I rolled a 6, what is the probability of having chosen a manipulated dice?

Question

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.

Answer 1

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,

$$P(X_2=6\mid X_1=1)=\frac{\frac5{36}}{\frac13}=\frac5{12}.$$

Answer 2

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: $$\frac34\frac12+\frac14\frac16=\frac5{12}$$

Answer 3

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