A man claims he can predict the outcome of a dice roll with 90% accuracy. If he predicts correctly on his first attempt, what is the probability that he actually has the ability to predict dice rolls with 90% accuracy? Assume there's a 1% chance that someone can have this ability.
Let's use Bayes' theorem again to solve this problem:

Let A represent the event that the man actually has the ability to predict dice rolls with 90% accuracy, and C represent the event of predicting correctly on the first attempt.

We want to find P(A|C), the probability that the man actually has the ability given that he predicted correctly on his first attempt.

Bayes' theorem states that P(A|C) = P(C|A) * P(A) / P(C)

First, let's find P(C|A): the probability of predicting correctly on the first attempt if the man actually has the ability. Since he claims 90% accuracy, this probability is 0.9.

Next, let's find P(A): the probability that someone actually has the ability to predict dice rolls with 90% accuracy. We are told this is 1%, so P(A) = 0.01.

Now we need to find P(C): the overall probability of predicting correctly on the first attempt. This can be calculated as the sum of probabilities for each case: P(C) = P(C|A) * P(A) + P(C|¬A) * P(¬A), where ¬A represents not having the ability and P(¬A) = 1 - P(A) = 0.99.

To find P(C|¬A), the probability of predicting correctly on the first attempt without the ability, we use the fact that there's a 1/6 chance of guessing correctly by random chance: P(C|¬A) = 1/6.

So, P(C) = (0.9)*(0.01) + (1/6)*(0.99) = 0.009 + 0.165 = 0.174.

Finally, we can calculate P(A|C) using Bayes' theorem:

P(A|C) = P(C|A) * P(A) / P(C) = (0.9)*(0.01) / (0.174) ≈ 0.0517.

Therefore, the probability that the man actually has the ability to predict dice rolls with 90% accuracy is approximately 5.17%.