Probability of Choosing a Loaded Die I was hoping to get some help on this problem.  I am doing a Discrete Mathematics course right now and am finding it quite difficult.  Anyways here is the problem I am having problems with:

Assume that you have two dice, one of which is fair, and the other is biased toward landing on six, so that 1/4 of the time it lands on six, and 1/6 of the time it lands on each of 2, 3, 4 and 5, and 1/12 of the time on 1. You choose a die at random, and spin it six times, getting the values 4, 3, 6, 6, 5, 5. What is the probability that the die you chose is the fair die?

Also the book we are using is crap and an example they gave in the book relating to dice I think is wrong:
Is This Correct?
it is showing the $\frac17$, shouldn't it be $\frac16$ for a die sample space?
 A: For your added question: the book does it just fine: if 6 is twice as likely to come up as the others, then the probabilities do indeed become $\frac{2}{7}$ for the 6, and $\frac{1}{7}$ for the others.
For your original question:
You need to calculate the conditional probability that
$P(F|O)$
Where $F$ is the event of having picked the fair die, and $O$ the event of getting the values you got. That is given the outcomes that you got, what is the probability that you did pick the fair die?
And for that, you want to use Bayes' Theorem:
$$P(F|O) = \frac{P(O|F) * P(F)}{P(O)}$$
where:
$P(O) = P(O|F) * P(F) + P(O|F') * P(F')$
Now, $P(F)$ is the probability of picking the fair die without knowing the outcomes that you got. So this is the probability of picking the fair die when you picked one die initially and randomly. So, $P(F) = \frac{1}{2}$. Likewise, $P(F')$ is the probability of picking the unfair die, again without knowing the outcomes that you got. So $P(F') = \frac{1}{2}$ as well.
But again: what you want is to compute $P(F|O)$, because you do know that the outcomes are, and that will effect the probability of you having picked the fair die. For example, suppose the unfair die was such that it had a $0$ probability of getting a 6. Then knowing that you did get a 6 in your throws, the probability of having picked the unfair die becomes $P(F'|O) = 0$, and hence the probability of having picked the fair die becomes $P(F|O) = 1$.
Now, in your case, both the fair and the unfair die could have resulted in the outcomes you got, but which is more likely? Intuitively, the fact that you got a couple of 6's and no 1's all seems quite compatible with the unfair die, since that one has a preference for 6, and against 1. Now, the devil is in the details, and you will have to go through the actual calculations, but I bet $P(F|O) < P(F'|O)$!
Anyway, to finish your calculation for Bayes' formula, you still need to calculate the following:
$P(O|F)$: the probability of getting the values you got with the fair die
$P(O|F')$: the probability of getting the values you got with the unfair die
But I'll leave you to that!
