Discover where Bob is sleeping using hidden Markov chains Bob lives in four different houses $A, B, C$ and $D$ that are connected like the following graph shows:



Bob likes to sleep in any of his houses, but they are far apart so he only sleeps in a house adjacent to the one in which he slept the previous night. To clarify, this means that if Bob slept in house $A$ on night $1$, he may sleep in house $A$, $B$ or $C$ on night $2$ (not house $D$). The probability of each case is the same (one third); on each day, Bob takes a random walk from where stayed the previous night (and he might stay put).

Now Bob is a wanted criminal so on a given night the FBI would like to estimate where Bob is sleeping. Data from a satellite gives us the following probabilities of where Bob is sleeping on night $1$ and night $2$ (and any subsequent nights):

         Night 1   Night 2   ...
House A  0.8       0.05      ...
House B  0.1       0.4       ...
House C  0.05      0.05      ...
House D  0.05      0.5       ...


How can we use this data to calculate the probability of where Bob was sleeping on night $2$, for example calculating the probability Bob slept in house A? Could we use that method iteratively to calculate where Bob was sleeping on night $n$ if we continue to receive satellite data for each night?

Note: I made up this problem to understand better how hidden Markov chains work because I am interested in seeing the calculations on a concrete example. Many thanks for any input.
The goal is to update probability of Bob given satellite measurements.

The state of Bob at time $k$ is given by $x_{k}$. The measurement of a satellite is $y_{k}$. Since no prior information is given, we initialize with a uniform distribution
$$ \text{for } i={A,B,C,D},\,\, p(x_{0}=i|y_{0})=1/4.$$

Now that we have initialized, let's use Baye's rule to get $p(x_{1}=i|y_{0},y_{1})$. I'm only going to do this for i=A. Baye's rule yields

$$ p(x_{1}=A|y_{0},y_{1}) = \frac{p(x_{1}=A \cap y_{1}|y_{0})}{p(y_{1}/y_{0})}. $$
Now we just fill in the blanks. First, I like to calculate my "prediction", which is

$$ p(x_{1}=A | y_{0}) = \sum_{x_{0}\in{A,B,C,D}} P(x_{1}=A|x_{0})P(x_{0}|y_{0}).$$

Assuming bob is equally likely to stay in the house, or go to an adjacent house, this value is  3*(1/3)*1/4 = 1/4, and for this iteration at least, the prediction for other states is the same value. Now let's take a look at the Baye's Rule equation we outlined earlier

$$ p(x_{1}=A \cap y_{1}|y_{0}) = p(y_{1}|x_{1}=A)p(x_{1}=A|y_{0})=1/5$$

Finally, 

$$ p(y_{1}/y_{0}) =  \sum_{x_{1}\in{A,B,C,D}} p(y_{1}/x_{1})p(x_{1}/y_{0}). $$

This comes out to be 1/4. Finally,

$$ p(x_{1}=A|y_{0},y_{1}) = \frac{1/5}{1/4} = 4/5. $$
You may want to look at this.
http://en.wikipedia.org/wiki/Viterbi_algorithm