Consider that we have two sets of dices. One set contains fair dices while the other set contains dices in which the probability of $6$ is $1/2$. The fair dices correspond to $99\%$ of the whole sample, which means the loaded dices occur with a frequency of $1\%$.
If I pick a dice at random and throw 3 sixes in a row, what is the probability that I have a loaded dice?
It is clear that Bayes rule can be applied:
$$ P( \mathrm{Loaded} \, | \, \mathrm{3 six} ) = \frac{ P( \mathrm{3 six} \, | \, \mathrm{Loaded} ) P ( \mathrm{Loaded} ) }{ P( \mathrm{3 six} )} $$
my question however is about the prior probability. After I've thrown one six, I can update my beliefs whether the die is loaded or not. Do I use $P(\mathrm{Loaded}) = 1/100$ or $P(\mathrm{Loaded}) = P( \mathrm{Loaded} \, |\, \mathrm{2 six})$, which to be computed needs $P(\mathrm{Loaded} \, | \, \mathrm{1 six})$?
These give different results. What is their interpretation? Which has the correct interpretation as probability of having a loaded die?
This calculation is an example in the book Richard Durbin, Sean R. Eddy, Anders Krogh, Graeme Mitchison - Biological Sequence Analysis - Probabilistic Models of Proteins and Nucleic Acids - Cambridge University Press (1999) in section 1.3. They compute $P(\mathrm{Loaded} \, | \, \mathrm{3 six})$ using directly $P(\mathrm{Loaded}) = 0.01$. They do not compute the probability of obtaining one six, then proceed to use that as a prior for the probability of obtaining two sixes, and so on. Is their procedure wrong? What is exactly what they are computing?
First, let's update it three times. Let $L$ be the event that we chose the loaded die, and let $S$ be the event that we roll a $6$.
For the first update:
$$\Pr(L|S)=\frac{\Pr(L\wedge S)}{\Pr(S)}=\frac{\frac12\cdot\frac1{100}}{\frac1{100}\cdot\frac12+\frac{99}{100}\cdot\frac16}=\frac1{34}$$
In the second update, we use $\frac1{34}$ as the prior.
$$\Pr(L|S)=\frac{\Pr(L\wedge S)}{\Pr(S)}=\frac{\frac12\cdot\frac1{34}}{\frac1{34}\cdot\frac12+\frac{33}{34}\cdot\frac16}=\frac1{12}$$
We do the same thing a third time, using $\frac1{12}$ as the prior:
$$\Pr(L|S)=\frac{\Pr(L\wedge S)}{\Pr(S)}=\frac{\frac12\cdot\frac1{12}} {\frac1{12}\cdot\frac12+\frac{11}{12}\cdot\frac16}=\frac3{14}$$
The posterior probability that the die is loaded is $\frac3{14}.$
Alternatively, we just update once. Let $S_3$ be the event that we roll three consecutive sixes.
$$\Pr(L|S_3)=\frac{\Pr(L\wedge S_3)}{\Pr(S)}=\frac{\frac18\cdot\frac1{100}}{\frac1{100}\cdot\frac18+\frac{99}{100}\cdot\frac1{216}}=\frac3{14},$$
the same answer as before.