What is the probability that she chose the "cheat" die?

Question

A gambler mixed a "cheat" die with all sixes into a box of eight normal dice. She chooses one at random, rolls it twice, and gets six both times. What is the probability that she chose the "cheat" die?

Answer 1

This is an example of "conditional probability," a tool that tells us the probability of one event given that another event has occurred. In this case, we know that a gambler rolled two sixes, and we want to know the probability that she chose the loaded die.

First, consider the two possibilities. If she chose the loaded die, then there is a probability of 1 that she would roll two sixes, because the die only has sixes on it. If she chose a fair die, then the probability of rolling two sixes is 1/36.

A common notation for this is P(A|B), "the probability of A given B." In this case, A is the event in which she chose the loaded die, and B is the event rolling two consecutive sixes.

the formula for conditional probability reads:

P(A|B)=P(A and B)/P(B)=P(B|A)*P(A)/P(B)

(the second equality is using something called "Bayes Theorem")

For this problem, we need to know the probability of rolling two consecutive sixes, which is P(rolling two sixes with a loaded die)*P(picking the cheat die)+P(rolling two sixes without a loaded die)*P(not picking a loaded die)

P(B)=1*1/9+1/36*8/9=1/9+2/81=11/81

P(A|B)=(1*1/9)/(11/81)=81/99

So we get 81/99