The Bizzare Dice

This is a really fantastic problem! As for the solution, there are two ways to tackle this problem. We can apply Bayes' Law, which is a really cool theorem that you can read about here . I have a much simpler solution that is a bit more intuitive:

First let us calculate what the probability that if Dice 1 is picked that that particular sequence occurs. This is simply $\frac{1}{5184}$ . Notice that if Dice 2 is picked, the probability that the sequence occurs is $\frac{1}{1728}$ . It is obvious that Dice 2 is three times more likely to roll that particular sequence! Read that line again: Dice 2 is three times more likely to roll that particular sequence! So if we've established that, then the probability that it is indeed Dice 2 is $\frac{3}{4}$ , thus the answer is $3+4=\boxed{7}$ and we're done! No messiness, no weird manipulations, not a lot of LaTeX! Always simplify problems, they produce beautiful results. :D

Let $\text{D}_1$ be the die number $1$ , $\text{D}_2$ be the die number $2$ , and $\text{N}$ be the first five rolls of the die. $\text{D}_1$ , $\text{D}_2$ , and $\text{N}$ are random variables. We use Bayesian approach. $\begin{aligned} &\text{Pr}[\text{D}_1]=\text{Pr}[\text{D}_2]=\frac{1}{2},\\ &\text{Pr}[\text{N}=2|\text{D}_1]=\frac{3}{6}=\frac{1}{2},\\ &\text{Pr}[\text{N}=1|\text{D}_1]=\text{Pr}[\text{N}=3|\text{D}_1]=\text{Pr}[\text{N}=4|\text{D}_1]=\frac{1}{6},\\ &\text{Pr}[\text{N}=4|\text{D}_2]=\frac{3}{6}=\frac{1}{2},\\ &\text{Pr}[\text{N}=1|\text{D}_2]=\text{Pr}[\text{N}=2|\text{D}_2]=\text{Pr}[\text{N}=3|\text{D}_2]=\frac{1}{6}, \end{aligned}$ and $\text{N}=\text{{2, 3, 4, 1, 4}}.$ Therefore $\begin{aligned} \text{Pr}[\text{D}_2|\text{N}=2, 3, 4, 1, 4]&=\frac{\text{Pr}[\text{N}=2, 3, 4, 1, 4|\text{D}_2]\cdot\text{Pr}[\text{D}_2]}{\text{Pr}[\text{N}=2, 3, 4, 1, 4|\text{D}_1]\cdot\text{Pr}[\text{D}_1]+\text{Pr}[\text{N}=2, 3, 4, 1, 4|\text{D}_2]\cdot\text{Pr}[\text{D}_2]}\\ &=\frac{\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{1}{2}}{\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{2}+\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{1}{2}}\\ &=\frac{3}{4}. \end{aligned}$ Thus, $p+q=3+4=\boxed{\color{#3D99F6}{7}}$ .

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$\Large\color{#3D99F6}{\text{\# }\mathbb{Q.E.D.}\text{ \#}}$

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There are three equally-likely ways to get this roll with die one (3x1x1x1x1), and nine equally-likely ways to get it with die two (1x1x3x1x3). We exclude all cases but these 12. 9/12, or 3/4 of them used die 2.

Its a very easy problem based on Bayes theorem