required, we can use the normal approximation method you mentioned. This gives a probability of approximately 92.4%, which is close to the result obtained through simulation. It is important to note that this is an approximation and may not give an exact result, especially for smaller sample sizes.

In terms of your different results, it is possible that there may have been a small error in your calculations or simulation. It is always good to double check your work and make sure all assumptions and formulas are correct. Additionally, the normal approximation method is just an approximation and may not give an exact result, especially for smaller sample sizes.

I hope this helps clarify the concept and calculations for you. Let me know if you have any further questions.