Question

Estimate the probability that in 1,200 launches of a
given honest, 6-sided, the sum total of the points obtained is greater
at 4000.

Answer 1

We know that we must find the probability that the sum of 1200 rolls is greater than 4000. This means that on average, we must find the probability of the average of each roll to be greater than 4000/1200= 3.33

We know that a die roll follows a uniform distribution, with

mean= 3.5

standard deviation= sqrt(35/12)

Thus, for a 1200 rolls, given the central limit theorem,

mean= 3.5

standard deviation= sqrt(35/12)/ sqrt(1000)

Thus, the required probability= 1- P(average roll less than 3.33)

= 1- P(Z< (3.33-3.5)/sqrt(35/12)/ sqrt(1000))

= 1- P(Z< -1.7/ 1.71/31.62)

= 1- P(Z< -1.7/0.054)

= 1- P(Z< -31)

= 1- 0

= 1

Thus, there is ~100% probability that the sum of 1200 rolls will be greater than 4000.