Dice probability: Rolling randomly selected weighted dice twice

Question

Problem Statement

An urn contains $4$ fair dice. Two of the dice have standard number schemes. One of the dice has numbered faces $2,2,4,4,6,6$. The last die has all faces numbered with $6$. One of the dice is selected from the urn and rolled. The same die is then rolled again. What is the probability that a $6$ is rolled both times?

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My Attempt

I started by laying out my probabilities: For the standard die we have:$$D_1 = [1,2,3,4,5,6]\\ P(D_1) = \frac{1}{2}\\ P(i\vert D_1) = \frac{1}{6}\ \text{Where i runs from 1 to 6}$$ For the $3$rd dice we have:$$D_2 = [2,2,4,4,6,6]\\ P(D_2) = \frac{1}{4}\\ P(k\vert D_2) = \frac{1}{3}\, k \in [2,4,6]$$ For the last dice we have: $$D_3 = [6]\\ P(D_3) = \frac{1}{4}\\ P(6\vert D_3) =1$$

From here I noted that for one roll we have the probability of getting a six to be: $$P(6) = P(6\vert D_3)P(D_3) + P(6\vert D_2)P(D_2) + P(6\vert D_1)P(D_1) = (1)(\frac{1}{4}) + (\frac{1}{3})(\frac{1}{4}) + (\frac{1}{6})(\frac{1}{2}) = \frac{5}{12}$$

From here I wasn't exactly sure how to proceed so I simply squared the answer and got $\frac{25}{144}$. Which is incorrect as the correct answer is $\frac{7}{24}$

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Reflection/Questions

I believe my answer for the single roll situation is correct as it's a simple use of the law of total probability. However, I am struggling to see how to incorporate the constraint that "The same dice is rolled again" mathematically. The way I did it - by squaring the answer - would be the answer, I believe, for the situation in which the die is replaced and then randomly selected and rolled for again. My question is then: how does one mathematically account for the constraint that the same die that is selected is rolled for again, without knowing which die has been rolled again. I understand that when given $n$ choices each with their own respective probability then we are to multiply the $n$ probabilities together to get the overall probability of the entire gambit.

I appreciate any help rendered.

Answer 1

Your scheme of separating the cases is correct. We have a two-steps experiment, the first step is the choice of the type $D1$, or $D2$, or $D3$, the second step being the double roll in the considered type. So the needed probability is: $$ \frac 24\cdot\left(\frac 16\right)^2 + \frac 14\cdot\left(\frac 13\right)^2 + \frac 14\cdot\left(\frac 11\right)^2 =\frac 1{4\cdot 6^2}(2+4+36) =\frac {42}{4\cdot 6^2} =\frac {7}{4\cdot 6} \ . $$

Answer 2

Problem with the OP's (i.e. original poster's) approach:

Your computation of $~\dfrac{5}{12}~$ for the probability of getting a 6 on the 1st roll is correct. However, your approach makes it difficult (but not impossible) to accommodate that it is guaranteed that you use the same die on both the 1st and 2nd roll.

Dan Fulea's answer shows the elegant Math needed to answer the actual question. The point of my response is to broaden your intuition. I regard intuition as indispensable for this type of problem.

For my inelegant approach, some way is needed to accommodate that the same die is required to be used for the 1st and 2nd roll. Consider the problem intuitively. If the 1st roll is a 6, it is much more likely than not that the die being used was the 3rd die (probability of 6 = 1/3) or the 4th die (probability of 6 = 1/1) rather than the 1st or 2nd die.

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I am going to show an inferior alternative to Dan Fulea's method which will (however) expand your intuition.

If the 1st roll is a 6, this can happen in one of 3 ways:

• Case 1
  Probability of Case 1 is $~(1/2).$
  If Case 1 occurs, then probability of 6 is $~1/6.~$
  Therefore, probability of 6 as a result of Case 1 is $~(1/2) \times (1/6) = 1/12.$

• Case 2
  Probability of Case 2 is $~(1/4).$
  If Case 2 occurs, then probability of 6 is $~1/3.~$
  Therefore, probability of 6 as a result of Case 2 is $~(1/4) \times (1/3) = 1/12.$

• Case 3
  Probability of Case 3 is $~(1/4).$
  If Case 3 occurs, then probability of 6 is $~1/1.~$
  Therefore, probability of 6 as a result of Case 3 is $~(1/4) \times (1/1) = 1/4.$

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So, your computation of $~(1/12) + (1/12) + (1/4) = 5/12~$ for the probability of a 6 on the first roll is correct.

Before reading this answer further, I suggest that you expand your intuition by reading Wikipedia : Conditional Probability.

Assume that the first die roll was a 6.

Let $~p_1~$ denote the probability that this occurred as a result of Case 1.

Let $~p_2~$ denote the probability that this occurred as a result of Case 2.

Let $~p_3~$ denote the probability that this occurred as a result of Case 3.

Then, given that the first die roll was a 6, the probability of a 6 on the second die roll is

$$A = \left[p_1 \times \frac{1}{6}\right] + \left[p_2 \times \frac{1}{3}\right] + \left[p_3 \times \frac{1}{1}\right]. \tag1 $$

So, once $~p_1, ~p_2, ~$ and $~p_3~$ are determined, then $~A~$ from (1) above can be computed.

Then, the final computation of the probability of a 6 on both rolls of the same die will be

$$\frac{5}{12} \times A.$$

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• $\displaystyle p_1 = \frac{\frac{1}{12}}{\frac{1}{12} + \frac{1}{12} + \frac{1}{4}} = \frac{1}{5}.$

• $\displaystyle p_2 = \frac{\frac{1}{12}}{\frac{1}{12} + \frac{1}{12} + \frac{1}{4}} = \frac{1}{5}.$

• $\displaystyle p_3 = \frac{\frac{1}{4}}{\frac{1}{12} + \frac{1}{12} + \frac{1}{4}} = \frac{3}{5}.$

Therefore,

$$A = \left[\frac{1}{5} \times \frac{1}{6}\right] + \left[\frac{1}{5} \times \frac{1}{3}\right] + \left[\frac{3}{5} \times \frac{1}{1}\right] = \frac{1 + 2 + 18}{30} = \frac{21}{30} = \frac{7}{10}. $$

Therefore, the probability of a 6 on both rolls of the die is

$$\frac{5}{12} \times \frac{7}{10} = \frac{35}{120} = \frac{7}{24}.$$