Gambler's Ruin Calculator
When it comes to the best takeout New Years Eve parties in the East Bay, Spicy Hideaway is definitely the place to go.
You're just rehashing the gambler's fallacy. If the ball landed in odd 100 times in a row on a fair wheel the odds that the next spin would be even are still the same as every spin, 47.37 on a double zero wheel. So it does not help that you can spin without betting. The ball does not have a memory. Star Wars - D&D 5th Edition Conversion (PF) - Free download as PDF File (.pdf), Text File (.txt) or read online for free. Star Wars gaming conversion using the D&D 5th Edition system. Nov 16, 2020 Actually, keeping a constant leverage ratio is the key to avoid gambler's ruin. See Ed Thorp's paper on applying Kelly criterion on blackjack. You can think the market is a blackjack dealer who reveals one card per day. Investing becomes how to size your bet everyday before the market reveals his card.
We handle all parts of your cocktail party, owner Noel Crist told Berkeleyside of Spicy Hideaways Chinatown location, which is closed for the holidays.
A proper New Years Eve, Crist said, is when we take you out to a fun dinner at a nice neighborhood restaurant.
Perfect for the first few hours of 2019, Spicy Hideaway serves a light, yet reasonably priced first course such as braised short ribs thats seasoned with fresh, wild mushrooms.
The second course is a traditional Chinese New Years offering called scallion soup with red pepper vinaigrette served over steamed rice.
Pro tip: Spicy Hideaways takeout takeout party is not only fast its heavily recommended because of the 20 cocktails Crist mentions before the crew settles down to a fitting meal. If youre truly looking to impress a friend with your revelry, Crist said that his takeout-only restaurant has a couple of hot takeout spots that are always hopping Jello Man, 112th Street and Bathulia, for example, that do a couple of two-hour pizzas on New Years Eve.
More of an extravaganza for the partygoing doyenne in the East Bay, Spicy Hideaways New Years Eve event is happening Friday, Dec. 28. And like anything else, there are some logistical and financial issues to consider. To be fair, Crist insisted, his New Years Eve event is uncomfortable.
The only thing you can do is come in early and we will take over, he said. We will take over the drinks, appetizers, new gala table and a big party room.
Spicy Hideaway has a real mission of celebrating Chinese New Year: We are here to honor, thank and cherish our first year, Crist said. Hopefully in 2018 well have another fruitful year.
Did you know? Here are five places to score a bargain or save some money at authentic Chinese New Years party or other New Years celebration.
Learning Objectives
In this section, you will learn to:
Identify absorbing states and absorbing Markov chains
Solve and interpret absorbing Markov chains.
In this section, we will study a type of Markov chain in which when a certain state is reached, it is impossible to leave that state. Such states are called absorbing states , and a Markov Chain that has at least one such state is called an Absorbing Markov chain.
Suppose you have the following transition matrix.
The state S 2 is an absorbing state, because the probability of moving from state S 2 to state S 2 is 1. Which is another way of saying that if you are in state S 2 , you will remain in state S 2 .
In fact, this is the way to identify an absorbing state. If the probability in row (i) and column (i), (p_ii), is 1, then state S i is an absorbing state.
Example (PageIndex1)
Consider transition matrices A, B, C for Markov chains shown below. Which of the following Markov chains have an absorbing state?
[A=left[beginarraylll
.3 .7 0
0 1 0
.2 .3 .5
endarrayright] quad B=left[beginarraylll
0 1 0
0 0 1
1 0 0
endarrayright] quad C=left[beginarraycccc
.1 .3 .4 .2
0 .2 .1 .7
0 0 1 0
0 0 0 1
endarrayright] nonumber]
Solution
has S 2 as an absorbing state.
If we are in state S 2 , we can not leave it. From state S 2 , we can not transition to state S 1 or S 3 ; the probabilities are 0.
The probability of transition from state S 2 to state S 2 is 1.
does not have any absorbing states.
From state S 1 , we always transition to state S 2 . From state S 2 we always transition to state S 3 . From state S 3 we always transition to state S 1 . In this matrix, it is never possible to stay in the same state during a transition.
has two absorbing states, S 3 and S 4 .
From state S 3 , you can only remain in state S 3 , and never transition to any other states. Similarly from state S 4 , you can only remain in state S 4 , and never transition to any other states.
We summarize how to identify absorbing states.
Summary
A state S is an absorbing state in a Markov chain in the transition matrix if
The row for state S has one 1 and all other entries are 0
AND
The entry that is 1 is on the main diagonal (row = column for that entry), indicating that we can never leave that state once it is entered.
Next we define an absorbing Markov Chain Royal panda contact number uk .
Absorbing Markov Chain
A Markov chain is an absorbing Markov Chain if
It has at least one absorbing state
AND
From any non-absorbing state in the Markov chain, it is possible to eventually move to some absorbing state (in one or more transitions).
Example (PageIndex2)
Consider transition matrices C and D for Markov chains shown below. Which of the following Markov chains is an absorbing Markov Chain?
[mathrmC=left[beginarrayllll
.1 .3 .4 .2
0 .2 .1 .7
0 0 1 0
0 0 0 1
endarrayright] quad mathrmD=left[beginarraylllll
1 0 0 0 0
0 1 0 0 0
.2 .2 .2 .2 .2
0 0 0 .3 .7
0 0 0 .6 .4
endarrayright] nonumber]
Solution
C is an absorbing Markov Chain but D is not an absorbing Markov chain.
Matrix C has two absorbing states, S 3 and S 4 , and it is possible to get to state S 3 and S 4 from S 1 and S 2 .
Matrix D is not an absorbing Markov chain.has two absorbing states, S 1 and S 2 , but it is never possible to get to either of those absorbing states from either S 4 or S 5 . If you are in state S 4 or S 5, you always remain transitioning between states S 4 or S 5m and can never get absorbed into either state S 1 or S 2
In the remainder of this section, well examine absorbing Markov chains with two classic problems: the random drunkards walk problem and the gambler's ruin problem. And finally well conclude with an absorbing Markov model applied to a real world situation. Drunkard's Random Walk
In this example we briefly examine the basic ideas concerning absorbing Markov chains.
Example (PageIndex3)
A man walks along a three block portion of Main St. His house is at one end of the three block section. A bar is at the other end of the three block section. Each time he reaches a corner he randomly either goes forward one block or turns around and goes back one block. If he reaches home or the bar, he stays there. The four states are Home (H), Corner 1(C 1 ), Corner 2 (C 2 ) and Bar (B).
Write the transition matrix and identify the absorbing states. Find the probabilities of ending up in each absorbing state depending on the initial state.
Solution Gambler's Ruin Calculator Formula
The transition matrix is written below.
Home and the Bar are absorbing states. If the man arrives home, he does not leave. If the man arrives at the bar, he does not leave. Since it is possible to reach home or the bar from each of the other two corners on his walk, this is an absorbing Markov chain.
We can raise the transition matrix T to a high power, (n). One we find a power T n that remains stable, it will tell us the probability of ending up in each absorbing state depending on the initial state.
T 91 = T 90 ; the matrix does not change as we continue to examine higher powers. We see that in the long-run, the Markov chain must end up in an absorbing state. In the long run, the man must eventually end up at either his home or the bar.
The second row tells us that if the man is at corner C 1 , then there is a 2/3 chance he will end up at home and a 1/3 chance he will end up at the bar. Gambler's Ruin Calculator Estimate
The third row tells us that if the man is at corner C 2 , then there is a 1/3 chance he will end up at home and a 2/3 chance he will end up at the bar.
Once he reaches home or the bar, he never leaves that absorbing state.
Note that while the matrix T n for sufficiently large (n) has become stable and is not changing, it does not represent an equilibrium matrix. The rows are not all identical, as we found in the regular Markov chains that reached an equilibrium.
We can write a smaller solution matrix by retaining only rows that relate to the non-absorbing states and retaining only the columns that relate to the absorbing states.
Then the solution matrix will have rows C 1 and C 2 , and columns H and B. The solution matrix is
The first row of the solution matrix shows that if the man is at corner C 1 , then there is a 2/3 chance he will end up at home and a 1/3 chance he will end up at the bar.
The second row of the solution matrix shows that if the man is at corner C 2 , then there is a 1/3 chance he will end up at home and a 2/3 chance he will end up at the bar.
The solution matrix does not show that eventually there is 0 probability of ending up in C1 or C2, or that if you start in an absorbing state H or B, you stay there. The smaller solution matrix assumes that we understand these outcomes and does not include that information.
The next example is another classic example of an absorbing Markov chain. In the next example we examine more of the mathematical details behind the concept of the solution matrix. Gamber's Ruin Problem
Example (PageIndex4)
A gambler has $3,000, and she decides to gamble $1,000 at a time at a Black Jack table in a casino in Las Vegas. She has told herself that she will continue playing until she goes broke or has $5,000. Her probability of winning at Black Jack is .40. Write the transition matrix, identify the absorbing states, find the solution matrix, and determine the probability that the gambler will be financially ruined at a stage when she has $2,000.
Solution
The transition matrix is written below. Clearly the state 0 and state 5K are the absorbing states. This makes sense because as soon as the gambler reaches 0, she is financially ruined and the game is over. Similarly, if the gambler reaches $5,000, she has promised herself to quit and, again, the game is over. The reader should note that p 00 = 1, and p 55 = 1.
Further observe that since the gambler bets only $1,000 at a time, she can raise or lower her money only by $1,000 at a time. In other words, if she has $2,000 now, after the next bet she can have $3,000 with a probability of .40 and $1,000 with a probability of .60.
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To determine the long term trend, we raise the matrix to higher powers until all the non-absorbing states are absorbed. This is the called the solution matrix. Gambler's Ruin Calculator Estimate
The solution matrix is often written in the following form, where the non-absorbing states are written as rows on the side, and the absorbing states as columns on the top. Gambler's Ruin Calculator Value
The table lists the probabilities of getting absorbed in state 0 or state 5K starting from any of the four non-absorbing states. For example, if at any instance the gambler has $3,000, then her probability of financial ruin is 135/211 and her probability reaching 5K is 76/211.
Example (PageIndex5)
Solve the Gambler's Ruin Problem of Example (PageIndex4) without raising the matrix to higher powers, and determine the number of bets the gambler makes before the game is over.
Solution
In solving absorbing states, it is often convenient to rearrange the matrix so that the rows and columns corresponding to the absorbing states are listed first. This is called the Canonical form. The transition matrix of Example 1 in the canonical form is listed below.
The canonical form divides the transition matrix into four sub-matrices as listed below.
The matrix (F = (I_n- B)-1) is called the fundamental matrix for the absorbing Markov chain, where In is an identity matrix of the same size as B. The (i), (j)-th entry of this matrix tells us the average number of times the process is in the non-absorbing state (j) before absorption if it started in the non-absorbing state (i).
The matrix (F = (I_n- B)-1) for our problem is listed below.
You can use your calculator, or a computer, to calculate matrix F.
The Fundamental matrix F helps us determine the average number of games played before absorption.
According to the matrix, the entry 1.78 in the row 3, column 2 position says that the gambler will play the game 1.78 times before she goes from $3K to $2K. The entry 2.25 in row 3, column 3 says that if the gambler now has $3K, she will have $3K on the average 2.25 times before the game is over.
We now address the question of how many bets will she have to make before she is absorbed, if the gambler begins with $3K? Gambler's Ruin Calculator Online
If we add the number of games the gambler plays in each non-absorbing state, we get the average number of games before absorption from that state. Therefore, if the gambler starts with $3K, the average number of Black Jack games she will play before absorption is
[1.07 + 1.78 + 2.25 + 0.90 = 6.0 nonumber]
That is, we expect the gambler will either have $5,000 or nothing on the 7th bet.
Lastly, we find the solution matrix without raising the transition matrix to higher powers. The matrix FA gives us the solution matrix.
[mathrmFA=left[beginarraycccc
1.54 .90 .47 .19
1.35 2.25 1.18 .47
1.07 1.78 2.25 .90
.64 1.07 1.35 1.54
endarrayright]left[beginarraycc
.6 0
0 0
0 0
0 .4
endarrayright]=left[beginarraycc
.92 .08
.81 .19
.64 .36
.38 .62
endarrayright] nonumber]
which is the same as the following matrix we obtained by raising the transition matrix to higher powers.
Example (PageIndex6)
At a professional school, students need to take and pass an English writing/speech class in order to get their professional degree. Students must take the class during the first quarter that they enroll. If they do not pass the class they take it again in the second semester. If they fail twice, they are not permitted to retake it again, and so would be unable to earn their degree.
Students can be in one of 4 states: passed the class (P), enrolled in the class for the first time (C), retaking the class (R) or failed twice and can not retake (F). Experience shows 70 of students taking the class for the first time pass and 80 of students taking the class for the second time pass.
Write the transition matrix and identify the absorbing states. Find the probability of being absorbed eventually in each of the absorbing states.
Solution
The absorbing states are P (pass) and F (fail repeatedly and can not retake). The transition matrix T is shown below.
If we raise the transition matrix T to a high power, we find that it remains stable and gives us the long-term probabilities of ending up in each of the absorbing states.
Of students currently taking the class for the first time, 94 will eventually pass. 6 will eventually fail twice and be unable to earn their degree.
Of students currently taking the class for the second time time, 80 will eventually pass. 20 will eventually fail twice and be unable to earn their degree.
The solution matrix contains the same information in abbreviated form
Note that in this particular problem, we dont need to raise T to a very high power. If we find T 2 , we see that it is actually equal to T n for higher powers (n). T n becomes stable after two transitions; this makes sense in this problem because after taking the class twice, the student must have passed or is not permitted to retake it any longer. Therefore the probabilities should not change any more after two transitions; by the end of two transitions, every student has reached an absorbing state.
Absorbing Markov Chains
A Markov chain is an absorbing Markov chain if it has at least one absorbing state. A state i is an absorbing state if once the system reaches state i, it stays in that state; that is, (p_ii = 1).
If a transition matrix T for an absorbing Markov chain is raised to higher powers, it reaches an absorbing state called the solution matrix and stays there. The (i), (j)-th entry of this matrix gives the probability of absorption in state (j) while starting in state (i).
Alternately, the solution matrix can be found in the following manner.
Express the transition matrix in the canonical form as below. [mathrmT=left[beginarrayll
mathbfI_mathbfn mathbf0
mathbfA mathbfB
endarrayright] nonumber] where (I_n) is an identity matrix, and 0 is a matrix of all zeros.
The fundamental matrix (F = (I - B)-1). The fundamental matrix helps us find the number of games played before absorption.
FA is the solution matrix, whose (i), (j)-th entry gives the probability of absorption in state (j) while starting in state (i).
The sum of the entries of a row of the fundamental matrix gives us the expected number of steps before absorption for the non-absorbing state associated with that row.
You're just rehashing the gambler's fallacy. If the ball landed in odd 100 times in a row on a fair wheel the odds that the next spin would be even are still the same as every spin, 47.37 on a double zero wheel. So it does not help that you can spin without betting. The ball does not have a memory. Star Wars - D&D 5th Edition Conversion (PF) - Free download as PDF File (.pdf), Text File (.txt) or read online for free. Star Wars gaming conversion using the D&D 5th Edition system. Nov 16, 2020 Actually, keeping a constant leverage ratio is the key to avoid gambler's ruin. See Ed Thorp's paper on applying Kelly criterion on blackjack. You can think the market is a blackjack dealer who reveals one card per day. Investing becomes how to size your bet everyday before the market reveals his card.
We handle all parts of your cocktail party, owner Noel Crist told Berkeleyside of Spicy Hideaways Chinatown location, which is closed for the holidays.
A proper New Years Eve, Crist said, is when we take you out to a fun dinner at a nice neighborhood restaurant.
Perfect for the first few hours of 2019, Spicy Hideaway serves a light, yet reasonably priced first course such as braised short ribs thats seasoned with fresh, wild mushrooms.
The second course is a traditional Chinese New Years offering called scallion soup with red pepper vinaigrette served over steamed rice.
Pro tip: Spicy Hideaways takeout takeout party is not only fast its heavily recommended because of the 20 cocktails Crist mentions before the crew settles down to a fitting meal. If youre truly looking to impress a friend with your revelry, Crist said that his takeout-only restaurant has a couple of hot takeout spots that are always hopping Jello Man, 112th Street and Bathulia, for example, that do a couple of two-hour pizzas on New Years Eve.
More of an extravaganza for the partygoing doyenne in the East Bay, Spicy Hideaways New Years Eve event is happening Friday, Dec. 28. And like anything else, there are some logistical and financial issues to consider. To be fair, Crist insisted, his New Years Eve event is uncomfortable.
The only thing you can do is come in early and we will take over, he said. We will take over the drinks, appetizers, new gala table and a big party room.
Spicy Hideaway has a real mission of celebrating Chinese New Year: We are here to honor, thank and cherish our first year, Crist said. Hopefully in 2018 well have another fruitful year.
Did you know? Here are five places to score a bargain or save some money at authentic Chinese New Years party or other New Years celebration.
Learning Objectives
In this section, you will learn to:
Identify absorbing states and absorbing Markov chains
Solve and interpret absorbing Markov chains.
In this section, we will study a type of Markov chain in which when a certain state is reached, it is impossible to leave that state. Such states are called absorbing states , and a Markov Chain that has at least one such state is called an Absorbing Markov chain.
Suppose you have the following transition matrix.
The state S 2 is an absorbing state, because the probability of moving from state S 2 to state S 2 is 1. Which is another way of saying that if you are in state S 2 , you will remain in state S 2 .
In fact, this is the way to identify an absorbing state. If the probability in row (i) and column (i), (p_ii), is 1, then state S i is an absorbing state.
Example (PageIndex1)
Consider transition matrices A, B, C for Markov chains shown below. Which of the following Markov chains have an absorbing state?
[A=left[beginarraylll
.3 .7 0
0 1 0
.2 .3 .5
endarrayright] quad B=left[beginarraylll
0 1 0
0 0 1
1 0 0
endarrayright] quad C=left[beginarraycccc
.1 .3 .4 .2
0 .2 .1 .7
0 0 1 0
0 0 0 1
endarrayright] nonumber]
Solution
has S 2 as an absorbing state.
If we are in state S 2 , we can not leave it. From state S 2 , we can not transition to state S 1 or S 3 ; the probabilities are 0.
The probability of transition from state S 2 to state S 2 is 1.
does not have any absorbing states.
From state S 1 , we always transition to state S 2 . From state S 2 we always transition to state S 3 . From state S 3 we always transition to state S 1 . In this matrix, it is never possible to stay in the same state during a transition.
has two absorbing states, S 3 and S 4 .
From state S 3 , you can only remain in state S 3 , and never transition to any other states. Similarly from state S 4 , you can only remain in state S 4 , and never transition to any other states.
We summarize how to identify absorbing states.
Summary
A state S is an absorbing state in a Markov chain in the transition matrix if
The row for state S has one 1 and all other entries are 0
AND
The entry that is 1 is on the main diagonal (row = column for that entry), indicating that we can never leave that state once it is entered.
Next we define an absorbing Markov Chain Royal panda contact number uk .
Absorbing Markov Chain
A Markov chain is an absorbing Markov Chain if
It has at least one absorbing state
AND
From any non-absorbing state in the Markov chain, it is possible to eventually move to some absorbing state (in one or more transitions).
Example (PageIndex2)
Consider transition matrices C and D for Markov chains shown below. Which of the following Markov chains is an absorbing Markov Chain?
[mathrmC=left[beginarrayllll
.1 .3 .4 .2
0 .2 .1 .7
0 0 1 0
0 0 0 1
endarrayright] quad mathrmD=left[beginarraylllll
1 0 0 0 0
0 1 0 0 0
.2 .2 .2 .2 .2
0 0 0 .3 .7
0 0 0 .6 .4
endarrayright] nonumber]
Solution
C is an absorbing Markov Chain but D is not an absorbing Markov chain.
Matrix C has two absorbing states, S 3 and S 4 , and it is possible to get to state S 3 and S 4 from S 1 and S 2 .
Matrix D is not an absorbing Markov chain.has two absorbing states, S 1 and S 2 , but it is never possible to get to either of those absorbing states from either S 4 or S 5 . If you are in state S 4 or S 5, you always remain transitioning between states S 4 or S 5m and can never get absorbed into either state S 1 or S 2
In the remainder of this section, well examine absorbing Markov chains with two classic problems: the random drunkards walk problem and the gambler's ruin problem. And finally well conclude with an absorbing Markov model applied to a real world situation. Drunkard's Random Walk
In this example we briefly examine the basic ideas concerning absorbing Markov chains.
Example (PageIndex3)
A man walks along a three block portion of Main St. His house is at one end of the three block section. A bar is at the other end of the three block section. Each time he reaches a corner he randomly either goes forward one block or turns around and goes back one block. If he reaches home or the bar, he stays there. The four states are Home (H), Corner 1(C 1 ), Corner 2 (C 2 ) and Bar (B).
Write the transition matrix and identify the absorbing states. Find the probabilities of ending up in each absorbing state depending on the initial state.
Solution Gambler's Ruin Calculator Formula
The transition matrix is written below.
Home and the Bar are absorbing states. If the man arrives home, he does not leave. If the man arrives at the bar, he does not leave. Since it is possible to reach home or the bar from each of the other two corners on his walk, this is an absorbing Markov chain.
We can raise the transition matrix T to a high power, (n). One we find a power T n that remains stable, it will tell us the probability of ending up in each absorbing state depending on the initial state.
T 91 = T 90 ; the matrix does not change as we continue to examine higher powers. We see that in the long-run, the Markov chain must end up in an absorbing state. In the long run, the man must eventually end up at either his home or the bar.
The second row tells us that if the man is at corner C 1 , then there is a 2/3 chance he will end up at home and a 1/3 chance he will end up at the bar. Gambler's Ruin Calculator Estimate
The third row tells us that if the man is at corner C 2 , then there is a 1/3 chance he will end up at home and a 2/3 chance he will end up at the bar.
Once he reaches home or the bar, he never leaves that absorbing state.
Note that while the matrix T n for sufficiently large (n) has become stable and is not changing, it does not represent an equilibrium matrix. The rows are not all identical, as we found in the regular Markov chains that reached an equilibrium.
We can write a smaller solution matrix by retaining only rows that relate to the non-absorbing states and retaining only the columns that relate to the absorbing states.
Then the solution matrix will have rows C 1 and C 2 , and columns H and B. The solution matrix is
The first row of the solution matrix shows that if the man is at corner C 1 , then there is a 2/3 chance he will end up at home and a 1/3 chance he will end up at the bar.
The second row of the solution matrix shows that if the man is at corner C 2 , then there is a 1/3 chance he will end up at home and a 2/3 chance he will end up at the bar.
The solution matrix does not show that eventually there is 0 probability of ending up in C1 or C2, or that if you start in an absorbing state H or B, you stay there. The smaller solution matrix assumes that we understand these outcomes and does not include that information.
The next example is another classic example of an absorbing Markov chain. In the next example we examine more of the mathematical details behind the concept of the solution matrix. Gamber's Ruin Problem
Example (PageIndex4)
A gambler has $3,000, and she decides to gamble $1,000 at a time at a Black Jack table in a casino in Las Vegas. She has told herself that she will continue playing until she goes broke or has $5,000. Her probability of winning at Black Jack is .40. Write the transition matrix, identify the absorbing states, find the solution matrix, and determine the probability that the gambler will be financially ruined at a stage when she has $2,000.
Solution
The transition matrix is written below. Clearly the state 0 and state 5K are the absorbing states. This makes sense because as soon as the gambler reaches 0, she is financially ruined and the game is over. Similarly, if the gambler reaches $5,000, she has promised herself to quit and, again, the game is over. The reader should note that p 00 = 1, and p 55 = 1.
Further observe that since the gambler bets only $1,000 at a time, she can raise or lower her money only by $1,000 at a time. In other words, if she has $2,000 now, after the next bet she can have $3,000 with a probability of .40 and $1,000 with a probability of .60.
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To determine the long term trend, we raise the matrix to higher powers until all the non-absorbing states are absorbed. This is the called the solution matrix. Gambler's Ruin Calculator Estimate
The solution matrix is often written in the following form, where the non-absorbing states are written as rows on the side, and the absorbing states as columns on the top. Gambler's Ruin Calculator Value
The table lists the probabilities of getting absorbed in state 0 or state 5K starting from any of the four non-absorbing states. For example, if at any instance the gambler has $3,000, then her probability of financial ruin is 135/211 and her probability reaching 5K is 76/211.
Example (PageIndex5)
Solve the Gambler's Ruin Problem of Example (PageIndex4) without raising the matrix to higher powers, and determine the number of bets the gambler makes before the game is over.
Solution
In solving absorbing states, it is often convenient to rearrange the matrix so that the rows and columns corresponding to the absorbing states are listed first. This is called the Canonical form. The transition matrix of Example 1 in the canonical form is listed below.
The canonical form divides the transition matrix into four sub-matrices as listed below.
The matrix (F = (I_n- B)-1) is called the fundamental matrix for the absorbing Markov chain, where In is an identity matrix of the same size as B. The (i), (j)-th entry of this matrix tells us the average number of times the process is in the non-absorbing state (j) before absorption if it started in the non-absorbing state (i).
The matrix (F = (I_n- B)-1) for our problem is listed below.
You can use your calculator, or a computer, to calculate matrix F.
The Fundamental matrix F helps us determine the average number of games played before absorption.
According to the matrix, the entry 1.78 in the row 3, column 2 position says that the gambler will play the game 1.78 times before she goes from $3K to $2K. The entry 2.25 in row 3, column 3 says that if the gambler now has $3K, she will have $3K on the average 2.25 times before the game is over.
We now address the question of how many bets will she have to make before she is absorbed, if the gambler begins with $3K? Gambler's Ruin Calculator Online
If we add the number of games the gambler plays in each non-absorbing state, we get the average number of games before absorption from that state. Therefore, if the gambler starts with $3K, the average number of Black Jack games she will play before absorption is
[1.07 + 1.78 + 2.25 + 0.90 = 6.0 nonumber]
That is, we expect the gambler will either have $5,000 or nothing on the 7th bet.
Lastly, we find the solution matrix without raising the transition matrix to higher powers. The matrix FA gives us the solution matrix.
[mathrmFA=left[beginarraycccc
1.54 .90 .47 .19
1.35 2.25 1.18 .47
1.07 1.78 2.25 .90
.64 1.07 1.35 1.54
endarrayright]left[beginarraycc
.6 0
0 0
0 0
0 .4
endarrayright]=left[beginarraycc
.92 .08
.81 .19
.64 .36
.38 .62
endarrayright] nonumber]
which is the same as the following matrix we obtained by raising the transition matrix to higher powers.
Example (PageIndex6)
At a professional school, students need to take and pass an English writing/speech class in order to get their professional degree. Students must take the class during the first quarter that they enroll. If they do not pass the class they take it again in the second semester. If they fail twice, they are not permitted to retake it again, and so would be unable to earn their degree.
Students can be in one of 4 states: passed the class (P), enrolled in the class for the first time (C), retaking the class (R) or failed twice and can not retake (F). Experience shows 70 of students taking the class for the first time pass and 80 of students taking the class for the second time pass.
Write the transition matrix and identify the absorbing states. Find the probability of being absorbed eventually in each of the absorbing states.
Solution
The absorbing states are P (pass) and F (fail repeatedly and can not retake). The transition matrix T is shown below.
If we raise the transition matrix T to a high power, we find that it remains stable and gives us the long-term probabilities of ending up in each of the absorbing states.
Of students currently taking the class for the first time, 94 will eventually pass. 6 will eventually fail twice and be unable to earn their degree.
Of students currently taking the class for the second time time, 80 will eventually pass. 20 will eventually fail twice and be unable to earn their degree.
The solution matrix contains the same information in abbreviated form
Note that in this particular problem, we dont need to raise T to a very high power. If we find T 2 , we see that it is actually equal to T n for higher powers (n). T n becomes stable after two transitions; this makes sense in this problem because after taking the class twice, the student must have passed or is not permitted to retake it any longer. Therefore the probabilities should not change any more after two transitions; by the end of two transitions, every student has reached an absorbing state.
Absorbing Markov Chains
A Markov chain is an absorbing Markov chain if it has at least one absorbing state. A state i is an absorbing state if once the system reaches state i, it stays in that state; that is, (p_ii = 1).
If a transition matrix T for an absorbing Markov chain is raised to higher powers, it reaches an absorbing state called the solution matrix and stays there. The (i), (j)-th entry of this matrix gives the probability of absorption in state (j) while starting in state (i).
Alternately, the solution matrix can be found in the following manner.
Express the transition matrix in the canonical form as below. [mathrmT=left[beginarrayll
mathbfI_mathbfn mathbf0
mathbfA mathbfB
endarrayright] nonumber] where (I_n) is an identity matrix, and 0 is a matrix of all zeros.
The fundamental matrix (F = (I - B)-1). The fundamental matrix helps us find the number of games played before absorption.
FA is the solution matrix, whose (i), (j)-th entry gives the probability of absorption in state (j) while starting in state (i).
The sum of the entries of a row of the fundamental matrix gives us the expected number of steps before absorption for the non-absorbing state associated with that row.
