Probability - You have three 6-sided dice: Two regular dice, and one unbalanced

Question

You have three 6-sided dice: Two regular dice, and one unbalanced die which comes up as 6 all of the time. You roll a die and it comes up as 6. What is the probability that the die rolled was the unbalanced one?

I'm thinking about : each regular is 1/6 unbalanced 1, part of the game is 1-(1/6) = 0.8333

please let me know if it is the right way?

Answer 1

Hints:

$$Pr(A\mid B) = \dfrac{Pr(A\cap B)}{Pr(B)}$$

Let $B$ be the event that you roll a $6$. Let $A$ be the event that the die you rolled was the unbalanced one.

Approach directly.

---

Hint for an indirect approach: Imagine the unfair die is such that while all faces are equally likely to appear it is such that all faces show the number six. Further, imagine all dice were differently colored. We have then $18$ equally likely to occur faces that could have been rolled, $8$ of which show the result $6$ and $6$ of which came from the unfair die.

Answer 2

Bayes' Theorem for Three Dice--Two of Them Fair

Pedagogic Comment: Focused as much on the next such problem, as on this one.

For a beginner,

• It may be best to start by implementing Bayes' Theorem---with three partition sets. (This important result needs all the reinforcement it can get.)

• Also, to use notation with mnemonic value. (Always a good idea.)

• Then, because the three partition sets are equally likely, the clever 'indirect' approach mentioned by @Raoul and @JMoravitz (+1) may become obvious.

Let $H$ mean getting a 6. Let $U$ be rolling the unbalanced die. Let $B_1, B_2$ be rolling the first or second balanced die.

[Partition sets are $U,B_1,B_2.$ You're looking for $P(U|H),$ knowing probabilities such as $P(H|U), P(H|B_i).$ Switching the order of conditioning always ought to bring Bayes' Theorem to mind.]

Bayes' Theorem becomes:

$$P(U|H) = \frac{P(UH)}{P(H)} = \frac{P(UH)}{P(UH)+P(B_1H)+P(B_2H)}\\ =\frac{P(U)P(H|U)}{P(U)P(H|U)+P(B_1)P(H|B_1)+P(B_2)P(H|B_2)}\\ = \frac{(1/3)(1)}{(1/3)(1)+(1/3)(1/6)+(1/3)(1/6)} = \frac{6}{6+1+1}=\frac{3}{4}.$$

Note: In a formal Bayesian setting, probabilities $P(U)$ and $P(B_i)$ are called prior probabilities, information (data) is that $H$ occurred, probability $P(U|H),$ (with order of conditioning reversed) is called a posterior probability. The key part is $P(U|H) = P(U)P(H|U)/K.$ Sometimes it is not necessary to evaluate $K.$