Science:Math Exam Resources/Courses/MATH307/April 2009/Question 08 (d)

MATH307 April 2009

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Question 08 (d)

A player begins a game of chance by placing a marker in box 2, marked start. A die is rolled, and the marker is moved one square to the left if a 1 or 2 is rolled and one square to the right if a 3, 4, 5, or 6 is rolled. This process continues until the marker lands in square 1, in which case the player wins the game, or in square 4, in which case the player loses the game. (c) Given that when the initial state $x_{0}$ is written in terms of the eigenvectors, $v_{i}$, the coefficients $(c_{1},c_{2},c_{3},c_{4})=(0.43,0.57,0.28,0.93)$, determine the probability of winning the game. Explain your solution by writing a expression for the long-time behaviour of the game in terms of $c_{i}$, $\lambda _{i}$, and $v_{i}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Since $v_{1}$ and $v_{2}$ are the end states of the game (landing on the winning and losing square, respectively), $c_{1}$ and $c_{2}$ represent the probabilities of winning and losing, respectively. (They sum to 1, showing that if the game goes on long enough, it will end at one of these two states.) So the probability of winning the game is 0.43. Since two of the $\lambda _{i}$ have magnitude 1 (these being $\lambda _{1}$ and $\lambda _{2}$), the long-term behaviour of this matrix is a linear combination of the corresponding $v_{i}$, with coefficients being the corresponding $c_{i}$ (this depends on $x_{0}$ because there is more than one $\lambda _{i}$ with magnitude 1). In this case, these are $c_{1}$, $v_{1}$, $c_{2}$, and $v_{2}$. Thus the long-term behaviour of the game is $c_{1}*v_{1}+c_{2}*v_{2}$.