If 8 pairs of fair dice are rolled, approximate the probability that the sum of the values obtained (which ranges from 16 to 96) is between 30 and 40 inclusive.

Question

If 8 pairs of fair dice are rolled, approximate the probability that the sum of the values obtained (which ranges from 16 to 96) is between 30 and 40 inclusive.

Answer 1

Consider 16 rolls of a die.

The mean is (1+2+3+4+5+6)/2 =7/2

The variance is (1²+2²+3²+4²+5²+6²)/6 - (7/2)^2 = 35/12

Then, for16 rolls, the mean is 7/2*16 = 56

The variance is 35/12 * 16 = 140/3

Using the continuity correction, the probability of a roll between 30 and 40 equal

P((29.5-56)/(sqrt(140/3) <= z <= (40.5-56)/(sqrt(140/3) ) =

P(-3.87920278997042 <= z<= -2.26896766960534) =

from Excel, norn.s.dist(-2.26896766960534,1) - norm.s.dist(-3.87920278997042,1) =

.0116

The exact answer may be rounded off to 0.01103594

(You could actually get this via for a short-cut. Calculate the probability for 2 rolls. Apply these values to 2 sets of 2 rolls; apply these to 2 sets of 4 rolls; apply these to 2 sets of 8 rolls; this way, your biggest loop has 41^2 calculations)

Answer 2

probability = number of sums between 30 and 40/total possible sums

since the range of sums is between 16 to 96, there are 81 possible sums

there are 11 possible sums between 30 and 40 inclusive

so the probability of sums between 30 and 40 is 11/81