Question

A bag contains 1000 identical looking six-sided dice. All are fair dice except one which produces a ‘6’ with probability 0.4 with all other outcomes being equally likely. I put my hand in the bag and select a die at random. I roll it n times and every roll is a ‘6’. What is the probability (in terms of n) that I chose the unfair die? What is the least value of n such that the probability I actually have a fair die is less than one in a million? (Compute this with an equation: there is no credit for trying multiple values of n.)

Answer 1

For the unfair die, we are given here that:
P(X = 6) = 0.4, and therefore P(X = 1) = P(X = 2) = P(X = 3) = P(X = 4) = P(X = 5) = (1 - 0.4)/5 = 0.12

P(every roll is a '6' | fair die) = (1/6)ⁿ
P( every roll is a '6' | unfair die) = 0.4ⁿ

Using law of total addition of probability, we get here:

P( every roll is a '6' ) = P(every roll is a '6' | fair die)P(fair die) + P( every roll is a '6' | unfair die)P(unfair die)

P( every roll is a '6' ) = (1/6)ⁿ *(999/1000) + 0.4ⁿ*(1/1000)

Probability that the die is unfair is computed using bayes theorem here as:

P(unfair die | every roll is a '6') = P( every roll is a '6' | unfair die)P(unfair die) / P( every roll is a '6' )

P(unfair die | every roll is a '6') = 0.4ⁿ*(1/1000) / [ (1/6)ⁿ *(999/1000) + 0.4ⁿ*(1/1000) ]

This is the required probability here.

Probability of having a fair is given here as < 1/10⁶

0.999(1/6)ⁿ = (1/10⁶)*[ (1/6)ⁿ *(999/1000) + 0.4ⁿ*(1/1000) ]
(999000)(1/6)ⁿ = 0.999(1/6)ⁿ + 0.4ⁿ*0.001
(998999.001)(1/2.4)ⁿ = 0.001
2.4ⁿ = 998999001

Taking natural log both sides, we get here:

n = Ln(998999001) / 2.4 = 8.63

Therefore the minimum value of n required here is n = 9