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Got given this question:

In a football tournament a player never gets 2 yellow cards (YCs) in the same match but can get issued with one YC (with probability p) independently of other matches. The YC stays with him through any subsequent matches in the tournament, until he gets a 2nd YC when he is suspended for one match and then his YC count reverts to zero. Let X be the random variable denoting the number of YCs a player currently holds. Set this up as a Markov chain and find the long term proportion of matches that the player misses through suspension.

How would I draw this transition diagram? Do I need to represent the suspended games in the diagram? And how would I go about finding the long term proportion of suspended matches?

Thank you!

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Updating the answer, as per David's comments.

Assume the following state space, which represents the state a player is in before the game.

State 0: No yellow cards

State 1: 1 yellow card

State 2: Suspended

Then the transition matrix can be written as follows.

\begin{equation} P = \begin{bmatrix} 1-p & p & 0 \\ 0 & 1-p & p \\ 1 & 0 & 0 \end{bmatrix} \end{equation}

This assumes that if a player picks up a second YC during a match, he isn't technically suspended until the next match.

If the long run proportion of being in state $j$ is denoted by $\pi_j$, we can find the long run proportions by solving the following system of equations.

\begin{align} 1 &= \pi_0 + \pi_1 + \pi_2 \\ \pi_0 &= (1-p)\pi_0 + \pi_2 \\ \pi_1 &= p\pi_0 + (1-p)\pi_1 \\ \pi_2 &= p\pi_1 \end{align}

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    $\begingroup$ I'm not certain that this is correct. Can you define the state-space more specifically. If we define the state to be your 'status' at the beginning of a game. Then starting the game in state $2$ is equivalent to being suspended during that game. More generally, if you start in state $0$ you can either begin the next game in state $0$ or $1$, if you are in state $1$, you can begin the next game in state $1$ or $2$, and if you are in state $2$ you must begin the next game in state $0$ (w.p.1). ... 1/2 $\endgroup$
    – David
    Commented Mar 9, 2017 at 17:30
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    $\begingroup$ You can then form the transition probability matrix as $$P = \left[\begin{array} &1-p&p & 0\\0&1-p&p\\1&0&0\end{array}\right]$$ ... 2/2 $\endgroup$
    – David
    Commented Mar 9, 2017 at 17:31
  • $\begingroup$ There is more than one way to define the state space. I suppose I am defining it to be the state at the end of the game. There is certainly something to be said for having a simpler state space like you suggest. $\endgroup$
    – knrumsey
    Commented Mar 9, 2017 at 17:52
  • $\begingroup$ The issue isn't whether one is simpler or not, the issue is that they yield different solutions. I believe that your model is double counting the suspensions. If I end a game with $2$ cards, then I end the following game with $0$ cards. I don't end the subsequent game with a 'suspended' status. Edit: I just saw your edit after this comment. +1 $\endgroup$
    – David
    Commented Mar 9, 2017 at 18:00
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    $\begingroup$ I believe the issue is you've set up the system of equations 'row-wise' when they should be setup 'column-wise.' I think that's the issue. $\endgroup$
    – David
    Commented Mar 9, 2017 at 18:21

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