Select a die from a bag containing 2 dice, one die has 6 on all faces, and the other is a fair sided die.

Choosing one die at random, roll it, and get a 6. If you roll the same die, what is the probability that the next roll is also a 6?

Would this be (0.5)(1) + (.5)(1/6) = 0.5833, as picking one of the two dice is equally likely and the probability of obtaining the result is what follows.

  • $\begingroup$ What you have is indeed the probability of throwing a 6 if each for is equally likely to be the chosen one. However, the fact that the first throw is a 6 gives you information that makes it more likely that the die you have is the fudged die. $\endgroup$ – Arthur Oct 30 '19 at 16:33
  • $\begingroup$ A priori, the probability of choosing the fair die is $1/2$, but after you've rolled a $6$, it's rather more likely that you picked the biased die, isn't it? So you need to update the probabilities. $\endgroup$ – saulspatz Oct 30 '19 at 16:34

We are told that the first chosen die gave us a $6$, let this event be $A$.

Let $B$ be the event that the chosen die has only $6$.

$$P(B|A)=\frac{P(B)P(A|B)}{P(A)}=\frac{\frac12}{\frac12+\frac12\cdot \frac16}=\frac67$$

Let $C$ be the event that $6$ is observed again.

$$P(C|A)=P(C|BA)P(B|A)+P(C|B^CA)P(B^c|A)=\frac67 + \frac{1}{6}\frac{1}{7}=\frac{37}{42}$$


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