What is the Probability of Needing More Than 400 Rolls to Reach a Sum of 1380?

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SUMMARY

The probability of needing more than 400 rolls to reach a sum of 1380 when rolling a fair die can be analyzed using the standard normal distribution. The expected number of rolls is calculated as 1380 divided by the average outcome of a die throw, which is 3.5, resulting in a mean of 400 rolls. The discussion highlights that the probability of falling just 0.05 points short of the expected value in each throw is crucial, as it relates to the likelihood of rolling numbers 2 through 6 while avoiding a 1. This analysis leads to a definitive conclusion about the distribution of outcomes in relation to the target sum.

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A fair die is repeatedly rolled until a total of 1380 show up.What is the probability of this needs more than 400 rolls..

I think the number of rolls varies from 1380/6=230 to 1380/1=1380.

So can I use standard normal distribution of 1-P(X<400) and how..

please guide me in this problem
 
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1380 is 20 short of 1400.

Note that 1400/400 = 3.5, which is the mean (expected) value of a die throw = (1+...+6)/6. Since the uniform distribution is a symmetric distribution, its mean = its median so the probability of x > 3.5 is exactly 50%.

The question then becomes: what is the probability of falling just 0.05 points short of the expected value in each throw (20/400 = 0.05 = 1/20), on average?

Since 1 + ... + 6 = 21, 1/20 is very close to missing a "one" in each throw; so the probability of falling just 0.05 points short of 3.5 is nearly equal to the probability of obtaining {2, ..., 6} but not a {1}, on average.
 

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