Question

Consider the sum of 2D6.
(a) What is the probability that the sum of 2D6 never equals 7, in 12 rolls?
(b) On average, how many times do we need to roll 2D6 until their sum is first equal to 7?
(c) Suppose we keep rolling 2D6 until we get a sum of 7 exactly 3 times. What is the probability that we need to roll 15 times?

Answer 1

2d6 would mean, "roll two six-sided dice."

A.

P(Sum 7) = 6/36 = p

= 1/6

The given scenario becomes a binomial distribution with probability of success being 1/6 and n = 12. Let the number of success be x. For x = 0:

P(x = 0) = 0.112

B.

This is a case of the Geometric distribution.

So, the expected value for the first success is = 1/p

= 6

C.

This is a case of Pascal distribution where x = 15 and r = 3 and p = 1/6.

Formula for Pascal distribution:

P(x=15) =0.0473

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