I'm struck with an exercise. I tried, but the results don't seem to fit to those proposed.

Exercise: Two players play the following game. The one who begins draws two cards from a deck of 40 cards (10 cards per suit):

  • if they are both clubs the player wins;
  • if they are of the same suit but not clubs, the player shuffles the cards and start again;
  • else the player shuffles the cards and let the other player play.

1) Model the game with a Markov Chain.

2) What is the probability that who started wins?

My attempt: If $X_n$ is the player in the current turn, we have a three state MC whose state space is $\{A,\, B,\, \text{exit}\}$. The exit state is the one reached when the game ends. The transition matrix is

$$P=\begin{pmatrix}9/52 & 10/13 & 3/52 \\ 10/13 & 9/52 & 3/52 \\ 0 & 0 & 1 \end{pmatrix}$$

since the probability of staying is $\frac{3 \cdot 9}{4 \cdot 39}$ and so on.

For the second question I thought I could use the Strong Markov Property (the one which states that a Markov Chain can start afresh in a Stopping Time) by using the last time I see A (or B) before the game ends. Starting from that point, the probability is just the jump from A (or B) to exit times two (to consider both the A and the B case).

What's wrong with this last point?

  • $\begingroup$ $40$ cards or $52$? $\endgroup$ – Henry Jun 30 '15 at 21:54
  • $\begingroup$ 40. It's an Italian deck :) $\endgroup$ – Phugo Jun 30 '15 at 21:55
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    $\begingroup$ Would you, please, explain the meaning of the states $A,B,exit$? $\endgroup$ – zoli Jun 30 '15 at 22:01
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    $\begingroup$ I only read this quick but usually, the ``last time A happens before B'' types of variables are not stopping times. Edit : the time variable you take is definitely not a stopping time $\endgroup$ – hHhh Jul 1 '15 at 15:18
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    $\begingroup$ @Phugo intuitively a stopping time is a time at which you can stop knowing only the past and present. In your case you need to know what happens after $T$ to determine the value of $T$. see en.wikipedia.org/wiki/Stopping_time $\endgroup$ – hHhh Jul 1 '15 at 18:37

First, I wouldn't modelise the markov chain like that , I would consider four states 1 = "player A play", 2 = "player B play", 3 = "player A had won" and 4 = "player B had won"

The transition matrix would be

$$M = \begin{pmatrix} 9/52 & 10/13 & 3/52 & 0 \\ 10/13 & 9/52 & 0 & 3/52 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$

Then the probability to be at each state after n step knowing that we started in state 1) is given by

$$\pi^{(n)} = (1,0,0,0)M^n$$

And the probability we're looking for is

$$p = \lim_{n\to +\infty} \pi_3^{(n)}$$

To calculate $M^n$, you can diagonalize it, and it should give you the answer (I didn't do the calcul)

  • $\begingroup$ Thank you! It's more or less the same procedure outlined in my handout, but I was really interested in understanding why the method I proposed is not alright. Probably I didn't understand what's the Markov Property is about and I want to know where I have been wrong. Thank you, again :) $\endgroup$ – Phugo Jun 30 '15 at 22:30

To solve question 2 you don't need Markov chains, only first step analysis. In the first step there are 3 possibilities:

  1. player $A$ wins,
  2. the game starts over,
  3. the deck goes to player $B$.

Letting $p_A$ be the probability that player $A$ is the eventual winner, and taking these 3 cases into account gives equation (1) below.

With $p={10\choose 2}/{40\choose 2}$ we get $$p_A=p\cdot 1 +3p\cdot p_A+(1-4p)\cdot(1-p_A).\tag 1$$ Solving (1) gives $p_A={43\over 83}= 0.51808$.


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