I rolled a 6, what is the probability of having chosen a manipulated dice? I have two dice, $F$ and $M$.  $F$ is a fair die and thus $P_F(6)=\frac{1}{6}$.  $M$ is manipulated and $P_M(6)=\frac{1}{2}$.  I randomly choose a die and throw it.
I already know  $P(6)=\frac{1}{2}*\frac{1}{6}+\frac{1}{2}*\frac{1}{2}\approx 0.33 = 33\%$.
Now the question is: If I already rolled a dice and got a 6, what is the probability of rolling another 6? I understand that the question would be almost answered if I knew the probability of having chosen a fair or manipulated dice after I got a 6.  I can't really grasp the problem because it's about something that has already happened connected to something that will happen.  I drew a decision tree but that didn't help me so far.
 A: I will assume that having chosen a die that die will be rolled twice.
Let $X_1$ and $X_2$ denote the results of the two rolls, respectively. The question is the value of the following conditional probability:
$$P(X_2=6\mid X_1=6)=\frac{P(X_2=6\cap X_1=6)}{P(X_1=6)}.$$
Now,
$$P(X_1=6)=\frac12\left(\frac16+\frac12\right)=\frac13$$
and
$$P(X_2=6\cap X_1=6)=\frac12\left(\frac1{36}+\frac14\right)=\frac5{36}.$$
So,
$$P(X_2=6\mid X_1=1)=\frac{\frac5{36}}{\frac13}=\frac5{12}.$$
A: An intuitive answer.
Think of the manipulated die as a normal one with $3$ faces with a $6$ on it.
$\begin{array}{ccccccc}
F & 6 & x & x & x & x & x\\
M & 6 & 6 & 6 & x & x & x\end{array}$
The $x$ stands for: "no $6$".
At your first throw a $6$ turned up so you are in one of the $4$ "spots" with a $6$ in it. In $3$ of these spots you are dealing with the manipulated die so the probability on a second $6$ will be $\frac12$. In $1$ of the $4$ spots you are dealing with the fair die so the probability on a second $6$ will be $\frac16$.
This results in a probability: $$\frac34\frac12+\frac14\frac16=\frac5{12}$$
A: Think about how you calculated the probability of getting a 6 on your first roll:

1/2 * 1/6 + 1/2 * 1/2 = 1/12 + 1/4 = 1/12 + 3/12 = 4/12 = 1/3

Now think about how much each possible die "contributed" to the final result. The fair die contributed 1/12, the loaded die contributed 1/4 or 3/12. So of the 4/12 total probability, we have (1/12) / (4/12) from the fair die and (3/12) / (4/12) from the loaded die, or 1/4 and 3/4.
So the probability that the chosen die was the fair die is 1/4 and the probability that it was the loaded die is 3/4.
From there it's a simple matter to calculate the probability of the next roll:

1/4 * 1/6 + 3/4 * 1/2 = 1/24 + 3/8 = 1/24 + 9/24 = 10/24 = 5/12

