Question

How many dice must be thrown so that there is a better than even chance of obtaining a 6?

Answer 1

Let the number of dice thrown be n such that there is a better than even chance of obtaining a 6

When 1 die is thrown, P(getting a 6) = 1/6

P(not getting a 6) = 1 - 1/6 = 5/6

When n dice are thrown, to have even chance of getting a 6,

P(getting at least one 6) = 0.5

1 - P(not getting any 6) = 0.5

= 1 - (5/6)ⁿ = 0.5

(5/6)ⁿ = 0.5

n log(5/6) = log 0.5

n = log 0.5 / log(5/6)

n = 3.8

4

4 dice must be thrown so that there is a better than even chance of obtaining a 6