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

[victor007]

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

 

[EnumaElish]

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.