Conditional probability - coloured dice

Question

I am struggling with the below question and am obviously making a mistake somewhere:

An urn contains four (fair, 6-sided) dice, one green, one red and two blue. Two dice are thrown. Then, the remaining dice are thrown. What is the probability they show the same number and are the same colour, given the first two did not show the same number and colour.

The answer is meant to be $\frac{1}{42}$.

Attempt: Let $N_i$ and $C_i$ be the events the numbers and colours are the same on the $i$th role, respectively. Let $RB$, $RG$, $BG$ be the events that red and blue, red and green, blue and green are rolled on the first toss, respectively.

We want to calculate $P(N_2\cap C_2\mid (N_1\cap C_1)^c)$

Note that $(N_1\cap C_1)^c=N_1\cap C_1^c \bigcup N_1^c\cap C_1 \bigcup N_1^c\cap C_1^c$ and $P((N_1\cap C_1)^c)=1-\frac{1}{36}=\frac{35}{36}$.

Therefore, we have:

\begin{align*} P(N_2\cap C_2\mid (N_1\cap C_1)^c)&=\frac{P(N_2\cap C_2 \bigcap \left(N_1\cap C_1^c \bigcup N_1^c\cap C_1 \bigcup N_1^c\cap C_1^c\right)}{P(N_1\cap C_1)^c}\\\\ &=\frac{P(N_2\cap C_2 \cap N_1\cap C_1^c) + P(N_2\cap C_2 \cap N_1^c\cap C_1) + P(N_2\cap C_2 \cap N_1^c\cap C_1^c)}{\frac{35}{36}}\\\\ &=\frac{36}{35}\left[P(N_2\cap C_2 \cap N_1\cap C_1^c) + P(N_2\cap C_2 \cap N_1^c\cap C_1^c)\right] \end{align*}

Since each event in the union is disjoint and the middle term vanishes as it is impossible to roll same colour die on both the first and second roll.

Since $C_1^c=RB\cup RG\cup BG$ and it is impossible to roll the same colour (2 blue) on the second role if any blue is rolled on the first roll, we have the above quantity equal to \begin{align*} &=\frac{36}{35}\left[P(N_2\cap C_2 \cap N_1\cap RG )) + P(N_2\cap C_2 \cap N_1^c\cap RG ))\right]\\\\ &=\frac{36}{35}\left[P(N_2\cap C_2 \mid N_1\cap RG ))\cdot P(N_1\cap RG) + P(N_2\cap C_2 \mid N_1^c\cap RG ))\cdot P(N_1^c\cap RG)\right] \end{align*} If red and green are drawn on the first roll, then there is a 100% chance the remaining dice are both blue. Therefore, we have, by independence of rolling numbers/colours \begin{align*} &=\frac{36}{35}\left[\frac{1}{6}\cdot 1\cdot \frac{1}{6}\cdot \frac{1}{6} + \frac{1}{6}\cdot 1 \cdot \frac{5}{6}\cdot \frac{1}{6}\right]\\\\ &=\frac{36}{35}\cdot \frac{1}{36}\\\\ &=\frac{1}{35} \end{align*} But this means I am out by a factor of $\frac{5}{6}$. I am sure my mistake is obvious/am missing something. I am also guessing there is probably a more intuitive/easy way to tackle the problem, but I still want to use a brute force approach and arrive at the correct answer.

Answer 1

This is a badly formulated problem with a purported answer that’s wrong no matter how we resolve the ambiguity in the problem. If possible, you should toss this text and use a different one.

There’s insufficient information in the problem to answer it, since we don’t know how the two dice that were thrown first were chosen. We can only assume (and you seem to have assumed) that there’s meant to be an implication that they were uniformly randomly selected among all $6$ pairs of dice. If so, the problem should have made that assumption explicit.

But no matter how exactly the dice were chosen, it’s clear that conditioning on an event involving the first two dice not having the same colour must make it more, not less probable that the second two dice have the same colour.

Since the unconditional probability for the second pair to show the same number and colour is $\frac1{36}$, your answer satisfies this basic plausibility check, and the answer from the source doesn’t. I checked your calculations and couldn’t find a mistake, so your result is probably OK.

Answer 2

Let's solve it another way.

Let $A$ denote the event in which the first two rolls do not produce the same color and number.

To find $P(A)$, we can focus on $A^c$ - the event in which the first two rolls produce the same color and the same number.

$$\begin{align*} P(A^c) &= \left(\frac{2}{4} \times \frac{1}{3}\right) \times \frac{1}{6} \\[0.3cm] &= \frac{1}{36} \end{align*}$$

That gives us,

$$P(A) = 1 - P(A^c) = \frac{35}{36}$$

Now, let B be the even in which the last two rolls produce the same color and same number.

Let's find $P(A \cap B)$.

$A \cap B$ implies that the first two dice are not of the same color while also showing the same number. And the last two dice have the same color and the same number.

For the last two dice to be of the same color, they must be blue and blue. So the first two dice have to be green and red (in any order). Also, we don't care what numbers show up because, as long as the colors are different event A has occurred.

Then for the last two rolls, the dice will inevitably be the same color (only two blues are available). So all we care about is them showing the same number.

Effectively, $A \cap B$ reduces to this. The first two dice must be green and red (any order) and the last two dice must show the same number. So,

$$\begin{align*} P(A \cap B) &= \left(\frac{1}{2} \times \frac{1}{3}\right) \times \frac{1}{6} \\[0.3cm] &= \frac{1}{36} \end{align*}$$

So far, so good. Next, using the Bayes' theorem, we have

$$\begin{align*} P(B | A) &= \frac{P(A \cap B)}{P(A)} \\[0.3cm] &= \frac{\frac{1}{36}}{\frac{35}{36}} \\[0.3cm] &= \frac{1}{35} \end{align*}$$

So yes, as @joriki says, your answer is probably correct.