Respective chances that each of $3$ dice is loaded

Question

Here's a question from my probability textbook:

There are three dice A, B, C, two of which are true and one is loaded so that in twelve throws it turns up six $3$ times, ace once, and each of the other faces twice. Each of the dice is thrown three times and A turns up $6$, $6$, $1$; B turns up $6$, $5$, $4$; and C turns up $3$, $2$, $1$. What are now the respective chances that A, B or C is loaded?

Here's what I did. The relative proportions of loadedness are$${{\left({3\over{12}}\right)^2\left({1\over12}\right)}\over{\left({3\over{12}}\right)^2\left({1\over12}\right) + 2\left({1\over6}\right)^3}} = {9\over{25}} = {{63}\over{175}}, \quad {{\left({3\over{12}}\right)\left({2\over{12}}\right)^2}\over{\left({3\over{12}}\right)\left({2\over{12}}\right)^2 + 2\left({1\over6}\right)^3}} = {3\over{7}} = {{75}\over{175}}, \quad {{\left({2\over{12}}\right)^2\left({1\over{12}}\right)}\over{\left({2\over{12}}\right)^2\left({1\over{12}}\right) + 2\left({1\over6}\right)^3}} = {1\over{5}} = {{35}\over{175}},$$and so the respective chances are ${{63}\over{173}}$, ${{75}\over{173}}$, ${{35}\over{173}}$.

However, the answer in the back of my book is as follows:

A priori it is equally likely that the loaded die is A, B, or C.

The consequent chances of the observed event are as$${1\over4}\cdot{1\over4}\cdot{1\over{12}}: {1\over4} \cdot {1\over6} \cdot{1\over6} : {1\over6} \cdot {1\over6} \cdot {1\over{12}} = 9 : 12: 4$$The a posteriori chances are ${9\over{25}}$, ${{12}\over{25}}$, ${4\over{25}}$.

I don't really understand why I'm wrong, nor do I understand the solution in the back of the book. Could anyone explain:

1. Where specifically did I go wrong?
2. How did they come up with the answer in the back of the book?

Answer 1

A calculation such as $${{\left({3\over{12}}\right)^2\left({1\over12}\right)}\over{\left({3\over{12}}\right)^2\left({1\over12}\right) + 2\left({1\over6}\right)^3}} = {9\over{25}} = {{63}\over{175}} $$ makes sense when you only roll $A$, it comes up $6,6,1$, and you want to know the probability that $A$ is loaded. Formally, let $H_A, H_B, H_C$ be the three hypotheses "$A$ is loaded", "$B$ is loaded", and "$C$ is loaded". Let $E$ be the evidence we have: that $A$ came up $6,6,1$. Then we would write $$ \Pr[H_A \mid E] = \frac{\Pr[E \mid H_A] \Pr[H_A]}{\Pr[E\mid H_A] \Pr[H_A] + \Pr[E\mid H_B] \Pr[H_B] + \Pr[E\mid H_C] \Pr[H_C]} $$ which gives us the calculation you did. (The prior probabilities $\Pr[H_A]$, $\Pr[H_B]$, $\Pr[H_C]$ cancel because they are all $\frac13$.)

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In this problem, we didn't just roll $A$; we also rolled $B$ and $C$. Those outcomes are also informative of whether $A$ is loaded (if you imagine that $B$'s rolls were very very suspicious, then that should make us less suspicious of $A$). So we should take $E$ to be the entire set of $9$ rolls. Now we get

• $\Pr[E \mid H_A] = \color{blue}{(\frac3{12})^2 (\frac1{12})} \cdot \color{orange}{(\frac16)^3} \cdot \color{purple}{(\frac16)^3} = \frac{3^2 \cdot 1}{12^3 \cdot 6^6}$
• $\Pr[E \mid H_B] = \color{blue}{(\frac16)^3} \cdot \color{orange}{(\frac3{12}) (\frac{2}{12})^2} \cdot \color{purple}{(\frac16)^3} = \frac{3 \cdot 2^2}{12^3 \cdot 6^6}$
• $\Pr[E \mid H_C] = \color{blue}{(\frac16)^3}\cdot \color{orange}{(\frac16)^3} \cdot \color{purple}{(\frac2{12})^2 (\frac1{12})} = \frac{2^2 \cdot 1}{12^3 \cdot 6^6}$

Here, the three colors correspond to the probabilities of the results of $A$, $B$, and $C$ respectively.

If you use these probabilities, you will get answers matching the textbook. (It also takes the shortcut of using odds, which lets us ignore the denominators of $12^3 \cdot 6^6$ and simplify the calculations.)