How Do I Create a Transition Matrix for This Markov Chain Scenario?

[spence1]

I just discovered this website and want to thank everyone who is willing to contribute some of their time to help me. I appreciate it more than you know

First off, assume that state 1 is Chinese and that state 2 is Greek, and state 3 is Italian.

A student never eats the same kind of food for 2 consecutive weeks. If she eats a Chinese restaurant one week, then she is equally likely to have Greek as Italian food the next week. If she eats a Greek restaurant one week, then she is four times as likely to have Chinese as Italian food the next week. If she eats a Italian restaurant one week, then she is twice as likely to have Chinese as Greek food the next week.

I feel like I am on the right track, but I'm having trouble translating the words to notation (the most important part).
Chinese could be represented by x, Greek by y, and Italian by z, correct? And that has to add up to 1?

"If she eats a Greek restaurant one week, then she is four times as likely to have Chinese as Italian food the next week."
(I'm using ... to mean I think that there is something more in the equation)

y=4x...

"If she eats an Italian restaurant one week, then she is twice as likely to have Chinese as Greek food the next week."

z=2x...
So yeah. I kinda get the idea, I kinda don't. I think

 

[Monoxdifly]

"If she eats a Greek restaurant one week, then she is four times as likely to have Chinese as Italian food the next week."
(I'm using ... to mean I think that there is something more in the equation)

y=4x...

"If she eats an Italian restaurant one week, then she is twice as likely to have Chinese as Greek food the next week."

z=2x...

She eats restaurants?

 

[Opalg]

I just discovered this website and want to thank everyone who is willing to contribute some of their time to help me. I appreciate it more than you know

First off, assume that state 1 is Chinese and that state 2 is Greek, and state 3 is Italian.

A student never eats the same kind of food for 2 consecutive weeks. If she eats a Chinese restaurant one week, then she is equally likely to have Greek as Italian food the next week. If she eats a Greek restaurant one week, then she is four times as likely to have Chinese as Italian food the next week. If she eats a Italian restaurant one week, then she is twice as likely to have Chinese as Greek food the next week.

I feel like I am on the right track, but I'm having trouble translating the words to notation (the most important part).
Chinese could be represented by x, Greek by y, and Italian by z, correct? And that has to add up to 1?

"If she eats a Greek restaurant one week, then she is four times as likely to have Chinese as Italian food the next week."
(I'm using ... to mean I think that there is something more in the equation)

y=4x...

"If she eats an Italian restaurant one week, then she is twice as likely to have Chinese as Greek food the next week."

z=2x...
So yeah. I kinda get the idea, I kinda don't. I think

In this problem there are three states: Chinese, Greek and Italian. The transition matrix tells you the probability of changing from one state to another. So it will be a $3\times3$ matrix. The rows and columns of the matrix will correspond to the states. So row 1 corresponds to Chinese, row 2 to Greek, and row 3 to Italian, and similarly for the columns. The $(i,j)$-element of the matrix give the probability of changing from state $j$ to state $i$. So for example the $(1,1)$-element of the matrix is the probability of changing from Chinese to Chinese. But the student never eats the same kind of food for 2 consecutive weeks. That tells you that the $(1,1)$-element of the matrix is $0$.

The $(2,1)$-element is the probability of changing from Chinese to Greek, and the $(3,1)$-element is the probability of changing from Chinese to Italian. Those probabilities must add up to $1$, and you are told that they are both equally llikely. So they must both be $\frac12$. The first column of the matrix is therefore $\begin{bmatrix}0 \\ \frac12 \\ \frac12 \end{bmatrix}$.

Now you have to do the same thing for the other two columns.