Most likely dice sequence given weights and switching probabilities

Question

An occasionally dishonest casino uses two kinds of dice: a fair die that has $\frac16$ probability of rolling any number, and a loaded die that has $\frac12$ probability to roll a $6$ and a $\frac1{10}$ probability to roll each of the remaining numbers. The probability that the casino switches from the fair to loaded die is $0.1$ and the probability of switching back from loaded to fair is $0.05$. What is the most likely sequence of dice used if you observe the following numbers: $$4,6,6,6.$$ Assume the casino always starts with the fair die. Show your work.

Answer 1

Hint:

There are $8$ different possibilities for the sequence of dice types ($\text{F}$ is fair, $\text{L}$ is loaded):

$$\text{FFFF}, \text{FFFL}, \text{FFLF}, \text{FFLL}, \text{FLFF}, \text{FLFL}, \text{FLLF}, \text{FLLL}$$

Let $A_i$, with $1 \leq i \leq 8$ denote the event that the sequence of dice is the $i^{th}$ of the $8$ configurations above. Find the probability $P(A_i)$ for each $i$.

Let $B$ be the event that the numbers shown are $4$, $6$, $6$, and $6$. Find $P(B)$. You have to be a bit careful at this step. $P(B)$ isn't just $(1/6)^4$ since switching the dice is possible.

Now you can use Bayes' Theorem to find $$P(A_i \mid B) = \frac{P(B \mid A_i) \cdot P(A_i)}{P(B)}$$

for each $1 \leq i \leq 8$.

---

Intuitively, I'd guess it's one of $\text{FFFF}$ or $\text{FLLL}$. You might be able to just apply Bayes' Theorem to those two cases; whichever gives the higher $P(A_i \mid B)$ is your answer.

It shouldn't be any of the others because switching multiple times is unlikely, and switching later in the sequence makes the second and/or third $6$ less likely.

Answer 2

So the most probable dice sequence is $\{Fair, Fair, Fair, Fair\}$. (T = Fair, F = Loaded)